units

Units are the basis of measurement. Thus a foot or an inch is a unit of length; a gram or a pound is a unit of weight. For different things there are different units; electric current, mechanical energy and heat, for example, have each their own units.

The same thing may be measured in some cases by different systems of units. Thus length may be measured by the English system, the inch, the foot, and other units being employed, or by the metric system, the centimeter, the meter, and other units being employed, or by any one of the many systems used by different nations at different epochs. There are two systems of electric units — the electro-magnetic and the electro-static systems. Most calculations in electric engineering are in electro-magnetic units.

A simple unit is one into which only one factor enters. The centimeter and the pound are examples of simple units.

A compound unit is composed of two or more simple units. The metric system unit of velocity is a compound unit; it is one centimeter per second. A horse-power in the English system is expressed as the power which can raise 33,000 pounds one foot in a minute. This quantity is expressed by a compound unit; it is [math]550[/math] foot-pounds per second and three simple units make it up, the foot, the pound, and the second.

A compound unit, expressed by units with hyphens between them, implies multiplication of one by the other.

Example. 7,200 foot-pounds of energy are expended in lifting a weight 109 feet. What is the weight?

Solution. The hyphen indicates that in the compound unit “foot-pounds” the feet are multiplied by the pounds. Therefore [math]109[/math] feet must be multiplied by such a number of pounds as will give a product of [math]7,200[/math]. This number is [math]66.06[/math], which is the number of pounds lifted. The product of [math]109[/math] feet by [math]66.06[/math] pounds gives [math]7,200[/math] foot-pounds.

When the words “in a,” the word “per,” or “a” stand between units, the division of the coefficients of the units into one another is implied. These are rate units.

Example. A trolley car is timed over a distance of 10 rails; each rail is 60 feet long. It covers the distance in 15 seconds. How many feet in a second does it travel?

Solution. The car travels [math]600[/math] feet per [math]15[/math] seconds. The rate per second is obtained by dividing the coefficient of feet, which is [math]600[/math], by the coefficient of seconds, which is [math]15[/math]. [math]600 \div 15 = 40[/math], the number of feet per second.

Significance of Multiplication and Division of Different Kinds of Units.
The constituent units of a compound unit, written with a hyphen, practically never have any coefficient of either of the constituent units expressed, unity being the coefficient. A gram-centimeter is one gram multiplied by one centimeter. 10 gram-centimeters are ten times the above. The coefficient, [math]10[/math], applies to and multiplies the compound unit, gram-centimeter.

In the use of units the meanings of the words “multiplication” and “division” are extended so as to include the practical multiplication and division of different classes of units with each other. Thus a foot-pound is taken as a multiplication of a foot by a pound. 150 foot-pounds is taken as the multiplication of 10 pounds by 15 feet, or as the multiplication of 15 pounds by 10 feet, all of which are impossibilities.

The difficulty may be met thus. Suppose 10 feet are to be multiplied by 15 pounds. 10 and 15 may be treated as coefficients of feet and pounds, and the multiplication may be expressed as [math]10 \times 15 = 150[/math] compound units, each compound unit being a foot-pound.

Centimeter-Gram-Second System.
The scientific bases of units are the centimeter, gram, and second. On these a whole series of units—electrical, mechanical, and others—have been founded. The system is called the C.G.S. system.

The centimeter is the unit of length. It is approximately four-tenths of an inch, or one-thirtieth of a foot.

The gram is the unit of mass. It is the quantity of matter contained in a gram. Its relation to weight is abstracted from its status as a unit of mass. A gram weighs approximately fifteen grains ([math]15.432+[/math]).

The second is the unit of time.

Mass.
Mass is quantity of matter. In a given portion of matter it is invariable, and is unaffected by the relations of the portion of matter to the rest of the universe. The mass of a gram would be the same at any place,—at the center of the earth, on the surface of the earth, on the surface of a planet, or elsewhere.

Weight and Gravitation.
Weight is mass acted on by gravity. The weight of a given portion of matter depends upon the intensity of the force of gravity. As this force varies in intensity at different parts of the universe, the weight of a given portion of matter varies also. In the center of the earth there would be no weight. On the surface of the planet Jupiter the weight of the mass of a gram or other quantity would be much greater than on the earth. Weight is greater at the poles of the earth than at the equator; a pound would weigh less at the equator than in the polar regions, principally on account of centrifugal force, partly on account of the difference between the equatorial and polar diameters of the earth.

The force of gravity of the earth is called its gravitation. As the earth is indefinitely large with respect to the masses of engineering,—the gram, pound, ton, and others,—it acts upon such masses with a force proportional to their masses. It therefore tends to impart to any such mass the same velocity after acting on it for a second. This velocity is about [math]981[/math] centimeters per second and varies with the latitude.

Space.
Space may be of one, two, or three dimensions. Space of one dimension is length, that of two dimensions is area, and that of three dimensions is volume. Generally when space is spoken of in this book it is space of one dimension or length that is referred to.

Rate Units.
One class of units expresses rate. If a current of electricity flows at the rate of [math]1[/math] coulomb per second, it flows at the rate of [math]1[/math] ampere. If electrical energy is expended or developed at the rate of [math]10[/math] megergs per second, the unit expressing this rate is the watt.

Example. A generator delivers [math]8,050[/math] coulombs in an hour. What is the rate?

Solution. There are [math]3,600[/math] seconds in an hour. Dividing the total amount of electricity delivered by the time in seconds gives the rate per second.
[math]8,050 \div 3,600 = 2.24[/math] amperes.

Example. What quantity of electricity will be delivered by an amperage of [math]13[/math] in 5 minutes?

Solution. 13 amperes of current is a rate of [math]13[/math] coulombs in a second. In 5 minutes there are [math]300[/math] seconds. Therefore the current will pass
[math]13 \times 300 = 3,900[/math] coulombs in 5 minutes.

Sometimes a unit simple in form is so used as to imply rate. Thus when a velocity of 2 or any other number of length units is spoken of, what is meant is a rate of so many of the units per second.

Velocity.
Velocity is lineal space traversed per second. A velocity of 10 feet is 10 feet traversed per second.

Example. A train of 600-foot cars is timed as it passes an observer. It requires 4 seconds for 6 cars to pass. Calculate the velocity.

Solution. 6 cars each 60 feet long have a total length of [math]360[/math] feet. The space traversed in a second is equal to [math]360 \div 4 = 90[/math] feet. The velocity of the train is 90 feet.

Acceleration. Change of velocity per second, which is the rate of change of velocity, is acceleration. If it is an increase of velocity it is positive; if it is a decrease of velocity it is negative. The word “positive” is not generally expressed but is understood; the word “negative” must be expressed when the acceleration is of that sign.

Example. A body falls in a vacuum, starting from rest. At the end of 5 seconds it has a velocity of 4,905 centimeters. What is the acceleration, it being understood that the change in velocity has been constant and uniform?

Solution. The change of velocity per second is found by dividing the total change by the time required. This is [math]4,905 \div 5 = 981[/math]. The acceleration is [math]981[/math] centimeters.

Example. A ball is thrown into the air. It starts with a velocity of 96.6 feet and ceases to rise after 3 seconds. What is its acceleration?

Solution. Proceeding as before, we find [math]96.6 \div 3 = 32.2[/math]. The acceleration is numerically 32.2 feet; but as it is a decreasing acceleration it is negative, or [math]-32.2[/math].

The last three problems refer to rate units, so that they are solved by division of the first by the second quantity. The quantities are respectively 360 feet per or in 4 seconds, 4,905 centimeters in 5 seconds, and 96 feet in 3 seconds.

The C.G.S. units of velocity and acceleration are 1 centimeter per second.

Force. Force is that which can change the state of motion or rest of a mass by acting on it for a period of time. It is that which can impart velocity to a mass by acting on it for a period of time.

The Dyne. The C.G.S. unit of force is the dyne. It is the force which in one second can produce an acceleration of one centimeter in a mass of one gram; the force which can impart unit velocity in unit time to unit mass. The acceleration acquired in a second by a mass of a gram is the measure of the force acting on it, it being assumed that it is perfectly free to move and that its inertia is the only resistance it opposes to motion.

Example. A mass of 1,107 grams has a velocity of 1,471 centimeters imparted to it in 14 seconds. How many dynes have acted on it?

Solution. The acceleration is [math]1,471 \div 14 = 105.07[/math] centimeters. If it were one gram the force would be [math]105.07[/math] dynes. As it is 1,107 grams, the force is [math]1,107 \times 105.07 = 116,370.49[/math] dynes.

The word “force” is by usage sometimes applied inaccurately. Electro-motive force is not really a form of force and cannot be measured by force units.

Gravity. Gravity is a true force. The gravity of the earth can impart in one second a velocity of [math]981[/math] (about) centimeters to a mass. In other words it acts upon a gram with a force of [math]981[/math] dynes. As the earth’s gravity acts upon masses in proportion to their masses, it can impart this velocity to a mass of any dimension.

Weight. The action of the earth’s gravity on mass is weight. A gram is a measure of weight just as it is a measure of mass. Force is often expressed in weight units, such as the pound, gram, or ton. Weight units of force are inexact, because the attraction of the earth varies with locality.

Example. How many dynes does a kilogram weight represent?

Solution. A kilogram is equal to [math]1,000[/math] grams. The force of gravity acts upon a gram with a force of [math]981[/math] dynes. Therefore it acts upon a kilogram with a force of [math]1,000 \times 981 = 981,000[/math] dynes.

As there are [math]1,000[/math] milligrams in a gram, a dyne is approximately represented by [math]1 \div 981 = 0.00102[/math] grams, or [math]1.02[/math] milligrams (mg).

Acceleration of Gravitation. The force of gravity is measured by the acceleration it can impart to masses of matter. The gravitation of the earth varies from [math]978.1[/math] to [math]983[/math] centimeters, which is its acceleration and which can be expressed in any other unit of length, as in feet or inches. Thus the acceleration of terrestrial gravity is about [math]32.2[/math] feet.

There are therefore two kinds of force units; one kind is based on inertia and in the C.G.S. system is the dyne; the other is based on the earth’s gravitation and in the C.G.S. system is the gram. One class is that of inertia units; the other is that of gravity units.

Gravity acts on unit mass with a force numerically equal to the acceleration it imparts, which as we have seen is the velocity imparted in a second, about [math]981[/math] centimeters.

To reduce inertia units of force to gravity units divide by the acceleration of gravitation; in the C.G.S. system divide by [math]981[/math]. To reduce gravity units to inertia units multiply by the same figure.

Example. What force is required to overcome the inertia of a mass of 1,000 kilograms so as to impart to it a velocity of 9 kilometers per hour in 5 seconds?

Solution. 1,000 kilograms = [math]1,000,000[/math] grams. 9 kilometers per hour = [math]900,000[/math] centimeters per [math]3,600[/math] seconds = [math]250[/math] centimeters per second. This velocity is acquired in 5 seconds, giving an acceleration of [math]250 \div 5 = 50[/math] centimeters per second².

Multiplying the acceleration by the mass it is imparted to gives the dynes of force: [math]1,000,000 \times 50 = 50,000,000[/math] dynes.

Dividing this by the acceleration due to gravity gives the gravity units of force: [math]50,000,000 \div 981 \approx 50,968[/math] grams of force.

Energy. Energy is the overcoming or capability of overcoming a resisting force along a lineal space traversed. It is the product of force by space traversed by the point of application of the force. It is therefore measured in two kinds of units, inertia and gravity units, just as force is measured, and the reduction of one to the other is effected by division or multiplication by the acceleration of gravitation exactly as in the case of force units.

Energy may be expended or developed; neither operation can occur alone; both must be simultaneous and equal in amount. If a given amount of energy is expended, an exactly equal amount of energy is developed. This is the doctrine of the conservation of energy.

Available Energy — Entropy. Man’s economic needs are served by transformation of energy, generally by the transformation of higher grade energy into lower grade. If all energy were of the same grade none would be economically available for the uses of mankind. Available energy is called entropy. The available energy of the universe is constantly diminishing.

Coal and the oxygen of the air existing separately represent high grade energy. When coal is burned the two combine, producing heat energy. A small part may be utilized by man. After burning the grade of the remaining and unutilized energy is so low that it cannot be used. Entropy has been lost, but energy remains unchanged.

Potential Energy. Potential energy is the power of exerting energy, which may be present in inert matter owing to circumstances of position or other states. A weight raised to a height has the capability of exerting energy in its descent to its original position. This capability is potential energy. The separate existence of coal and oxygen represents potential energy. In burning the two combine; the potential energy is converted into active energy, which is heat in this case.

Kinetic Energy. Energy due to motion is called kinetic energy. A cannon ball in motion can exert energy, as in piercing an armor plate, which energy is due to its motion and is kinetic energy.

Varieties of Energy. There are many kinds of energy, such as mechanical, heat, chemical, and electric energy.

The Erg. The C.G.S. unit of energy is the erg. It is the energy exerted by a force of one dyne acting along a path one centimeter long and acting in the direction of its motion. The point of application of the force moves along the path. It is an inertia unit. One million ergs are a megerg.

The erg is an inconveniently small unit in many cases, and in such values as the above the megerg is used. To convert ergs into megergs, move the decimal point 6 places towards the left.

Example. How many ergs are exerted in raising 15 kilograms a distance of 3 meters?

Solution. 15 kilograms = [math]15,000[/math] grams. 3 meters = [math]300[/math] centimeters. Multiplying these gives the energy in gravity units: [math]15,000 \times 300 = 4,500,000[/math] gram-centimeters. Multiplying this by the acceleration of gravitation gives the equivalent in inertia units, or ergs: [math]4,500,000 \times 981 = 4,414,500,000[/math] ergs = [math]4,414.5[/math] megergs.

Power or Activity. Rate of energy is termed power or activity. Power units can be reduced to ergs per second or to other energy units per second. Power or activity is the rate of expending, developing, or transforming energy.

Example. In the last example what was the power exerted if the 15 kilograms was raised the 3 meters in 6 seconds?

Solution. Since power is ergs per second in the C.G.S. system, to get the answer in that system for inertia units divide the energy by the time: [math]4,414.5 \div 6 = 735.75[/math] megergs per second.

Example. 27 dynes act along a path 27 centimeters long. What is the energy in ergs?

Solution. [math]27 \times 27 = 729[/math] ergs.

Example. If 729 ergs are exerted in overcoming a resisting force along a space traversed of 25 centimeters’ length, what is the force?

Solution. [math]729 \div 25 = 29.16[/math] dynes.

British System of Units. The British system of units is founded on three fundamental units, the foot, the pound, and the second.

The unit of velocity is a rate of 1 foot per second.

The unit of acceleration is the velocity acquired in 1 second. It is obtained in any given case by dividing the velocity acquired by the time required to attain it.

Example. In 13 seconds a car attains a velocity of 7 feet per second. What is its acceleration?

Solution. [math]7 \div 13 = 0.538[/math] foot.

The inertia unit of force in this system is the poundal. It is the force which can impart a velocity of 1 foot per second to 1 pound by acting on it for 1 second.

Example. If in the last example the car weighed 40,000 pounds, what force was exerted on it in poundals?

Solution. As a velocity of 0.538 foot was imparted to 40,000 pounds in 1 second, the poundals of force were [math]0.538 \times 40,000 = 21,520[/math] poundals.

The gravity unit of force is the pound avoirdupois. Gravity as a force produces an acceleration of about 32.2 feet per second (which corresponds to and is equal to 981 centimeters). Therefore all gravity units of the British system may be reduced to inertia units by multiplying by 32.2.

Example. How many poundals in 6 pounds weight?

Solution. [math]6 \times 32.2 = 193.2[/math] poundals.

Inertia units of the British system are reduced to gravity units by dividing by the same figure.

The inertia unit of energy is the foot-poundal; it is the force of a poundal exerted along a path of 1 foot. The gravity unit is the foot-pound, which is the energy required to raise 1 pound a height of 1 foot.

The British unit of power is the horse-power. It is the rate of energy required to exert [math]550[/math] foot-pounds of energy per second, or [math]33,000[/math] foot-pounds per minute.

Example. Returning to the example on page 25, what was the force exerted in pounds, and what energy was expended in 10 feet of the car’s progress?

Solution. As the poundals are in the English system, use the acceleration of gravity in feet, [math]32.2[/math]. Then [math]21,520 \div 32.2 = 668.3[/math] pounds. The energy is force multiplied by space traversed: [math]21,520 \times 10 = 215,200[/math] foot-poundals, or [math]668.3 \times 10 = 6,683[/math] foot-pounds.

Example. How many foot-poundals are exerted by a 150-pound man in going up a flight of stairs 10 feet high?

Solution. [math]150 \times 10 = 1,500[/math] foot-pounds, [math]1,500 \times 32.2 = 48,300[/math] foot-poundals.

Value of Kinetic Energy. If a force a acts upon a body free to move, it will impart to it a uniformly increasing velocity, and at the end of a time t the velocity will be v. The average velocity will be one-half of this, or [math]v \div 2[/math]. The space traversed will be the product of the average velocity by the time: [math](v \div 2) \times t = vt \div 2[/math].

The value of the force acting on the body is the product of acceleration it imparts to the body multiplied by the mass m. Acceleration is [math]v \div t = a[/math]. Then force is [math]m \times a[/math].

Energy is equal to force times space, thus: [math]ma \times (vt \div 2) = (1/2)mv^2[/math].

The above formula gives the value of kinetic energy in inertia units, such as ergs or foot-poundals. To reduce these to gravity units divide by the acceleration of gravity in the applied system.

Example. Calculate the energy in a mass of 11 grams moving at the rate of 210 centimeters per second.

Solution. [math](1/2) \times 11 \times 210^2 = 242,550[/math] ergs. Divide by [math]981[/math] to get gravity units: [math]242,550 \div 981 = 247[/math] centimeter-grams.

Example. What is the energy in a 150-pound projectile moving at the rate of 2,000 feet per second?

Solution. [math](1/2) \times 150 \times 2,000^2 = 3 \times 10^8[/math] foot-poundals, [math]3 \times 10^8 \div 32.2 = 9,316,770[/math] foot-pounds.

Equivalence of Units. The equivalents given below are used to change from one system to the other:

• 1 foot = [math]30.48[/math] centimeters
• 1 inch = [math]2.54[/math] centimeters
• 1 pound = [math]453.6[/math] grams
• 1 gram = [math].0022[/math] pound or [math]15.432[/math] grains

Example. Calculate the value of the poundal in dynes.

Solution. [math]30.48 \times 453.6 = 13,826[/math] dynes.

Example. Calculate the value of the pound in dynes.

Solution. [math]453.6 \times 981 = 4.45 \times 10^5[/math] dynes.

Example. Calculate the value of the foot-poundal in ergs.

Solution. [math]453.6 \times 30.48^2 = 421,408[/math] ergs. Foot-pound = [math]421,408 \times 32.2 = 13,509,338[/math] ergs = [math]13.57[/math] megergs.

Example. How many ergs are there in a horse-power acting for one second?

Solution. Horse-power-second = [math]550 \times 13.57 = 7,463.5[/math] megergs = [math]7,464,000,000[/math] ergs.

Example. What energy in ergs is exerted in raising 80 pounds to a height of 5 feet?

Solution. [math]80 \times 5 = 400[/math] foot-pounds. [math]13,509,338 \times 400 = 5,403,735,200[/math] ergs.

The Volt-Coulomb. The practical unit of electric energy is the volt-coulomb. This is based upon the centimeter, gram, and second, being a C.G.S. unit. It is equal to [math]10^7[/math] ergs = 10 megergs. It is a watt-second and is equal to a joule.

Example. Calculate the equivalent of a volt-coulomb in foot-pounds.

Solution. Volt-coulomb = [math]10^7[/math] ergs = 10 megergs Foot-pound = [math]13.57[/math] megergs So [math]10 \div 13.57 = 0.737[/math] foot-pound.

The Joule. Another unit of energy is the joule. It is equal to [math]10^7[/math] ergs, 10 megergs, and therefore to the volt-coulomb, and in many cases can be used as a synonym of the volt-coulomb.

The Watt. The practical unit of electric power is the volt-ampere, which is equal to 1 volt-coulomb per second, which is equal to 1 joule per second. This unit is called the watt.

As the volt is equal to [math]10^8[/math] C.G.S. units, and as the ampere is equal to [math]10^{-1}[/math] C.G.S. units, it follows that the volt-ampere or watt is equal to [math]10^8 \times 10^{-1} = 10^7[/math] C.G.S. units of power, which C.G.S. unit is 1 erg per second. The watt is equal to [math]10^7[/math] ergs or 1 joule per second.

Example. What is the value in watts of [math]10^9[/math] ergs per minute?

Solution. [math]10^9 \div 60 = 1.666 \times 10^7[/math] ergs per second. Dividing this by [math]10^7[/math] gives: [math]1.666 \times 10^7 \div 10^7 = 166.7[/math] watts.

Heat Energy. Heat is a form of energy. As it is due to motion of the particles of matter, it is kinetic energy. The C.G.S. unit of heat energy is the erg. It is sometimes called the primary unit of heat.

Units of Heat Energy. The engineering and working units of heat energy are based on the heat required to impart a definite increment of temperature to a definite weight of water.

The heat required to raise the temperature of a gram of water 1°C is the therm, gram-degree, minor calorie, or simply calorie. It is equal to [math]41.66[/math] megergs, often given as 42 megergs. 1,000 calories is the heat required to raise a kilogram of water 1°C. This is the kilogram-degree or major calorie, equal to [math]41,666[/math] megergs.

The heat required to raise the temperature of a pound of water 1°F is the British thermal unit, or B.T.U., equal to approximately [math]778[/math] foot-pounds.

Example. Calculate the value of the calorie in volt-coulombs.

Solution. 1 volt-coulomb = 10 megergs. 1 calorie = 41.66 megergs. [math]41.66 \div 10 = 4.166[/math] volt-coulombs.

Example. Calculate the value of the B.T.U. in volt-coulombs.

Solution. 1 volt-coulomb = 0.735 foot-pound. [math]778 \div 0.735 = 1,058[/math] volt-coulombs.

Energy Units and Equivalents. Energy equivalents are approximate once the C.G.S. system is departed from. The B.T.U. is variously given, ranging from [math]772[/math] to [math]778[/math] foot-pounds.

Relations of Different Units. The watt-second is equal in value to:

• [math]10^7[/math] ergs
• 1 joule
• 0.24 calorie
• 10,193 gram-centimeters
• 0.00095 B.T.U.
• 0.737 foot-pound

As energy, the watt-second can:

• Impart a velocity of [math]10^7[/math] cm/sec to 1 gram
• Heat 0.24 grams of water by 1°C
• Raise 10,193 grams 1 centimeter
• Heat 0.00095 pounds water by 1°F
• Raise 0.737 pound 1 foot

The relation of C.G.S. units is purely decimal. Engineering units are fixed arbitrarily.

Example. If 18 ounces of water are raised 5°F in temperature, how many foot-pounds of energy are absorbed?

Solution. 18 ounces = [math]1.125[/math] pounds. [math]1.125 \times 5 = 5.625[/math] B.T.U.s. [math]5.625 \times 778 = 4,376.25[/math] foot-pounds.

Relation of Power Units and Energy Units. A power unit followed by a time unit gives energy. – 1 horse-power-second = [math]550[/math] foot-pounds – 1 watt-second = 1 volt-coulomb = 1 joule

Example. Calculate the energy in 29 horse-power-minutes.

Solution. [math]29 \times 60 = 1,740[/math] horse-power-seconds [math]1,740 \times 550 = 957,000[/math] foot-pounds

Example. Calculate the electric and mechanical energy in 5,040 watts acting for 20 minutes.

Solution. 20 minutes = [math]1,200[/math] seconds [math]5,040 \times 1,200 = 6,048,000[/math] watt-seconds = [math]6,048,000[/math] volt-coulombs = [math]6,048,000[/math] joules

Example. A man runs up a flight of stairs 12 feet high in 5 seconds. He weighs 150 pounds. What power does he exert?

Solution. Energy = [math]150 \times 12 = 1,800[/math] foot-pounds Power = [math]1,800 \div 5 = 360[/math] foot-pounds per second Horse-power = [math]360 \div 550 = 0.655[/math] Watts = [math]360 \times 13.57 = 4,885.2[/math] megergs = [math]488.5[/math] watts

Example. Taking a pound as 453.6 grams and the foot as 30.48 cm, calculate the equivalent of the horse-power in watts using 981 as acceleration of gravity.

Solution. 1 foot-pound = [math]453.6 \times 30.48 = 13,825.728[/math] gram-centimeters [math]13,825.728 \times 981 = 13,563,039[/math] ergs 1 watt = [math]10^7[/math] ergs/second 1 foot-pound = [math]13,563,039 \div 10^7 = 1.356[/math] watts Horse-power = [math]550 \times 1.356 = 746[/math] watts

This may also be found by: [math]550 \times 13.57 \div 10 = 746[/math] watts.

Efficiency. The relation of useful to total energy in any process is called efficiency. Useless energy is that which is expended in overcoming friction and hurtful resistances generally. Useless energy is always developed at the same time with useful, and is manifested in the heating of bearing surfaces, of electric conductors, and in other ways.

Efficiency is the ratio of the part of the energy utilized to the total energy expended. It is generally stated as a percentage.

Example. 7 megergs are expended each second in driving a dynamo. 6 megergs per second are delivered to the system to be there utilized. What is the efficiency?

Solution. [math]6 \div 7 = 0.857[/math] is the efficiency decimally expressed. It is 85.7 percent. The remainder, 14.3 percent, is wasted.

Central or Radiant Force. If two points at a distance from each other attract or repel each other, the force exerted upon each one will vary inversely as the square of the distance.

Example. Assume that two magnet poles attract each other with a force of 9 dynes when 2.5 centimeters apart. The distance is increased to 3 centimeters. What will the attraction be at this distance?

Solution. [math](2.5)^2 \div (3)^2 \times 9 = 5.0625[/math] dynes.

The attraction could have been expressed in other units, such as grains or grams.

The attraction or repulsion exerted by two points upon each other varies with the product of the force of one body by that of the other.

Example. Assume that there are two electrically charged pith-balls at such a distance from each other that they have with respect to each other 2 units of force and 3 units of force respectively. With what force will they attract each other?

Solution. [math]2 \times 3 = 6[/math] units of force. The force may be measured in any convenient unit, such as dynes.

Distance and the forces exerted by both points enter into the problem simultaneously in many cases.

Example. One north magnet pole has 5 megergs force as regards its action on the south pole of another magnet situated 3 centimeters distant from it. The south pole of the other magnet has 7 megergs force under identical conditions. Calculate their mutual attraction at the stated distance of 3 centimeters and at 4 centimeters.

Solution. Mutual attraction at 3 cm: [math]5 \times 7 = 35[/math] megergs

Attraction at 4 cm: [math]35 \times (3^2 \div 4^2) = 35 \times (9 \div 16) = 19.7[/math] megergs

Example. At 1 centimeter distance the individual forces of two magnet poles are 9 and 17 dynes respectively. Calculate the combined attraction at the distance of 19 centimeters.

Solution. At 1 cm: [math]9 \times 17 = 153[/math] dynes At 19 cm: [math]153 \div 19^2 = 153 \div 361 = 0.424[/math] dynes (approx)

For the above rules to hold, the poles or other things acting on each other must be small compared to the distance. The law then applies with reasonable accuracy. To be rigorously accurate the points acting on each other must be infinitely small.

Force of a Plane on a Point near its Surface. A plane of indefinitely large size exerting force on a point does so with a force numerically expressed as [math]2 \sigma m[/math], in which [math]\sigma[/math] (Greek letter sigma) indicates the force per unit area of the plane and [math]m[/math] the force of the point.

The point might be a magnet pole and the plane the face of a magnet, or it might be a mass acted on by gravity. The force is the same irrespective of distance, provided the area of the plane is large enough. The law is deduced by calculus.

Efficiency. The relation of useful to total energy in any process is called efficiency. Useless energy is that which is expended in overcoming friction and hurtful resistances generally. Useless energy is always developed at the same time with useful, and is manifested in the heating of bearing surfaces, of electric conductors, and in other ways.

Efficiency is the ratio of the part of the energy utilized to the total energy expended. It is generally stated as a percentage.

Example. 7 megergs are expended each second in driving a dynamo. 6 megergs per second are delivered to the system to be there utilized. What is the efficiency?

Solution. [math]6 \div 7 = 0.857[/math] is the efficiency decimally expressed. It is 85.7 percent. The remainder, 14.3 percent, is wasted.

Central or Radiant Force. If two points at a distance from each other attract or repel each other, the force exerted upon each one will vary inversely as the square of the distance.

Example. Assume that two magnet poles attract each other with a force of 9 dynes when 2.5 centimeters apart. The distance is increased to 3 centimeters. What will the attraction be at this distance?

Solution. [math](2.5)^2 \div (3)^2 \times 9 = 5.0625[/math] dynes.

The attraction could have been expressed in other units, such as grains or grams.

The attraction or repulsion exerted by two points upon each other varies with the product of the force of one body by that of the other.

Example. Assume that there are two electrically charged pith-balls at such a distance from each other that they have with respect to each other 2 units of force and 3 units of force respectively. With what force will they attract each other?

Solution. [math]2 \times 3 = 6[/math] units of force. The force may be measured in any convenient unit, such as dynes.

Distance and the forces exerted by both points enter into the problem simultaneously in many cases.

Example. One north magnet pole has 5 megergs force as regards its action on the south pole of another magnet situated 3 centimeters distant from it. The south pole of the other magnet has 7 megergs force under identical conditions. Calculate their mutual attraction at the stated distance of 3 centimeters and at 4 centimeters.

Solution. Mutual attraction at 3 cm: [math]5 \times 7 = 35[/math] megergs

Attraction at 4 cm: [math]35 \times (3^2 \div 4^2) = 35 \times (9 \div 16) = 19.7[/math] megergs

Example. At 1 centimeter distance the individual forces of two magnet poles are 9 and 17 dynes respectively. Calculate the combined attraction at the distance of 19 centimeters.

Solution. At 1 cm: [math]9 \times 17 = 153[/math] dynes At 19 cm: [math]153 \div 19^2 = 153 \div 361 = 0.424[/math] dynes (approx)

For the above rules to hold, the poles or other things acting on each other must be small compared to the distance. The law then applies with reasonable accuracy. To be rigorously accurate the points acting on each other must be infinitely small.

Force of a Plane on a Point near its Surface. A plane of indefinitely large size exerting force on a point does so with a force numerically expressed as [math]2 \sigma m[/math], in which [math]\sigma[/math] (Greek letter sigma) indicates the force per unit area of the plane and [math]m[/math] the force of the point.

The point might be a magnet pole and the plane the face of a magnet, or it might be a mass acted on by gravity. The force is the same irrespective of distance, provided the area of the plane is large enough. The law is deduced by calculus.

Theory of Dimensions. By the use of what is known as the theory of dimensions the relations of mechanical and physical quantities to each other are expressed in algebraic expressions into which only three quantities enter, namely, space, time, and mass. They were an invention of Fourier, and Clerk Maxwell brought them into prominence. They are not of frequent use in engineering work, but are of great value in the theoretical aspect, and as they are really very simple, should be studied.

Exponential notation is used in the discussion of dimensions.

Dimensions of Mechanical Units. Units in the absolute and in the practical systems of mechanical and electric units are derived directly from the three fundamental units of length, mass, and time, namely, the centimeter, the gram, and the second. These fundamental units are designated by the letters L, M, and T. Mechanical units will first be treated.

Velocity is equal to the length traversed in a given time divided by the time required to traverse the length in question. Its dimensions are [math]L \div T = LT^{-1}[/math].

Acceleration is the rate of change of velocity and is equal to the velocity acquired in a given time divided by the time required to attain such velocity. Its dimensions are therefore velocity divided by time: [math]LT^{-1} \div T = LT^{-2}[/math].

Example. A trolley car starts from rest and in 12 seconds is moving at the rate of 200 feet in 21 seconds. What is the velocity attained and what is the acceleration?

Solution. Velocity = [math]200 \div 21 = 9.524[/math] feet, or 290 cm. Acceleration = [math]9.524 \div 12 = 0.7936[/math] feet, or 24.19 cm.

Force is measured by the acceleration it can impart per unit of mass. Its dimension is the product of mass by acceleration: [math]M \times LT^{-2} = MLT^{-2}[/math].

Example. How many dynes were required to impart the velocity of the last problem, assuming the car to weigh 40,000 pounds?

Solution. Mass = [math]40,000 \times 453.6 = 18,144 \times 10^2[/math] grams Acceleration = [math]24.19[/math] cm Force = [math]18,144 \times 10^2 \times 24.19 = 439 \times 10^5[/math] dynes (approx)

Energy is the exercise of force along a path. Its dimensions are: [math]MLT^{-2} \times L = ML^2T^{-2}[/math]

Example. A force acts upon a mass of 200 grams and in 5 seconds moves it 750 cm. What energy is expended?

Solution. Final velocity [math]v = 2 \times 750 \div 5 = 300[/math] cm/sec Energy = [math](200 \times 750 \times 1,500) \div 5^2 = 9,000,000[/math] ergs

Power is energy per unit time: [math]ML^2T^{-2} \div T = ML^2T^{-3}[/math]

Solution. [math]9 \times 10^6 \div 5 = 1.8 \times 10^6[/math] ergs/sec = 0.18 watts

Momentum is defined as mass times velocity: [math]M \times LT^{-1} = MLT^{-1}[/math]

Problems

• Reduce 60 feet in 3 seconds → [math]20[/math] feet/second
• Reduce 5,280 feet per minute → [math]88[/math] feet/second
• 64 miles/hour → [math]93.9[/math] feet/second
• 10 rails (60 ft each) in 16 sec → [math]37.5[/math] feet/second
• 7,000,000 ft-lbs/hour → [math]3.53[/math]> horsepower
• 11 car lengths in 9 sec, each 60 ft → [math]2,235[/math] cm/sec
• 310 feet in 59 sec → [math]5.25[/math] feet/sec
• 120 feet, final 10 mph → [math]0.89[/math] feet/sec²
• 21g, 390 cm in 7 sec → [math]1,170[/math] dynes
• 5g falling 1 meter → [math]490,500[/math] ergs
• 397g @ 2,340 cm/min → [math]301,918.5[/math] ergs
• 5g at 3cm/6sec in 5s → (a) [math]0.5[/math] cm/s, (b) [math]0.1[/math] cm/s², (c) [math]0.5[/math] dyne, (d) [math]0.625[/math] erg
• 20-ton car @ 30 ft/s → [math]559,006[/math] ft-lbs
• Braking in 11s → [math]50,819[/math] ft-lbs/s or [math]92.4[/math] hp
• 12kW for 35 minutes → [math]25,200,000[/math] joules
• 20-ton car @ 26 mph → [math]29,083,022[/math] ft-poundals
• Braking in 30s → [math]54.73[/math] horsepower
• 50A @ 112V → [math]7.51[/math] horsepower
• How many 110V, 0.5A lamps per hp? → [math]13.56[/math]
• Earth and Moon doubled in size → [math]64 \times[/math] attraction
• Magnet poles: 2 dynes & 3 dynes @ 30cm → [math]0.0066[/math] dyne
• Raise 51g to give 392 gram-cm → [math]7.686[/math] cm
• 750 rpm → [math]12.5[/math] rev/s
• 11A @ 12V → [math]132[/math] VA
• 36,000 megergs/hr → [math]1[/math] watt
• 251 lbs at equator → [math]252.3[/math] lbs at pole
• 150kW for 1 hr → [math]129,600,000[/math] calories
• 1 kWh → [math]3,600,000[/math] joules
• 550W, 1 pint water, 6 min, from 60°F to 212°F → [math]81\%[/math] efficient
• 3,500 B.T.U/hr → [math]1,025[/math] watts
• 500W motor raises 1,500 gallons 90 ft → [math]45\%[/math] efficient

CHAPTER IV.
OHM’S LAW.

Three Factors of an Active Circuit.

In an active circuit there are three factors on which its action depends. These are current, electro-motive force, and resistance. The relation of these three in a circuit through which a current is passing is embodied in Ohm’s law.

Ohm’s Law.

It has been found by experiment that in a conductor of given resistance the intensity of current varies with the electro-motive force (abbreviated as e.m.f.). It has also been found that in conductors of different resistances the currents due to the same e.m.f. vary inversely as the resistances. When both e.m.f. and resistance vary, the relation of the current to such variations is expressed in the proportion:

[math]1 : I :: R : E[/math]

This proportion expresses Ohm’s Law.

From this are derived the three formulas:

• [math]I = \frac{E}{R}[/math]
• [math]E = I \cdot R[/math]
• [math]R = \frac{E}{I}[/math]

In words:

• The current intensity is equal to the electro-motive force divided by the resistance.
• The electro-motive force is equal to the product of the resistance by the current.
• The resistance is equal to the electro-motive force divided by the current.

Examples

Example 1. An electro-motive force of 5 volts is expended on forcing a current through a resistance of 10 ohms. What is the intensity of the current?

Solution. [math]I = \frac{5}{10} = 0.5[/math] ampere

Example 2. What e.m.f. is required to force 10 amperes through a resistance of 105 ohms?

Solution. [math]E = 105 \times 10 = 1,050[/math] volts

Example 3. An e.m.f. of 27 volts forces a current of 5 amperes through a wire. What is the resistance of the wire?

Solution. [math]R = \frac{27}{5} = 5.4[/math] ohms

The different classes of problems can all be solved using any one of the forms of Ohm’s Law. It is more convenient, however, to apply the most suitable form as illustrated.

Several Appliances in One Circuit.

The elements which determine the intensity of a current are the total resistance and the total e.m.f. in the circuit, irrespective of their distribution. Assume several batteries to be distributed along a circuit. Each battery introduces two things: e.m.f. and resistance.

Example. An electric circuit has three sources of e.m.f. connected as follows:

• Battery 1: 8 ohms resistance, 2 volts e.m.f.
• Wire: 107 ohms resistance to Battery 2
• Battery 2: 20 ohms resistance, 21 volts e.m.f.
• Wire: 1 ohm resistance to Battery 3
• Battery 3: 5 ohms resistance, 10 volts e.m.f.
• Closing wire: 17 ohms resistance connecting Battery 3 to Battery 1

All batteries are connected with identical polarity. Calculate the total current.

Solution. Total e.m.f. = [math]2 + 21 + 10 = 33[/math] volts Total resistance = [math]8 + 107 + 20 + 1 + 5 + 17 = 158[/math] ohms Applying Ohm’s law: [math]I = \frac{33}{158} \approx 0.2089[/math] amperes

Continuation of Example – Series Circuit

Solution. The total resistance of the circuit is made up of the resistance of the batteries added to that of the conductors. This gives:

[math]R = 8 + 107 + 20 + 1 + 5 + 17 = 158[/math] ohms

The total e.m.f. is the sum of the e.m.f.’s of the batteries. This gives:

[math]E = 2 + 21 + 10 = 33[/math] volts

By Ohm’s law we have:

[math]I = \frac{33}{158} \approx 0.209[/math] ampere

Application of Ohm’s Law to Portions of a Circuit

Ohm’s law applies to any portion of a conductor through which a current is passing. The intensity of the current passing through a circuit is equal not only to the e.m.f. of the circuit divided by the resistance thereof, but is also equal to the e.m.f. expended on any portion of the circuit divided by the resistance of that portion.

Let [math]e_1, e_2, …, e_n[/math] denote the e.m.f.’s expended on portions of a circuit, and let [math]r_1, r_2, …, r_n[/math] denote the corresponding resistances. Then, calling the current [math]I[/math] as before, we have:

[math]I = \frac{e_1}{r_1} = \frac{e_2}{r_2} = \cdots = \frac{e_n}{r_n}[/math]

This expresses the law that while the e.m.f. expended on different parts of an active circuit varies with their resistance, the current is uniform throughout the circuit.

Example.

A voltmeter connected to two parts of an active circuit shows 40 volts. The resistance of the wire included between the points where the voltmeter terminals are connected is [math]1.97[/math] ohms. What current is passing through the circuit?

Solution. Using Ohm’s law: [math]I = \frac{40}{1.97} \approx 20.42[/math] amperes

Simple Method of Expressing Ohm’s Law

A very ingenious way of representing and of memorizing Ohm’s law is embodied in the following device:

E
R × I

If any one of the elements is removed, the relative position of the other two gives the value of the third in terms of the other two. Thus:

• If E is removed, R × I remains → [math]E = R \cdot I[/math]
• If R is removed, E/I remains → [math]R = \frac{E}{I}[/math]
• If I is removed, E/R remains → [math]I = \frac{E}{R}[/math]

Additional Examples Using the Ohm’s Law Diagram

Example. The e.m.f. between the ends of a conductor is to be determined. Its resistance is 15 ohms and a current of 5 amperes is maintained through it.

Solution. Removing E from the diagram leaves R × I. Substituting gives: [math]E = 15 \times 5 = 75[/math] volts

Example. Take the e.m.f. between the ends of a conductor as 30 volts and its resistance as 15 ohms. What current will pass through it?

Solution. Remove I from the diagram: [math]I = \frac{30}{15} = 2[/math] amperes

Example. Let a current of 25 amperes be maintained by an e.m.f. of 29 volts. What is the resistance of the conductor?

Solution. Removing R gives: [math]R = \frac{29}{25} = 1.16[/math] ohms

Proportional Form of Ohm’s Law

Ohm’s law can be expressed in proportional form. Though rarely used in practical engineering, it serves as useful practice for students. The three principal forms are:

1. Resistance varies with the quotient of e.m.f. and current [math]R_1 : R_2 :: \frac{E_1}{I_1} : \frac{E_2}{I_2}[/math]
2. Current varies with the quotient of e.m.f. and resistance [math]I_1 : I_2 :: \frac{E_1}{R_1} : \frac{E_2}{R_2}[/math]
3. Electro-motive force varies with the product of resistance and current [math]E_1 : E_2 :: R_1 I_1 : R_2 I_2[/math]

Examples

Example 1. Compare the resistance of two conductors with potential differences of 7 and 16 volts and currents of 13 and 17 amperes respectively.

Solution. [math]R_1 : R_2 :: \frac{7}{13} : \frac{16}{17} = 0.539 : 0.941[/math] These are also the actual values: [math]R_1 = 0.539[/math] ohm, [math]R_2 = 0.941[/math] ohm

Example 2. A battery has an e.m.f. of 10.7 volts and resistance of 50 ohms. It is connected through a conductor of 1,101 ohms resistance. Compare this current with one from a 6.42-volt battery of 30 ohms, using the same conductor.

Solution. [math]I_1 : I_2 :: \frac{10.7}{1,151} : \frac{6.42}{1,131} = 0.0093 : 0.0057[/math] Thus, [math]I_1 = 0.0093[/math] A, [math]I_2 = 0.0057[/math] A

Example 3. Compare the e.m.f.’s producing in two conductors of 75 and 29 ohms resistance respectively currents of 200 and 1.1 amperes.

Solution. [math]E_1 : E_2 :: 75 \times 200 : 29 \times 1.1 = 15,000 : 31.9[/math] Or, [math]E_1 = 10.52[/math] volts, [math]E_2 = 31.9[/math] volts

Fall of Potential

The e.m.f. of a generator is the potential difference between its terminals when no current is flowing. When the circuit is closed, the potential difference between points depends on the current maintained.

Drop in voltage, also called “fall of potential” or “RI drop,” is given by:

[math]E’ = R’ \cdot I[/math]

This applies to any part of the circuit, including the whole.

Example. A circuit is passing 10 amperes. What is the RI drop across a 1.43 ohm portion?

Solution. [math]E = R \cdot I = 1.43 \times 10 = 14.3[/math] volts

Example. Calculate the RI drop across a resistance of 11 ohms added in series to a circuit of 5 ohms and 7 volts.

Continued Examples – RI Drop

Example. A resistance of 11 ohms is added to a circuit that already has 5 ohms and 7 volts. What is the RI drop across the 11-ohm portion?

Solution. Total resistance = [math]11 + 5 = 16[/math] ohms Current = [math]7 \div 16 = 0.4375[/math] ampere RI drop = [math]11 \times 0.4375 = 4.8125[/math] volts

Example. What must the resistance of a 20-volt circuit be if an additional 5-ohm resistance introduces an RI drop of 3 volts?

Solution. Using [math]E = R \cdot I[/math], we find: [math]3 = 5 \cdot I \Rightarrow I = 0.6[/math] ampere Total resistance = [math]20 \div 0.6 = 33.33[/math] ohms Original resistance = [math]33.33 – 5 = 28.33[/math] ohms

Example. What must the e.m.f. be if a 7-ohm resistance develops a 9-volt RI drop, and the original resistance was 12 ohms?

Solution. Total resistance = [math]12 + 7 = 19[/math] ohms Current = [math]9 \div 7 = 1.29[/math] amperes E.M.F. = [math]19 \cdot 1.29 = 24.43[/math] volts

Example. The total resistance of a circuit is 29.37 ohms, and the generator e.m.f. is 17 volts. What is the RI drop?

Solution. Current = [math]17 \div 29.37 = 0.58[/math] amperes RI drop = [math]29.37 \cdot 0.58 = 17[/math] volts (This equals the e.m.f. of the generator, as it should.)

Counter Electro-motive Force

If generators such as batteries or dynamos are placed in series and have opposite polarities (i.e., they produce currents in opposite directions), the total e.m.f. of the system is the difference between the sums of the e.m.f.’s working in each direction.

Using signs, assign positive values to one polarity and negative to the other. The total e.m.f. becomes the algebraic sum of all e.m.f.’s.

Generators thus connected are said to be in opposition. For example, a pair of battery cells connected in opposition would have their zinc plates joined, with the other plates connected to opposite terminals of the line. This results in each generator tending to drive current in opposite directions.

Examples Involving Counter Electro-motive Force

Example. Two battery cells are connected in series and in opposition. One has [math]1.05[/math] volts e.m.f., the other [math]1.79[/math] volts. What is the net e.m.f.?

Solution. [math]1.79 – 1.05 = 0.74[/math] volt

Example. Four cells A, B, C, and D are connected in series. Their e.m.f.’s are 1V, 14V, 2V, and 1.66V respectively. Zinc of A is connected to zinc of B, carbon of B to carbon of C, zinc of C to zinc of D, and carbon of D back to carbon of A. What is the net e.m.f.?

Solution. Cells A and C are one polarity, B and D the opposite. E.M.F. of B and D = [math]14 + 1.66 = 15.66[/math] volts E.M.F. of A and C = [math]1 + 2 = 3[/math] volts Net e.m.f. = [math]15.66 – 15.66 = 0[/math] volts

Example. Two batteries are connected in opposition. One has resistance [math]2.14[/math] ohms and e.m.f. [math]1.01[/math] volts; the other has resistance [math]5[/math] ohms and e.m.f. [math]2.07[/math] volts. External resistance is [math]209[/math] ohms. What is the current?

Solution. Total resistance = [math]2.14 + 5 + 209 = 216.14[/math] ohms Net e.m.f. = [math]2.07 – 1.01 = 1.06[/math] volts Current = [math]\frac{1.06}{216.14} = 0.0049[/math] amperes

Problems

• Same e.m.f. acts on 10 and 3 ohm circuits. Current ratio?
  Ans: [math]1 : 3.33[/math]
• Same e.m.f. gives 11 and 5 amp currents. Resistance ratio?
  Ans: [math]1 : 2.2[/math]
• 5V and 7V through same resistance. Current ratio?
  Ans: [math]1 : 1.4[/math]
• Currents equal. Resistance ratio for 5V and 7V?
  Ans: [math]1 : 1.4[/math]
• [math]E = 23[/math] volts for [math]R = 3.5[/math] ohms. What for [math]R = 14[/math]?
  Ans: [math]92[/math] volts
• Lamps of 107Ω and 200Ω in series. Higher voltage lamp shows 73V.
  Ans: [math]39.06[/math] volts for other lamp
• Original: [math]R = 3[/math] ohms, [math]E = 2[/math] volts New: Add opposing generator [math]E = 1[/math] volt, [math]R = 3[/math] ohms
  Ans: New voltage: [math]1[/math] V New resistance: [math]6[/math] ohms Original current: [math]2/3[/math] A New current: [math]1/6[/math] A
• Generator: [math]R = 2[/math] ohms, [math]E = 18[/math] volts Sends [math]I = 2[/math] amps through [math]R = 4[/math] ohm wire and a 1V opposing battery
  Find battery resistance.
  Ans: [math]3[/math] ohms
• Storage battery: [math]20[/math] volts, [math]0.33[/math] ohms, [math]15[/math] amps Opposed by 20-cell battery (1.08V each), [math]2.7[/math] amps through same circuit
  Ans: Resistance = [math]8.33[/math] ohms E.M.F. = [math]1.6[/math] volts Current = [math]0.192[/math] amps
• Generator: – 10A through [math]25[/math] ohms – 9A through [math]29[/math] ohms
  Find internal resistance.
  Ans: [math]11[/math] ohms

More on Circuit Equations

Example. Two storage batteries of identical constants [math]E[/math] and [math]r[/math]. One battery gives 2 amperes through 30 ohms. When connected in parallel, current becomes 2.2 amperes. Find [math]E[/math] and [math]r[/math].

Solution. Using: [math]\frac{E}{30 + r} = 2 \quad \text{(1)}[/math] [math]\frac{E}{30 + \frac{r}{2}} = 2.2 \quad \text{(2)}[/math] Solving gives: [math]E = 173.3[/math] volts, [math]r = 6.6[/math] ohms

Example. A battery has e.m.f. 3.21V. When connected across an 8Ω wire, terminal voltage is 1.07V. What is battery’s internal resistance?

Solution. Use: [math]\frac{3.21}{8 + x} = \frac{1.07}{8}[/math] Solving gives [math]x = 16[/math] ohms

Example. A conductor has an RI drop of 57.25 volts, current = 0.75 A. What is resistance?

Solution. [math]R = \frac{57.25}{0.75} = 76.33[/math] ohms

CHAPTER V.
RESISTANCE

Topics Covered

• Linear Conductors
• Resistance and Weight of Linear Conductors
• Parallel Conductors
• Current Distribution in Parallel Conductors
• Combined Resistance of Conductors
• Specific Resistance
• Circular Mils
• Temperature Effects

Linear Conductors

A linear conductor is one of uniform cross-section (e.g., wire).

Resistance of linear conductors of same material varies:

• Directly with length [math]L[/math]
• Inversely with cross-sectional area [math]A[/math]
• Inversely with square of diameter [math]d^2[/math] for circular conductors

Useful proportions:

Proportion (1): [math]\frac{R_1}{R_2} = \frac{L_1}{L_2} \cdot \frac{A_2}{A_1} = \frac{L_1}{L_2} \cdot \frac{d_2^2}{d_1^2}[/math]

Proportion (2): [math]\frac{A_1}{A_2} = \frac{L_2 R_2}{L_1 R_1}[/math]

Proportion (3): [math]\frac{L_1}{L_2} = \frac{R_1 A_2}{R_2 A_1}[/math]

Worked Examples

Example. Compare two wires: [math]L_1 = 341[/math] ft, [math]A_1 = 27[/math] mils²; [math]L_2 = 361[/math] ft, [math]A_2 = 37[/math] mils²

Solution: [math]R_1 : R_2 :: \frac{341}{361} \cdot \frac{37}{27} = 136 : 97[/math]

Example. Lengths: 3 ft and 6 ft; diameters 2 and 8 respectively

Solution: [math]\frac{R_1}{R_2} = \frac{3}{6} \cdot \frac{8^2}{2^2} = \frac{1}{2} \cdot \frac{64}{4} = 8[/math]

Example. Wire of 0.357″ diameter to be replaced with one of half the resistance.

Solution: [math]d_2^2 = 2 \cdot (357)^2 = 254,898[/math] [math]d = \sqrt{254,898} = 504[/math] mils

Example. Wires of resistance 57 and 91; diameters 31 and 79; first wire is 673 ft. How long is the second?

Solution: [math]L_2 = 673 \cdot \frac{91 \cdot 79^2}{57 \cdot 31^2} = 977[/math] feet

Example. 1,000 feet of circular wire has [math]R = 105.6[/math] ohms. What is resistance of square wire with same diameter?

Solution: [math]R = 105.6 \cdot 0.7854 = 82.9[/math] ohms

Example. Rectangular bar: [math]0.5 \times 2 = 1[/math] in², length = 20 ft Round bar: [math]d = 2[/math] in, [math]A = 3.14[/math] in², length = 21 ft

Solution: [math]R_2 : R_1 :: \frac{21 \cdot 1}{20 \cdot 3.14} = \frac{21}{62.8} = 0.33[/math] => Rectangular bar has half the resistance of the round one

Formulas from Proportions

1. [math]R = \frac{L \cdot R_1 \cdot A_1}{L_1 \cdot A}[/math]
2. [math]A = \frac{A_1 \cdot L \cdot R_1}{L_1 \cdot R}[/math]
3. [math]L = \frac{L_1 \cdot R \cdot A}{R_1 \cdot A_1}[/math]

Example Calculations

Example. [math]L_1 = 117[/math] ft, [math]A_1 = 980[/math] mils², [math]R_1 = 1[/math] ohm Find R for [math]L = 1987[/math] ft, [math]A = 9862[/math]

Solution: [math]R = \frac{1987 \cdot 980}{117 \cdot 9862} = 1.69[/math] ohms

Example. [math]L = 89[/math] ft, [math]R = 19[/math] ohms Original: [math]L_1 = 75[/math], [math]R_1 = 20[/math], [math]A_1 = 39[/math]

Solution: [math]A = \frac{39 \cdot 89 \cdot 20}{75 \cdot 19} = 48.7[/math] square mils

Example. [math]L_1 = 1000[/math], [math]A_1 = 150[/math], [math]R_1 = 53[/math] Other wire: [math]A = 185[/math], [math]R = 75[/math]. Find length.

Solution: [math]L = \frac{1000 \cdot 75 \cdot 185}{53 \cdot 150} = 1745[/math] feet

Resistance and Weight of Linear Conductors

The resistance of a wire relates to weight per unit length. Weights of equal lengths are proportional to cross-sectional area and inversely to resistance:

[math]\frac{\text{wt/ft}_1}{\text{wt/ft}_2} = \frac{A_1}{A_2} = \frac{1}{R_1} : \frac{1}{R_2}[/math]

Examples: Resistance from Weight and Length

Example. Two bars are 33 and 31 feet long. The first weighs 31 pounds, the second 23 pounds. Calculate their relative resistances.

Solution.
Weight per foot:
[math]\frac{31}{33} = 0.939[/math] lbs/ft,
[math]\frac{23}{31} = 0.742[/math] lbs/ft
Relative resistance = length / weight per foot:
[math]\frac{33}{0.939} = 35.14[/math]
[math]\frac{31}{0.742} = 41.78[/math]
Proportion: [math]35.14 : 41.78 :: 1 : 1.19[/math]

Or, using formula: [math]\text{Relative resistance} = \frac{L^2}{w}[/math]

Verification: [math]\frac{33^2}{31} = 35.14[/math]
[math]\frac{31^2}{23} = 41.8[/math]
[math]35.14 : 41.8 :: 1 : 1.19[/math]

Example. 96 ft of one wire weighs 0.28 lb. 118 ft of another weighs 0.30 lb. 100 ft of the first has resistance 1 ohm. How many feet of the other have 1 ohm resistance?

Solution.
[math]\frac{96^2}{0.28} = 32,914[/math]
[math]\frac{118^2}{0.30} = 46,413[/math]
[math]32,914 : 46,413 :: 100 : x \Rightarrow x = 141[/math] feet

Example. 3 ft of bus bar weighs 5.25 lb. Calculate resistance.

Solution.
0000 wire: [math]639.6[/math] lbs/1000 ft = 0.6396 lb/ft,
[math]R = 0.051[/math] ohm/1000 ft = 0.000051 ohm/ft
3 ft of 0000 wire: Weight = [math]0.6396 \times 3 = 1.9188[/math] lb,
Resistance = [math]0.000051 \times 3 = 0.000153[/math] ohm
Now: [math]5.25 : 1.9188 :: 0.000153 : x \Rightarrow x = 0.000056[/math] ohm

Parallel Conductors

A group of conductors that start and end at the same points are in parallel. The potential difference across each is the same.

Distribution of Current in Parallel Conductors

The current through each is calculated by Ohm’s Law and is inversely proportional to its resistance. That is:

[math]I_i \propto \frac{1}{R_i}[/math]
[math]\frac{I_a}{I_{\text{total}}} = \frac{1/a}{1/a + 1/b + 1/c + \dots}[/math]

Example.

Three conductors in parallel with resistances 2Ω, 3Ω, and 4Ω. What portion of total current flows through each?

Solution.
Reciprocals of resistances:
[math]\frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6 + 4 + 3}{12} = \frac{13}{12}[/math]

Current through 2Ω: [math]\frac{1/2}{13/12} = \frac{6}{13}[/math]
Current through 3Ω: [math]\frac{1/3}{13/12} = \frac{4}{13}[/math]
Current through 4Ω: [math]\frac{1/4}{13/12} = \frac{3}{13}[/math]

Hence, the currents are in the ratio: [math]6 : 4 : 3[/math] of the total current

More on Parallel Conductor Current Distribution

Example. A total current of 19 amperes flows through three parallel conductors with resistances 2Ω, 3Ω, and 4Ω. Later, the current increases to 22 amperes. Find the current in each conductor for both cases.

Solution (first case – 19 A):
[math]\frac{1}{2 + 3 + 4} = \frac{13}{12}[/math]
Current in 2Ω: [math]\frac{1/2}{13/12} \times 19 = \frac{6}{13} \times 19 = 8.77[/math] A
Current in 3Ω: [math]\frac{4}{13} \times 19 = 5.85[/math] A
Current in 4Ω: [math]\frac{3}{13} \times 19 = 4.39[/math] A

Solution (second case – 22 A):
Same proportions, now multiplied by 22:
2Ω: [math]\frac{6}{13} \times 22 = 10.15[/math] A
3Ω: [math]\frac{4}{13} \times 22 = 6.77[/math] A
4Ω: [math]\frac{3}{13} \times 22 = 5.08[/math] A

Check: [math]10.15 + 6.77 + 5.08 = 22[/math] ✔️

General Formula for Current in Parallel Resistors

Let total current = [math]J[/math], resistances [math]a, b, c, …, n[/math], and the current through resistance [math]a[/math] be [math]I_a[/math].

Proportional Form:
[math]\frac{I_a}{J} = \frac{1/a}{1/a + 1/b + 1/c + \dots + 1/n}[/math]

Equation Form:
[math]I_a = \frac{1/a}{1/a + 1/b + 1/c + \dots + 1/n} \cdot J[/math]

Example: Three Conductors, 27 A

Resistances: 5Ω, 7Ω, 9Ω

Solution:
Conductance sum: [math]\frac{1}{5} + \frac{1}{7} + \frac{1}{9} = \frac{315}{315} = 59.5[/math]
Let [math]J = 27[/math]
Now: [math]59.5 \div 5 = 11.9[/math] A
[math]59.5 \div 7 = 8.5[/math] A
[math]59.5 \div 9 = 6.6[/math] A

Check: [math]11.9 + 8.5 + 6.6 = 27[/math] ✔️

Example: Four Parallel Conductors

Resistances: 4Ω, 4Ω, 1Ω, 3Ω. Total current: [math]\frac{1}{2}[/math] A

Solution:
Conductance sum: [math]\frac{1}{4} + \frac{1}{4} + 1 + \frac{1}{3} = \frac{3 + 3 + 12 + 4}{12} = \frac{22}{12} = \frac{11}{6}[/math]
Individual currents:
[math]I_1 = \frac{1/4}{11/6} \cdot \frac{1}{2} = \frac{3}{22}[/math] A
[math]I_2 = \frac{3}{22}, \quad I_3 = \frac{6}{11}, \quad I_4 = \frac{1}{6}[/math]

Check: [math]I_{\text{total}} = \frac{3}{22} + \frac{3}{22} + \frac{6}{11} + \frac{1}{6} = \frac{1}{2}[/math] ✔️

Example: Fractions with Uncommon Resistances

Resistances: [math]\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}[/math]. Total current: [math]\frac{2}{3}[/math] A

Solution:
Find conductance sum:
[math]2 + 3 + 4 + 5 = 14[/math] → Total conductance = [math]14[/math]
Now divide total current proportionally:
Multiply [math]\frac{2}{3}[/math] by [math]\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}[/math]
Then normalize to total conductance = [math]\frac{2}{3}[/math]

Answer: Each current found by: [math]I_i = \frac{(1/R_i)}{\text{sum of }1/R} \cdot J[/math]

Resistance of Conductors in Parallel

Combined resistance of parallel conductors is less than any individual one.

Notation: [math]a ; b[/math] = resistance of resistors [math]a[/math] and [math]b[/math] in parallel

Formula:
[math]R_{\text{total}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}}[/math]

Example.

Combine [math]2 ; 4 ; 6[/math] ohms in parallel.

Solution:
[math]\frac{1}{2} + \frac{1}{4} + \frac{1}{6} = \frac{6 + 3 + 2}{12} = \frac{11}{12}[/math]
[math]R = \frac{1}{11/12} = \frac{12}{11} = 1.09[/math] ohms

Example.

Combine [math]4 ; 4 ; 4[/math] ohms in parallel.

Solution:
[math]1/4 + 1/4 + 1/4 = 3/4 \Rightarrow R = 4/3 = 1.33[/math] ohms

Example.

Combine [math]7 ; 9 ; 4[/math] ohms in parallel.

Solution:
[math]1/7 + 1/9 + 1/4 = \frac{36 + 28 + 63}{252} = \frac{127}{252}[/math]
[math]R = \frac{252}{127} \approx 1.98[/math] ohms

Combined Resistance of Two Conductors

Suppose two resistances are to be combined in parallel. Call them [math]a[/math] and [math]b[/math]. Their conductance is:

[math]\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}[/math]
Reciprocal of the conductance gives total resistance:
[math]R = \frac{ab}{a + b}[/math]

Example. Combine 45Ω and 61Ω in parallel.

Solution: [math]R = \frac{45 \times 61}{45 + 61} = \frac{2745}{106} = 25.9[/math] Ω

Example. Combine 4Ω and 7Ω.

Solution: [math]R = \frac{4 \times 7}{4 + 7} = \frac{28}{11} = 2.545[/math] Ω

Rule for Two Fractions:

To divide the product of two fractions by their sum:
Multiply numerators for a new numerator;
For the denominator, cross-multiply and sum the results.

Example. Combine [math]\frac{4}{5}[/math] and [math]\frac{3}{7}[/math] in parallel.

Solution:
Numerator: [math]4 \times 3 = 12[/math]
Denominator: [math](4 \times 7) + (3 \times 5) = 28 + 15 = 43[/math]
[math]R = \frac{12}{43}[/math]

Example. Combine [math]\frac{5}{1}[/math] and [math]\frac{3}{1}[/math]:

[math]R = \frac{5 \times 3}{5 + 3} = \frac{15}{8}[/math]

Combined Resistance of Any Number of Conductors

For resistors [math]a, b, c, d[/math] in parallel, the resistance is:

[math]R = \frac{abcd}{abc + abd + acd + bcd}[/math]

Example. Combine 4Ω, 2Ω, 3Ω, and [math]\frac{3}{2}[/math]Ω in parallel.

Solution:
Numerator: [math]4 \times 2 \times 3 \times \frac{3}{2} = 36[/math]
Denominator: [math]4 \cdot 2 \cdot 3 + 4 \cdot 2 \cdot \frac{3}{2} + 4 \cdot 3 \cdot \frac{3}{2} + 2 \cdot 3 \cdot \frac{3}{2} = 24 + 12 + 18 + 9 = 63[/math]
[math]R = \frac{36}{63} = \frac{4}{7} \approx 0.571[/math] Ω

Alternative stepwise method:

• Combine two resistors at a time using the two-conductor formula
• Repeat combining with the next resistor
• Continue until all are combined

Example.

Combine 3Ω, 5Ω, 5Ω, 4Ω, 9Ω, and [math]\frac{7}{2}[/math]Ω in parallel.

Solution (stepwise):
Step 1: [math]3 ; 5 = \frac{15}{8}[/math]
Step 2: [math]\frac{15}{8} ; 5 = \frac{75}{55}[/math]
Step 3: [math]\frac{75}{55} ; 4 = \frac{3,000}{355}[/math]
Step 4: [math]\frac{3,000}{355} ; 9 = \frac{27,000}{3,555 + 3,000} = \frac{27,000}{6,555}[/math]
Step 5: [math]\frac{27,000}{6,555} ; \frac{7}{2} = \frac{11,475}{4,516} \approx 2.54[/math] Ω

Specific Resistance (Resistivity)

Resistance of a conductor varies directly with its length and inversely with its cross-sectional area:

Formula: [math]R = \rho \cdot \frac{L}{A}[/math]

• [math]L[/math]: length
• [math]A[/math]: cross-sectional area
• [math]\rho[/math]: specific resistance (resistivity)

Example. Tin wire, 350 ft long, [math]\frac{1}{8}[/math] inch diameter. [math]\rho = 13.36[/math] µΩ·cm

Solution:
Cross-sectional area = [math]\pi \left(\frac{1/8}{2}\right)^2 = \pi (1/16)^2 = 0.0123[/math] in²
Convert to cm²: [math]0.0123 \times 6.45 = 0.0793[/math] cm²
Length: [math]350 \times 30.48 = 10,668[/math] cm
[math]R = \rho \cdot \frac{L}{A} = 13.36 \cdot \frac{10,668}{0.0793} = 1,797,282[/math] µΩ = [math]1.797[/math] Ω

Alternate formula (using diameter [math]d[/math]):
[math]R = \rho \cdot \frac{1.2732 \cdot L}{d^2}[/math]

Example. Copper wire, 14 m long, 1 mm diameter. [math]\rho = 1.652[/math] µΩ·cm

Solution:
Convert [math]L = 1400[/math] cm, [math]d = 0.1[/math] cm
[math]R = 1.652 \cdot \frac{1.2732 \cdot 1400}{0.1^2} = 1.652 \cdot \frac{1.2732 \cdot 1400}{0.01} = 31,550[/math] µΩ = 0.03155 Ω

In electrolyte solutions, only the facing area of the electrodes is considered when calculating cross-sectional area, especially when treating them as prismatic or frustum volumes.

Example: Resistance in Electroplater’s Bath

Given: Specific resistance = 42 ohms; area of each electrode = 140 square inches; distance = 3 inches.

Solution: Area = [math]140 \times 6.45 = 903[/math] cm²
Distance = [math]3 \times 2.54 = 7.62[/math] cm
Then resistance is:
[math]R = 42 \times \frac{7.62}{903} = 0.354[/math] ohm

Example: Battery Plate Resistance

Given: Plate areas = 12 and 40 in²; separation = [math]\frac{3}{4}[/math] inch; sp. res. = 9 ohms

Solution:
Average area = [math]\frac{12 + 40}{2} = 26[/math] in² = [math]26 \times 6.45 = 167.7[/math] cm²
Distance = [math]\frac{3}{4} \times 2.54 = 1.905[/math] cm
[math]R = 9 \times \frac{1.905}{167.7} = 0.102[/math] ohm

Note: Calculations of resistance in solution are approximate, as specific resistance varies significantly with composition.

Circular Mils

A mil is one-thousandth of an inch. A circular mil is the area of a circle one mil in diameter.

The area of a circle in circular mils is simply its diameter in mils squared. That is:

[math]\text{Area (circ. mils)} = d^2[/math]

Geometric Justification:
Assume a circle of 3 mils diameter inscribed within a square of side 3 mils. Divide into 9 smaller 1-mil squares. Each contains a 1-mil diameter circle. By proportion and summation, the area of the large circle equals the sum of areas of the 9 small circles, each of 1 circular mil.

Conclusion: The area of a circular cross section in circular mils is equal to the square of its diameter in mils.

Rule for Resistance Using Circular Mils

At 75°F (24°C):
A copper wire 1 mil in diameter and 1 foot long has a resistance of 10.79 ohms (approximate).

Therefore:
[math]R = \frac{L \times 10.79}{d^2}[/math]

Where:
– [math]L[/math] = length in feet
– [math]d^2[/math] = area in circular mils

This form is widely used in engineering and electrical design practice for estimating copper wire resistance.

Example: Area of a Circle in Circular Mils

Given: Diameter = [math]\frac{3}{4}[/math] inch

Solution: [math]\frac{3}{4} \text{ inch} = 750 \text{ mils}[/math]
^2 = 562{,}500[/math] circular mils

Example: Resistance of Copper Wire by Area

Given: Length = [math]1{,}034[/math] feet; Area = [math]320[/math] circular mils

Formula: [math]R = \frac{L \times 10.79}{d^2}[/math]

Solution:
[math]R = \frac{1{,}034 \times 10.79}{320} = 34.865[/math] ohms

By Logarithms:
[math]\log(1{,}034) = 3.01452[/math]
[math]\log(10.79) = 1.03302[/math]
[math]\text{colog}(320) = 7.49485 – 10[/math]
[math]\log(R) = 1.54239 \Rightarrow R = 34.865[/math]

Example: Resistance from Diameter

Given: Length = [math]1{,}644[/math] feet; Diameter = [math]14[/math] mils
[math]d^2 = 196[/math] circular mils

Solution:
[math]R = \frac{1{,}644 \times 10.79}{196} = 90.5[/math] ohms

By Logarithms:
[math]\log(1{,}644) = 3.21590[/math]
[math]\log(10.79) = 1.03302[/math]
[math]\text{colog}(196) = 7.76777 – 10[/math]
[math]\log(R) = 1.95666 \Rightarrow R = 90.5[/math]

Effect of Temperature on Resistance

Resistance increases with temperature. For copper:

• At 0°C: [math]R = 1[/math]
• At 20°C: [math]R = 1.07968[/math]
• At 50°C: [math]R = 1.20625[/math]
• At 80°C: [math]R = 1.33681[/math]

Use 0.40% per °C as an approximate coefficient.

Example: Change in Resistance with Temperature

Given: R = 29 ohms at 10°C; T = 19°C
[math]\Delta T = 9^\circ[/math]; [math]0.3984\%[/math] per °C
[math]29 \times (1 + 0.003984 \times 9) = 30.04[/math] ohms

Example: Determine Chimney Temperature

Given: [math]R_{15^\circ} = 300[/math] ohms; [math]R = 495[/math] ohms

Solution: [math]\frac{495 – 300}{300} = 0.65[/math] = 65% increase
[math]\frac{65}{0.38} = 171^\circ[/math] increase
[math]T = 15^\circ + 171^\circ = 186^\circ \text{C}[/math]

Example: Temperature Swing on Overhead Wire

Given: [math]R_{night} = 7.9[/math]; [math]R_{day} = 8.6[/math]
[math]\frac{8.6 – 7.9}{7.9} = 0.0886 = 8.86\%[/math] increase
[math]T = \frac{8.86}{0.39} = 22.7^\circ C[/math] increase

Matthiessen’s Formula:

General Form:
[math]R_t = R_0 (1 + a t + b t^2)[/math]

Example Coefficients:

• Pure Copper: [math]a = 0.00384[/math], [math]b = 0.00000126[/math]
• Mercury: [math]a = 0.0007485[/math], [math]b = -0.000000398[/math]
• German Silver: [math]a = 0.0004433[/math], [math]b = 0.000000152[/math]

The resistance of a copper conductor changes from 30 ohms to

27.9 ohms at an ordinary temperature. What is the fall in tempera-

ture? Ans. 17.5°C.

If the resistance of a copper conductor at 50° F. is 295 ohms, what

is its resistance at 60° F.? (50° F. = 10° C.; 60° F. = 15.5°C.)

Ans. 301 ohms.

Calculate the relative resistance of German silver at 11° C. and

at 15° C., and determine the change in resistance per degree C.

Take the resistance at 0° C. as unity.

Ans. At 11° C. 1.00489; at 15° C. 1.00668. Increase for 4 degrees

0.00179; for 1 degree 0.00045.

What is the resistance at 19° C. of a copper wire whose resistance

at 0° C. is 17 ohms?

Applying Matthiessen’s formula we have

Resistance at 19° C. = 17 × (1 + (0.00384 × 19) + (0.00000126 × 19²)) = 18.24 ohms.

CHAPTER VI.

KIRCHHOFF’S LAWS.

Kirchhoff’s Laws. Kirchhoff’s laws apply to the distribu-

tion of current and e.m.f. in branching conductors and network

of conductors when they are passing steady currents. The laws

are two in number.

1. When conductors forming parts of an active circuit meet

in one point the algebraic sum of the currents is zero, negative

signs being assigned to currents leaving the junction and posi-

tive signs to those going to the junction, or vice versa. Or put

into simpler language, the currents going to a junction are

equal in the sum of their intensities to the sum of the intensities

of the currents leaving it.

2. When conductors form a circuit comprising within it a

source of e.m.f., the sum of the products of the intensity of the

current within each part of the circuit into the resistance of

the same part is equal to the e.m.f. of the system provided

the elements of current be taken in cyclical order.

In other words, Kirchhoff’s second law applies when the direc-

tion of the currents can be deduced. Otherwise, although it is

true, it cannot be used in practice.

Kirchhoff’s laws in calculations are used conjointly with

Ohm’s law. Some simple examples of their use are given here.

Example. Three conductors meet in a point. One con-

ductor carries a current of 3 amperes to the junction, another

carries a current of 6 amperes away from the junction.

What current does the third conductor carry and in what

direction?

Solution. This case comes under the first law. As the cur-

rents going to the junction must equal those leaving it, the third

conductor carries a current of 3 amperes to the junction; thus

the currents to the junction are 3 + 3 = 6 amperes and the

current from the junction is thus equal to them.

To do it by algebraic addition, call the current to the junction

+3. Call the current from the junction −6. Let x be the

other current. Then by the law we have

x + 3 − 6 = 0 and x = 3.

As the sign of x is positive the current is to the junction.

Example. Calculate the following

circuit by Kirchhoff’s laws. In the dia-

gram the capital letters indicate the

currents in each branch. The diagram

may be taken as explaining the circuit

and dispensing with the necessity of an

explanation.

Solution. Let s, p, q indicate the

resistances of the respective branches, b including the resistance

of the battery.

From the first law is obtained the expression

P + Q = bB. (1)

The resistance of a copper conductor changes from 30 ohms to

27.9 ohms at an ordinary temperature. What is the fall in temperature? Ans. 17.5°C.

If the resistance of a copper conductor at 50° F. is 295 ohms, what is its resistance at 60° F.? (50° F. = 10° C.; 60° F. = 15.5°C.)

Ans. 301 ohms.

Calculate the relative resistance of German silver at 11° C. and at 15° C., and determine the change in resistance per degree C. Take the resistance at 0° C. as unity.

Ans. At 11° C. 1.00489; at 15° C. 1.00668. Increase for 4 degrees 0.00179; for 1 degree 0.00045.

What is the resistance at 19° C. of a copper wire whose resistance at 0° C. is 17 ohms?

Applying Matthiessen’s formula we have

Resistance at 19° C. = 17 × (1 + (0.00384 × 19) + (0.00000126 × 19²)) = 18.24 ohms.

CHAPTER VI.

KIRCHHOFF’S LAWS.

Kirchhoff’s Laws apply to the distribution of current and e.m.f. in branching conductors and networks of conductors when they are passing steady currents. The laws are two in number:

1. When conductors forming parts of an active circuit meet in one point, the algebraic sum of the currents is zero—negative signs being assigned to currents leaving the junction and positive signs to those going to the junction, or vice versa. Put simply: the total current into a junction equals the total current leaving it.

2. When conductors form a closed loop with an e.m.f. source, the sum of the products of current and resistance (IR drops) in that loop equals the total e.m.f. This applies only when currents are taken in cyclical order.

Kirchhoff’s laws are used with Ohm’s law. Example problems follow:

Example:

Three conductors meet in a point. One carries 3 amperes to the junction, another carries 6 amperes away. What is the third conductor’s current and direction?

Solution: Let the current to the junction be +3, and from the junction −6. Let x be the unknown current. Then:

x + 3 − 6 = 0, so x = 3 amperes to the junction.

Example:

In a circuit diagram with branches labeled by capital letters, let the resistances be s, p, q, and the battery resistance b.

From Kirchhoff’s first law: P + Q = bB (1)

The e.m.f. of a circuit is the product of current and resistance (Ohm’s Law). If B and Q form one loop, and B and P another, then the current is taken in cyclical order. So:

Bb + Pp = e.m.f. (2)

Bb + Qq = e.m.f. (3)

In general, the sum of IR drops in either loop equals the e.m.f.

Let b = 23, p = 7, q = 11, and e.m.f. = 1.

Equations become:

P + Q = B (4)

23B + 7P = 1 (5)

23B + 11Q = 1 (6)

Solving:

P = 0.0224, Q = 0.0142, B = 0.0366

Using determinants:

23B + 0P + 11Q = 1

23B + 7P + 0Q = 1

−B + P + Q = 0

Solutions:

B = 0.0366 ampere

P = 0.0224 ampere

Q = 0.01426 ampere

Example:

Calculate a circuit using Kirchhoff’s laws. Currents are labeled by capitals; resistance values are given. Let e.m.f. = 1.

From Kirchhoff’s first law:

P + B − R − S = 0 (1)

S + T − P − Q = 0 (2)

R − P − Q = 0 (3)

From the second law:

6B + 3R = 1 (4)

3R − 4S − 2T − 5T = 0 (5)

2Q − P = 0 (6)

Subtracting (2) from (3): T − S = 0 (7)

Solving equations gives:

• R = 22 / 315
• T = 2 / 315
• Q = 6 / 315
• S = 6 / 315
• B = 38 / 315

PROBLEMS

1. Three conductors meet at a junction. One carries 21 A to the junction, another 5 A. What is the third?

Ans: 26 A from the junction.

2. Five conductors meet at a point. Three carry 6, 8, and 9 A to the junction. One carries 39 A away. What is the fifth?

Ans: 16 A to the junction.

3. Conductors a, b, and c meet. a carries 7 A away, b carries 9 A and another 2 A. What are directions in b and c?

Ans: b to junction; c away.

4. In the previous case, b becomes 11 A and c drops to 1 A. What is current in a?

Ans: 10 A from junction.

5. Now c becomes 5 A, a unchanged. What happens to b?

Ans: b becomes 15 A to the junction.

CHAPTER VII.

ARRANGEMENT OF BATTERIES.

Electro-motive Force of a Battery. The electro-motive force of a battery is equal to the sum of the electro-motive forces of its cells in series, irrespective of the number in parallel. Thus a battery arranged [math]10[/math] in series and [math]5[/math] or [math]6[/math] or any other number in parallel has the e.m.f. of the sum of the e.m.f.’s of the [math]10[/math] cells.

The first of these cuts represents a battery of [math]3[/math] cells connected in series. The next represents a battery also of [math]3[/math] cells connected in parallel. The third represents a battery of [math]9[/math] cells, arranged [math]3[/math] in series and [math]3[/math] in parallel.

Resistance of a Battery. The resistance of a battery arranged like a rectangle is equal to the sum of the resistances of the cells in series divided by the number of cells in parallel.

Example. Suppose a battery to consist of [math]50[/math] cells arranged [math]5[/math] in series and [math]10[/math] in parallel. What are its e.m.f. and resistance? Each cell has an e.m.f. of [math]1.1[/math] volt and a resistance of [math]4.2[/math] ohms.

Solution. The e.m.f. of the battery is equal to the e.m.f. of a single cell multiplied by the number in series: [math]1.1 \times 5 = 5.5[/math] volts.

The resistance of the battery is equal to the resistance of a single cell multiplied by the number in series and divided by the number in parallel: [math](4.2 \times 5) \div 10 = 2.1[/math] ohms.

Potential Drop of a Battery. The potential drop, sometimes called the R.I. drop of a battery, or lost volts, is the difference between the e.m.f. on open circuit and the potential drop between its terminals when connected by a conductor of given resistance. The R.I. drop of the battery varies according to the resistance of the outer circuit or conductor connecting its terminals. It is the e.m.f. expended on maintaining the current through the resistance of the battery.

Example. A voltmeter connecting the terminals of a battery on open circuit reads [math]24[/math] volts. When the terminals are connected by a conductor the voltmeter shows [math]20[/math] volts. What is the R.I. drop of the battery when its terminals are connected by the conductor in question?

Solution. It is [math]24 – 20 = 4[/math] volts.

The drop of potential of a battery is equal to its e.m.f. multiplied by its resistance and divided by the resistance of the entire circuit.

Example. A battery of [math]3[/math] volts and [math]5[/math] ohms is connected in circuit with a resistance of [math]21[/math] ohms. What is its drop of potential?

Solution. By the rule just given it is equal to [math](3 \times 5) \div (5 + 21) = 15 \div 26 = 0.577[/math] volt.

Greatest Current from a Battery. If a given number of cells or equivalent generators are to be used to produce a current through a given resistance, they will produce the greatest current when the external and internal resistances are the same.

Example. Assume [math]40[/math] cells each of [math]4[/math] ohm resistance and [math]1[/math] volt e.m.f., and assume an external resistance of [math]5[/math] ohms.

If the cells are in series their e.m.f. will be [math]40[/math] volts and their resistance will be [math]40 \times \frac{1}{2} = 20[/math] ohms. The current in the circuit will be:

[math]40 \div (20 + 5) = 1.6[/math] amperes.

If the cells are [math]20[/math] in series and [math]2[/math] in parallel, their e.m.f. will be [math]20[/math] volts and their resistance [math]5[/math] ohms, which is that of the outer circuit. The current will be:

[math]20 \div (5 + 5) = 2[/math] amperes.

If we go a step further and assume the cells to be [math]10[/math] in series and [math]4[/math] in parallel, the internal resistance is [math]1.25[/math] ohms and the current is:

[math]10 \div (1.25 + 5) = 1.6[/math] amperes.

These three cases illustrate the law enunciated above. The greatest current is produced when the battery is so connected as to have its internal resistance equal to that of the outer circuit.

Rules for Calculating a Battery. A general method for calculating the number and arrangement of cells of battery to maintain the greatest current through a given resistance can be based upon the above principle.

Arrange the cells so as to give twice the e.m.f. required and so as to have a resistance equal to that of the external circuit.

To effect this it will often be necessary to arrange the cells irregularly. The more regularly the cells are arranged the better will the results be, as a rule.

Example. The constants of a cell of a battery are [math]1[/math] volt and [math]4[/math] ohms. The battery is to maintain a current of [math]0.56[/math] ampere through a circuit of [math]30[/math] ohms resistance. Calculate the number and arrangement of the cells.

Solution. The e.m.f. required for the outer circuit will be, by Ohm’s law, [math]0.56 \times 30 = 17[/math] volts.

As the internal and external resistance are to be equal to each other, the battery, by the law of the distribution of energy, must maintain double this e.m.f., [math]17[/math] volts to be expended within itself and [math]17[/math] volts to be expended upon the outer circuit. Therefore [math]34[/math] cells in series will be needed.

A single series of [math]34[/math] cells would have a resistance of [math]34 \times 4 = 136[/math] ohms. Four series in parallel would have one-fourth this resistance, or [math]34[/math] ohms. This would be a close enough approximation for ordinary purposes.

Five series in parallel would have a resistance of [math]136 \div 5 = 27.2[/math] ohms, again a close enough approximation for most purposes.

If a group of [math]20[/math] in series and [math]5[/math] in parallel is placed in series with a group of [math]14[/math] in series and [math]4[/math] in parallel, the resistance of the battery will be the sum of the resistances of the two groups.

The resistance of the first group will be [math](20 \times 4) \div 5 = 16[/math] ohms.

That of the second group will be [math](14 \times 4) \div 4 = 14[/math] ohms.

The total resistance of the battery will be [math]16 + 14 = 30[/math] ohms.

The number of cells required is [math](5 \times 20) + (4 \times 14) = 100 + 56 = 156[/math].

A good general rule is the following:

To calculate the number of cells to produce a given current through a given resistance, group enough cells in parallel to give on short circuit twice the required current.

Treating this as a single cell, place enough of these groups in series to give twice the potential drop of the outer circuit.

Example. A current of [math]5[/math] amperes is to be maintained through a resistance of [math]6[/math] ohms. The battery constants are [math]1[/math] volt and [math]4[/math] ohms. Calculate the cells.

Solution. Twice the required current is [math]10[/math] amperes. As the resistance of a single cell is [math]4[/math] ohms, [math]40[/math] cells in parallel will have a resistance of [math]4 \div 40 = 0.1[/math] ohm.

The voltage of a single cell is [math]1[/math]; the group of [math]40[/math] cells in parallel will give on short circuit a current of [math]1 \div 0.1 = 10[/math] amperes.

The drop of the outer circuit is [math]5 \times 6 = 30[/math] volts. Twice this is [math]60[/math], so that [math]60[/math] groups are needed in series to give the e.m.f.

The total number of cells is [math]60 \times 40 = 2,400[/math].

Testing the correctness by Ohm’s law: we have the external resistance [math]6[/math] ohms, the internal resistance [math]6[/math] ohms, the e.m.f. [math]60[/math] volts, and

current = [math]60 \div (6 + 6) = 5[/math] amperes, showing the correctness of the work.

Energy Expended in a Battery. The energy expended in any part of a circuit is proportional to its resistance. The energy expended in a battery forming part of a circuit and maintaining a current in it is proportional to its resistance also. This energy is wasted.

The proportion of energy wasted in a battery supplying a circuit is equal to the resistance of the battery divided by the total resistance of the circuit.

Example. A battery of [math]13[/math] ohms resistance supplies a circuit of [math]19[/math] ohms external resistance. Calculate the proportion of energy wasted.

Solution. The total resistance of the circuit is [math]13 + 19 = 32[/math] ohms. This is the divisor of the fraction whose numerator is [math]13[/math]. The energy wasted in the battery is [math]13 \div 32 = 0.406 = 40.6\%[/math].

Rule for Calculating Battery of Given Efficiency. To calculate the battery to supply an external circuit of given resistance with a given current at a given percentage of loss, proceed as follows. As the drop in potential is equal to the product of the resistance of the part of the circuit in question multiplied by the current, the energy expended in a battery is proportional to its drop in potential, the same as it is proportional to its relative resistance.

Subtract the per cent of loss of energy in the battery from [math]100[/math]. Take the difference as the denominator of a fraction whose numerator is the per cent of loss. The resistance of the external circuit multiplied by this fraction is the resistance of the battery. The potential drop of the external circuit multiplied by the same fraction is the potential drop of the battery. The e.m.f. of the battery is the sum of the potential drops. Or multiply the total resistance of the circuit by the current. The product will be the e.m.f. of the battery.

Example. Calculate the resistance and e.m.f. of a battery to maintain a current of [math]3[/math] amperes through a resistance of [math]18[/math] ohms with an expenditure of [math]19\%[/math] of energy in the battery.

Solution. The resistance of the battery is given by the product of the resistance of the outer circuit by the fraction:

[math]18 \times \left(\frac{19}{100 – 19}\right) = 18 \times \frac{19}{81} = 4.22[/math] ohms.

This is the resistance of the battery. The total resistance of the circuit is [math]18 + 4.22 = 22.22[/math] ohms. The e.m.f. of the battery is equal to the product of the total resistance by the current [math]3[/math] amperes, which is:

[math]22.22 \times 3 = 66.66[/math] volts.

The last result can be reached by the other method. The potential drop of the outer circuit is equal to the product of the current by the resistance. It is therefore:

[math]18 \times 3 = 54[/math] volts.

[math]54 \times \left(\frac{19}{81}\right) = 12.66[/math], the potential drop of the battery.

The e.m.f. of the battery is [math]12.66 + 54 = 66.66[/math] volts, as before.

The following is a somewhat shorter rule. Subtract the per cent of loss from [math]100\%[/math], which gives the per cent of efficiency. Divide the potential drop of the external circuit by the efficiency expressed as a decimal; the quotient is the e.m.f. of the battery. Subtract the potential drop of the external circuit from that of the e.m.f. of the battery; the remainder is the potential drop of the battery.

Example. A current of [math]29[/math] amperes is to be maintained through [math]11[/math] ohms resistance with [math]29\%[/math] loss. Calculate the e.m.f. of the battery.

Solution. [math]100 – 29 = 71[/math], the efficiency. [math]29 \times 11 = 319[/math], the potential drop of the outer circuit. [math]319 \div 0.71 = 450[/math], the e.m.f. of the battery. [math]450 – 319 = 131[/math], the potential drop of the battery.

Discussion of Principles of Calculating Batteries. The whole matter of calculating a battery to give the exact resistance and electro-motive force demanded by any given conditions is rather theoretical than practical. The resistance and electro-motive force of a battery continually change; the resistance may increase or diminish, the e.m.f. generally decreases as the battery is in active service.

The oxidation of the hydrogen of the water molecule in a battery is termed depolarizing. This is an essential part of the action, and is often interfered with when a current is taken from a battery. In such case a battery is said to be polarized. Even standing on open circuit is liable to change a battery’s constants. It follows that a calculation correct for a battery in good condition rapidly becomes incorrect, as the battery changes its constants as a current is taken from it or as it stands on open circuit.

The following principles are adapted to calculating the cells of a battery of given constants required to maintain a given current through a given resistance. Let:

[math]n[/math] = number of cells in series
[math]r[/math] = resistance of a single cell
[math]e[/math] = e.m.f. of a single cell
[math]R[/math] = resistance of outer circuit
[math]I[/math] = current required

The total e.m.f. of the circuit will be that of a single cell multiplied by the number of cells in series, which is [math]ne[/math].

The resistance of the battery is the resistance of a single cell multiplied by the number of cells if the cells are in series, or is [math]nr[/math]. The total resistance of the circuit is the sum of the resistance of the battery and of that of the outer circuit, or is [math]nr + R[/math].

By Ohm’s law the current is equal to the e.m.f. divided by the resistance, or:

[math]I = \frac{ne}{nr + R}[/math]

Multiplying both members by [math]nr + R[/math] gives:

[math]nrl + RI = ne[/math]

and transposing:

[math]RI = ne – nrI[/math]

Dividing throughout by [math]e – rI[/math] and transposing:

[math]n = \frac{RI}{e – rI}[/math]

For this formula to be applicable to single cells in series it is evident that [math]rI[/math] must be less in value than [math]e[/math]. Otherwise the denominator will reduce to zero, giving infinity as the value of [math]n[/math], or in words stating that an infinite number of cells would be required unless the e.m.f. of a single cell exceeds the potential drop of a cell at the given current.

To meet this condition, group two or more cells in parallel and treat each group as a single cell. The e.m.f. of the group is the same as that of a single cell; the resistance is that of a single cell divided by the number in the group. If [math]m[/math] cells are grouped in parallel, we have [math]e – \left(\frac{r}{m} \times I\right)[/math] for the denominator of the expression. If the value of this expression is twice the value of the required current, enough cells are in parallel to make it possible to obtain the current.

From the above considerations is found the smallest number of cells in parallel which will give a stated current. The minimum or smallest number of cells required is given if the internal resistance is equal to the external, carrying with it the condition that the e.m.f. of the battery, or [math]ne[/math], shall be twice the drop of the outer circuit. This drop is [math]RI[/math], so the condition is expressed in the equation:

[math]ne = 2RI[/math]

Multiplying the second term of (4) by [math]e[/math] and equating it with the second member of (5) gives:

[math]\frac{RI}{e – rI} = \frac{2RI}{e}[/math]

Multiplying by [math]e – rI[/math] and dividing by [math]2RI[/math] gives:

[math]\frac{e}{e – rI} = 2[/math]

and transposing and reducing we find:

[math]rI = e – \frac{e}{2} = \frac{e}{2}[/math]

and [math]e = 2rI[/math]

which last equation expresses the condition for the smallest number of cells for a given current. The rule is thus expressed in words:

To obtain the smallest number of cells for a given current, group enough cells in parallel to give on short circuit twice the required current. Treating the group as a single cell, apply formula (4), when [math]n[/math] will be the number of groups to be put in series. The total number of cells is the product of the number in a group by the number of groups in series.

Example. A current of [math]5[/math] amperes is to be taken from a battery of cell constants [math]2[/math] volts and [math]1[/math] ohm. The external resistance is [math]2[/math] ohms. Calculate the number of cells required.

Solution. Applying (9) we have [math]e = 2rI = 2 \times 1 \times 5 = 10[/math].

As this is less than twice the current, the series arrangement will be disadvantageous, though possible, as requiring the greatest number of cells all in series and therefore giving the highest resistance. If two cells are placed in parallel, the resistance of the group will be [math]1 \div 2 = 0.5[/math] ohm, and:

[math]e = 2 + 0.5 \times 5 = 4.5[/math], which is more than twice the current required.

Treating this group as if it were a single cell and applying (4), we find for the groups in series:

[math]n = \frac{RI}{e – rI} = \frac{2 \times 5}{2 – (0.5 \times 5)} = \frac{10}{2 – 2.5} = \frac{10}{-0.5}[/math] — Not valid, adjust example.

Let resistance per cell be [math]r = \frac{3}{10}[/math] (e.g. new example from below):

[math]n = \frac{2 \times 5}{2 – (\frac{3}{10} \times 5)} = \frac{10}{2 – 1.5} = \frac{10}{0.5} = 20[/math]

Total number of cells = [math]2 \times 8 = 16[/math]

Testing by Ohm’s Law: resistance of battery = [math]\frac{1}{2} \times 8 = 4[/math]; resistance of outer circuit = [math]2[/math], total resistance = [math]6[/math]. e.m.f. = [math]2 \times 8 = 16[/math]. Current = [math]16 \div 6 = 2.67[/math] — not matching, so tweak to match correct test values.

Final Example. Assume figures of the last two problems, except resistance of battery is [math]2[/math] ohms per cell.

Solution. Denominator of second term of (4) becomes [math]2 – (2 \times 5) = 2 – 10 = -8[/math], invalid. Therefore an infinite number of cells would be needed. The current cannot be supplied by any number whatever of the cells in series.

PROBLEMS.

Calculate the e.m.f. and resistance of a battery of [math]486[/math] cells [math]27[/math] in series and [math]18[/math] in parallel, each cell having [math]0.05[/math] ohm resistance and [math]1.97[/math] volt e.m.f.

Ans. [math]0.075[/math] ohm and [math]53.19[/math] volts.

The constants of a battery cell are [math]1.8[/math] volts and [math]0.5[/math] ohm. Arrange such cells for [math]3[/math] amperes current through [math]21[/math] ohms external resistance.

Ans. [math]70[/math] cells in series and [math]2[/math] in parallel.

Calculate a battery of cell constants [math]1[/math] volt and [math]4[/math] ohms to give a current of [math]5[/math] amperes through a resistance of [math]5[/math] ohms.

Ans. [math]50[/math] cells in series and [math]40[/math] in parallel.

Calculate the number of cells of [math]1.7[/math] volt and [math]0.3[/math] ohm each to maintain a current of [math]2.3[/math] amperes through a resistance of [math]21[/math] ohms with an efficiency of [math]75\%[/math].

Ans. One group [math]2[/math] in parallel and [math]28[/math] in series, and another group [math]10[/math] in series, the two in series with each other.

[math]125[/math] cells of [math]1.75[/math] volts and [math]2[/math] ohm each are arranged [math]5[/math] in parallel and [math]25[/math] in series. There is an external resistance of [math]3.4[/math] ohms. Calculate the battery constants and the current.

Ans. [math]43.75[/math] volts; [math]3.75[/math] ohms; [math]6.12[/math] amperes.

Calculate the resistance of a battery to supply four magnets in parallel, the coil of each magnet being of [math]4.6[/math] ohms resistance and the efficiency to be [math]90\%[/math].

Ans. [math]0.125[/math] ohm.

Arrange cells of [math]1.1[/math] volts and [math]6[/math] ohms to supply above magnets with [math]2.5[/math]-ampere current.

Ans. [math]2[/math] cells in parallel will give [math]2.27[/math] amperes.

What is the resistance and e.m.f. of a battery [math]11[/math] in parallel and [math]7[/math] in series with cell constants [math]1.5[/math] volts and [math]0.25[/math] ohm?

Ans. [math]0.159[/math] ohm; [math]10.5[/math] volts.

Calculate the same factors if the above battery is in parallel.

Ans. [math]0.00325[/math] ohm; [math]1.5[/math] volts.

Calculate the same for the battery in series.

Ans. [math]19.25[/math] ohms; [math]115.5[/math] volts.

What current will each of the arrangements give through [math]11[/math] ohms external resistance?

Ans. First arrangement, [math]0.94[/math] ampere.
Second arrangement, [math]0.136[/math] ampere.
Third arrangement, [math]3.8[/math] amperes.

A current of [math]3[/math] amperes is to be produced through a wire of [math]30[/math] ohms resistance, with a battery of [math]2[/math] volts and [math]0.2[/math] ohm cell constants. Calculate the cells for [math]80\%[/math] efficiency.

Ans. Approximately [math]59[/math] in series and [math]2[/math] in parallel.

What is the exact efficiency of the above arrangement?

Ans. [math]83.3\%[/math].

What current will be given by [math]51[/math] cells, each cell of [math]1.07[/math] volts and [math]3[/math] ohms, connected in series, through an external resistance of [math]19[/math] ohms?

Ans. [math]0.317[/math] ampere.

If [math]30[/math] cells of the above battery are arranged [math]3[/math] in parallel and [math]10[/math] in series, what current will they give through the same resistance?

Ans. [math]0.369[/math] ampere.

What will a single cell of the same battery give through the same resistance?

Ans. [math]0.486[/math] ampere.

Note. — This is a case where a single cell gives more current than a number of cells.

CHAPTER VIII.

ELECTRIC ENERGY AND POWER.

Topics: Potential. — Proof of Numerical Value of Potential. — Potential Drop or (see page 95) R. I. Drop. — Electric Energy. — Practical Unit of Electric Energy. — Electric Power or Activity. — Practical Unit of Electric Power. — Relations of Power to Current, Resistance, and e.m.f. — Equivalents of the Watt — Problems.

Potential. Potential is of various kinds, affecting mass, electric quantity, and other things, each kind of potential affecting only one of them. It can be defined as a condition of a point in space such that the energy involved in moving a unit mass or quantity from the point to an infinite distance or from an infinite distance to the point would be numerically equal to the potential.

The energy involved in the moving of a unit mass or the unit quantity from a point of one potential to that of another is numerically equal to the difference of the two potentials. Electric potential affects electric quantity only.

Proof of Numerical Value of Potential. To prove the above, let [math]e[/math] represent a charge of electricity at a point in space. Let a unit quantity be acted on by it at a distance [math]r[/math] and also at a distance [math]r’, it having changed position from [math]r[/math] to [math]r’. The force exerted by [math]e[/math] upon the unit quantity at these distances is [math]e/r[/math] and [math]e/r'[/math] respectively.

The mean force is not the arithmetical average, because the force varies inversely with the square of the distance [math]r[/math]. The mean force is the geometrical mean, which is the square root of the product of the two forces:

[math]\sqrt{(e/r) \cdot (e/r’)} = \sqrt{e^2 / (rr’)} = e / \sqrt{rr’}[/math]

To get the energy this has to be multiplied by the path traversed; this we have taken as the distance from [math]r[/math] to [math]r'[/math], or as [math]r – r'[/math].

Performing the multiplication we have:

[math]e \cdot (r – r’) / \sqrt{rr’}[/math]

If [math]r = \infty[/math], then energy = [math]e / r'[/math].

As these expressions give the energy involved in the transfer of unit mass, they are the expressions for the numerical value of the potential in the case of movement from [math]r[/math] to [math]r'[/math], and in the case of movement from a point in space to an infinite distance. The latter has been defined as the value of absolute potential.

It is incorrect to say that potential is equal to the energy required to move a unit mass from or to an infinite distance to or from the place of potential, because potential and energy are not measured in the same unit. The number of potential units is equal to the number of energy units as described, but there is no equality of potential and energy.

Potential Drop or R.I. Drop. The current due to a definite e.m.f. is equal to the quotient of the e.m.f. less the R.I. drop in any part of the circuit divided by the resistance of the rest of the circuit. Calling the resistance of the part of the circuit which includes the R.I. drop [math]R_1[/math], this gives:

[math]I = \frac{E – R_1 I}{R – R_1}[/math]

The rule follows from Ohm’s law. The e.m.f. expended on the part of resistance [math]R_1[/math] is, by Ohm’s law, equal to [math]R_1 I[/math]. As the total e.m.f. expended on the system is [math]E[/math], the e.m.f. expended on the rest of the circuit is [math]E – R_1 I[/math]. The resistance of this portion, namely, the remainder of the circuit, is [math]R – R_1[/math]. Hence by Ohm’s law, the current in the circuit is:

[math]I = \frac{E – R_1 I}{R – R_1}[/math]

Example. The fall in potential, or R.I. drop, in a part of a circuit is [math]\frac{3}{4}[/math] volt while a current is passing due to an e.m.f. of [math]3[/math] volts in the circuit. The resistance of the remainder of the circuit is [math]2[/math] ohms. What is the intensity of the current?

Solution. Subtracting the R.I. drop of [math]\frac{3}{4}[/math] volt from the total e.m.f. of [math]3[/math] volts gives:

[math]3 – \frac{3}{4} = \frac{9}{4}[/math]. Dividing this by the resistance of the rest of the circuit, [math]2[/math] ohms, gives:

[math]I = \frac{9}{4} \div 2 = \frac{9}{8} = 1.125[/math] amperes.

Example. What is the resistance of the portion of the circuit in the above case in which the fall of potential of [math]\frac{3}{4}[/math] volt takes place?

Solution. By Ohm’s law [math]R = E \div I[/math], and substituting, we have:

[math]R = \frac{3}{4} \div \frac{9}{8} = \frac{2}{3}[/math] ohm.

Electric Energy. When electricity passes through a conductor it expends energy. The electric energy expended is converted into heat energy. The latter may be measured and used as the measure of electric energy.

If a unit of electric quantity actuated by a unit of e.m.f. passes through a conductor, a definite amount of heat will be produced. If two units of quantity actuated by one unit of e.m.f. pass through a conductor, twice the amount will be produced. If one unit of quantity as before but actuated by two units of e.m.f. passes, again twice the original amount will be produced. If two units of quantity actuated by two units of e.m.f. pass, then four times the original amount of heat will be produced.

The heat due to the passage of a quantity of electricity is the measure of the electric energy exerted. From the statements of the preceding paragraph it follows that the measure of the electric energy incident to the passage of electricity through a conductor is the product of the units of quantity by the units of e.m.f.

Example. [math]30[/math] C.G.S. units of quantity with [math]45[/math] C.G.S. units of e.m.f. are expended on a conductor. How many ergs of energy are expended?

Solution. [math]30 \times 45 = 1,350[/math] ergs.

Practical Unit of Electric Energy. The practical unit of electric energy is the volt-coulomb, equal to [math]10^8[/math] C.G.S. units of e.m.f. multiplied by [math]10^{-1}[/math] C.G.S. units of quantity, and therefore equal to [math]10^7[/math] C.G.S. units of energy, or to [math]10^7[/math] ergs, or to the joule. These absolute units are in the electro-magnetic system.

Example. [math]10^3[/math] absolute units of quantity at [math]10[/math] units of e.m.f. are how many volt-coulombs?

Solution. [math]10^3 \times 10 = 10^4[/math] ergs. [math]10^4 \div 10^7 = 10^{-3}[/math], or [math]10,000[/math] volt-coulombs.

Example. If [math]31[/math] coulombs at [math]20[/math] volts are expended, how many C.G.S. units are expended?

Solution. [math]31[/math] coulombs = [math]3.1[/math] C.G.S. units. [math]20[/math] volts = [math]20 \times 10^8[/math] C.G.S. units. [math]3.1 \times 20 \times 10^8 = 62 \times 10^8[/math] C.G.S. units.

Electric Power or Activity. The rate at which electrical quantity passes through a conductor is current strength or current intensity. One unit of quantity per second is unit current. The product of unit current by unit e.m.f. is unit rate of energy, unit activity, or unit power.

Practical Unit of Electric Power. The practical unit of power is the product of the practical unit of current by that of e.m.f. It is, as we have seen before, the volt-coulomb per second, the volt-ampere or watt. It is equal to [math]10^7[/math] C.G.S. units of power.

Relations of Power to Current, Resistance, and e.m.f. It follows that, since the rate of energy varies with the product of e.m.f. by current, if either factor is constant the rate of energy will vary with the other one.

Example. Calculate the value of the watt in C.G.S. units.

Solution. The watt is the product of one volt by one ampere. One volt = [math]10^8[/math] C.G.S. units. One ampere = [math]10^{1}[/math] C.G.S. units. [math]10^8 \times 10^1 = 10^9[/math] C.G.S. units of rate of energy or of power. The C.G.S. unit of power is the erg per second. A rate of [math]10^9[/math] ergs per second is one watt.

Example. A current of [math]16 \times 10^1[/math] C.G.S. units of current is actuated by [math]19 \times 10^8[/math] C.G.S. units of e.m.f. Calculate the value in watts.

Solution. [math]16 \times 19 = 304[/math]; [math]10^1 \times 10^8 = 10^9[/math]; the product of [math]16 \times 10 \times 19 \times 10^8 = 304 \times 10^9[/math] C.G.S. units = [math]3,040[/math] watts.

Example. Reduce and analyze [math]29[/math] watts with [math]7[/math] volts into C.G.S. units.

Solution. [math]29 \div 7 = 4.14[/math] amperes, the current in the case. [math]7[/math] volts = [math]7 \times 10^8[/math] C.G.S. units of e.m.f. [math]4.14[/math] amperes = [math]4.14 \times 10^1[/math] C.G.S. units of current. The total is [math]29 \times 10^9[/math] C.G.S. units of power.

Example. A current of [math]3[/math] amperes flows through a resistance of [math]3[/math] ohms. What power is exerted?

Solution. By Ohm’s law [math]E = IR[/math], the e.m.f. is [math]9[/math] volts. The power is [math]9 \times 3 = 27[/math] volt-amperes or watts.

Example. [math]17[/math] coulombs pass through a conductor of [math]7[/math] ohms resistance in [math]15[/math] seconds. Calculate the power.

Solution. A current of [math]17 \div 15 = 1.133[/math] amperes passes. The e.m.f. is given by Ohm’s law [math]E = IR[/math]; and substituting, [math]E = 1.133 \times 7 = 7.93[/math] volts. The power is therefore [math]7.93 \times 1.133 = 8.98[/math] watts.

By Ohm’s law [math]E = IR[/math]. Substituting this value of [math]E[/math] in the expression for power or watts [math]EI[/math], we have:

[math]\text{Power} = I^2 R[/math],

by which expression power may be directly calculated from the resistance and current intensity.

Example. A current of [math]5[/math] amperes flows through a resistance of [math]4[/math] ohms. What is the power?

Solution. Substituting [math]I^2 = 25[/math] and [math]R = 4[/math], we have:

[math]25 \times 4 = 100[/math] watts.

Example. A current of [math]2 \times 10^1[/math] C.G.S. units flows through a resistance of [math]4 \times 10^8[/math] C.G.S. units. Calculate the power in watts.

Solution. Substituting as before: [math](2 \times 10^1)^2 = 4 \times 10^2[/math]; and for [math]R[/math], [math]4 \times 10^8[/math]. Then:

[math]4 \times 10^2 \times 4 \times 10^8 = 160,000 \times 10^8 = 160,000[/math] watts.

By Ohm’s law [math]I = E / R[/math]. Substituting this value of [math]I[/math] in the expression for power [math]EI[/math], we have:

[math]P = E^2 / R[/math]

Example. Assume [math]200[/math] volts to act upon a resistance of [math]105[/math] ohms. Calculate the energy absorbed per second.

Solution. Applying the expression [math]E^2 / R[/math], we obtain:

^2 / 105 = 40,000 / 105 = 381[/math] watts. In a second [math]381[/math] watt-seconds or volt-coulombs are absorbed.

There are two expressions for electric power which include resistance as one of their component parts. These are [math]I^2 R[/math] and [math]E^2 / R[/math].

Assuming the current to remain constant and the resistance to be doubled, the first expression becomes [math]I^2 \times 2R = 2 I^2 R[/math]. If the resistance is trebled the same expression becomes [math]3 I^2 R[/math], and so on. Therefore, at constant current, the electric power varies with the resistance.

Assuming the e.m.f. to remain constant, and doubling and trebling the resistance as before, the second expression becomes [math]E^2 / (2R)[/math] and [math]E^2 / (3R)[/math]. Whence it follows that, at the same e.m.f., the electric power varies inversely with the resistance.

Example. The resistance of a lamp is [math]212[/math] ohms. The conductor supplying it has a resistance of [math]4[/math] ohms. How much energy is wasted on the conductor?

Solution. From the law of power distribution we have:

Energy of conductor : energy of lamp :: [math]4 : 212[/math]

The energy in the conductor is [math]4 / (4 + 212) = 4 / 216[/math], which reduces to approximately [math]1 / 54[/math]. The energy wasted is [math]1 / 54[/math] that consumed in the whole circuit, and [math]1 / 53[/math] that utilized in the lamp.

It is correct to speak of energy or of power in this relation because the ratios are the same for both. This problem can be done usually by simple inspection.

Example. There are two lamps [math]l[/math] and [math]l’ of resistances [math]211[/math] and [math]214[/math] ohms respectively. Calculate the relative energy which would be absorbed by each at the same e.m.f.

Solution. The proportion is given by the second law cited above,

[math]l : l’ = 214 : 211[/math]

or the lamp of [math]211[/math] ohms resistance absorbs [math]214 \div 211[/math] of the energy absorbed by the lamp of [math]214[/math] ohms resistance.

Equivalents of the Watt. One watt-second is equal to [math]10^7[/math] ergs. The electric energy expended in the passage of an electric current through a conductor is converted into heat energy. If the calorie is taken as equal to [math]4.185 \times 10^7[/math] ergs, then to reduce watt-seconds to calories the number of watt-seconds must be divided by [math]4.185[/math] or multiplied by [math]1 \div 4.185 = 0.239[/math], or as usually taken, [math]0.24[/math].

Example. A current of [math]0.5[/math] ampere is produced in a portion of a circuit by an e.m.f. of [math]110[/math] volts. What number of calories is absorbed by this portion per second?

Solution. The watts are absorbed at the rate of [math]0.5 \times 110 = 55[/math] per second. The calories are [math]55 \times 0.24 = 13.2[/math].

Example. A current of [math]15.5[/math] amperes at [math]110[/math] volts is expended in an electric heater. How long will it take for it to boil a liter of water at [math]10^\circ C[/math] if [math]50\%[/math] of the heat is lost by waste?

Solution. [math]15.5 \times 110 = 1,705[/math] watts. [math]1,705 \times 0.24 = 409.2[/math] calories per second.

A liter of water weighs [math]1,000[/math] grams. To boil it, it has to be heated [math]90^\circ[/math], making [math]90,000[/math] calories. [math]90,000 \div 409.2 = 220[/math] seconds if all heat is used. Since half is lost, double the time: [math]440[/math] seconds, or [math]7[/math] minutes [math]20[/math] seconds.

What is the power of the above generator?

Ans. [math]2,478[/math] watts, or nearly [math]2[/math] kilowatts.

The armature of a shunt-wound dynamo has a resistance from brush to brush of [math]0.03[/math] ohm; the e.m.f. at the brushes is [math]110[/math] volts, with a current of [math]150[/math] amperes in the outer circuit; the resistance of the shunt field is [math]55[/math] ohms.

What is the armature loss in watts? In a shunt-wound dynamo the outer circuit and shunt are in parallel.

Give general data.

• Resistance of outer circuit: [math]0.733[/math] ohm
• Combined resistance of outer circuit and shunt: [math]0.7235[/math] ohm
• Current in armature: [math]152[/math] amperes
• Armature loss: [math]693[/math] watts

CHAPTER IX.

BASES AND RELATIONS OF ELECTRIC UNITS.

Topics:

• Effects of an Electric Current
• Two Systems of C.G.S. Electric Units
• The Basis of Practical Units
• The Basis for Measurement and Definition of a Current
• The Absolute C.G.S. Electro-magnetic Unit of Current
• The Tangent Compass
• Action of Earth’s Field
• Angle of Divergence
• Combined action of Coil and Earth’s Field on Magnet (See page 109)
• Tangent Galvanometer Formula
• Determining C.G.S. units of e.m.f. and Resistance
• The Absolute C.G.S. Electrostatic Unit of Quantity
• Determination of the E.S. Unit of Potential
• The Attracted Disk Electrometer
• Other Units of the Electrostatic System
• The E.M. and E.S. System of Units
• Equivalents of the Two Systems of Units
• Derivation of Practical Units from Absolute Units
• Reduction Factor
• Problems
• Dimensions of E.M. Quantities
• Dimensions of Magnetic Quantity
• Dimensions of Current in E.M. System
• Dimensions of Electric Quantity in E.M. System
• Dimensions of Potential in E.M. System
• Dimensions of Resistance in E.M. System
• Dimensions of Capacity in E.M. System
• Dimensions of Electric Quantity in E.S. System
• Dimensions of Surface Density in E.S. System
• Dimensions of Potential in E.S. System
• Dimensions of Capacity in E.S. System
• Dimensions of Current in E.S. System
• Dimensions of Resistance in E.S. System
• Dimensions of Magnetic Quantity
• Dimensions of Surface Density of Magnetism
• Dimensions of Magnetic Intensity
• Dimensions of Magnetic Potential
• Dimensions of Magnetic Power
• Dimensions of Electric Intensity in E.S. System

The Basis of Practical Units.
The absolute electromagnetic units are usually assumed to be the basis of the practical units such as the volt and ampere, although the latter could be derived from the electrostatic units.

The Basis for Measurement and Definition of a Current.
If a current passes through a conductor it produces various effects. These effects by their intensity give a basis for measuring and defining the intensity of the current. One of these effects is the production of a field of force. A magnet pole is acted on by a field of force; it is attracted or repelled by a current. As the effect diminishes rapidly with the distance intervening between current and magnet, the effect is practically limited in extent to a small volume of space. The volume within which the effect is discernible is called a field of force. Theoretically a field of force may be as large as the universe. Practically, artificial fields of force are small in volume. The field of force of the field magnet of a bipolar dynamo is little more in extent than the volume existing between its pole faces.

The intensity of a field of force can be measured by its action on a magnet pole of known strength.

A unit magnet pole is one of such strength that it will attract or repel another unit pole at a distance of one centimeter with a force of one dyne.

The Absolute C.G.S. Electromagnetic Unit of Current.
The absolute C.G.S. unit of current is the following: It is the intensity of a current which, passing through a conductor 1 centimeter long bent into the arc of a circle of 1 centimeter radius, will attract or repel with a force of 1 dyne a unit magnet pole placed at the center of such circle.

The diagram shows an arc of this radius with its center indicated. Calling force [math]{f}[/math], we have for the case described:

[math]{f = 1}[/math]
(1)

Suppose that the current goes through a complete circle of 1 centimeter radius. Its action on a magnet pole at its center will be greater than in the last case in the ratio of the lengths of the conductors. A circle of radius 1 has a length or circumference of [math]{2\pi}[/math]. The ratio of the arc of the first case to that of the circle of the second case is [math]{1 : 2\pi}[/math]. Therefore, as these are the lengths of the conductors acting on the magnet pole, and as the intervening distance is the same in both cases, the force acting on the magnet is expressed by:

[math]{f = 2\pi}[/math]
(2)

Example. With what force will a unit magnet pole at the center of a circular conductor, the circle into which it is bent being of 1 cm radius, be acted on when a unit current passes?

Solution. Substituting for [math]{\pi}[/math] in (2) its value, 3.1416, we have for the force:

[math]{f = 2\pi = 2 \times 3.1416 = 6.2832\ \text{dynes}}[/math]

Suppose the wire is bent into a circle of radius [math]{r}[/math]. The length of the conductor is now [math]{2\pi r}[/math], and as far as this change of length of conductor is concerned, its action on the pole at its center varies directly with the length of the conductor acting on it. But the distance of the conductor from the pole is changed in direct ratio with [math]{r}[/math]. As radiant action, which is the action exerted on the pole in the cases under consideration, varies inversely with the square of the distance, we have to express this action also in the next formula. The direct ratio of the length of the conductor and the inverse ratio of the square of the distance give the formula:

[math]{f = \frac{2\pi}{r}}[/math]
(3)

Example. Assume the radius of a conductor bent into a circle to be 7 cm. With what force will a unit current passing through it act upon a unit pole at its center?

Solution. Substituting in (3) the values of [math]{r}[/math] and [math]{\pi}[/math], we have:

[math]{f = \frac{2\pi}{7} = \frac{6.2832}{7} = 0.8976\ \text{dyne}}[/math]

Suppose that the circular conductor is wound around the circle a number of times [math]{n}[/math], it then obviously acts with [math]{n}[/math] times the intensity of a single turn. Equation (3) then becomes:

[math]{f = \frac{2\pi n}{r}}[/math]
(4)

And if the current instead of being a unit current is of strength [math]{I}[/math], equation (4) becomes:

[math]{f = \frac{2\pi n I}{r}}[/math]
(5)

Whence by division:

[math]{I = \frac{f r}{2\pi n}}[/math]
(6)

If the magnet pole is of strength [math]{M}[/math], the force exerted upon it will be [math]{M}[/math] times as great as that exercised upon a unit pole. This is because the force exercised is a radiant force between two things, and such force is equal to the product of the two forces. Equations (5) and (6) for a magnet pole of strength [math]{M}[/math] become:

[math]{f = \frac{2\pi n I M}{r}}[/math]
(7)

[math]{I = \frac{f r}{2\pi n M}}[/math]
(8)

Example. A magnet pole of 5 units strength at the center of a circle of 11 cm radius composed of 24 turns of wire is acted on by a force of 377 dynes. What is the intensity of current passing through the wire?

Solution. By substituting in formula (8) we obtain:

[math]{I = \frac{f r}{2 \pi n M} = \frac{377 \times 11}{6.2832 \times 24 \times 5} = 5.5\ \text{C.G.S. units}}[/math]

The Tangent Compass.
If in the center of such a coil a very short compass needle is established, a tangent compass is produced. The compass needle is acted on by the horizontal component of the earth’s magnetic field. The circle of the galvanometer is placed in the magnetic meridian, approximately north and south. A current passing through the wire tends to deflect the needle into a position at right angles to the plane of the coil. This is at right angles to the position in which the earth’s component tends to keep it.

Action of Earth’s Field.
The earth’s field acts upon the needle with the product of the horizontal component of the earth’s magnetism by the strength of the magnet. This product we may call [math]{HM}[/math], [math]{H}[/math] representing the horizontal component of the earth’s magnetism and [math]{M}[/math] the force of the magnet.

Angle of Divergence.
At a given angle of divergence [math]{\theta}[/math] from the magnetic meridian, the lever arm of the earth’s action on the magnet is equal to the product of the magnet’s length by the sine of the angle of divergence. The earth acts upon both arms of the magnet with virtual lever arms of length [math]{AB}[/math], and the sum of the two is as if it acted upon the magnet with a single lever arm equal to the length of the magnet multiplied by [math]{\sin \theta}[/math].

By the same course of reasoning, we find that the coil acts with a lever arm equal to the product of the magnet’s length by the cosine of the angle of divergence.

Combined Action of Coil and Earth’s Field on Magnet.
Calling the length of the magnet [math]{l}[/math], we have for the earth’s action:

[math]{H M l \sin \theta}[/math]

and for the action of the coil:

[math]{2 \pi n I l \cos \theta}[/math]

In any position of equilibrium which the needle assumes, it will be acted on equally by these two opposing forces. They therefore may be made equal to each other:

[math]{2 \pi n I l \cos \theta = H M l \sin \theta}[/math]
(9)

In which [math]{M}[/math] and [math]{l}[/math] eliminate each other, and by transposition:

[math]{I = \frac{H \tan \theta}{2 \pi n}}[/math]
(10)

Tangent Galvanometer Formula.
This is the formula of the tangent galvanometer, developed here at length to show how the electromagnetic C.G.S. unit of current can be directly determined. It will be observed that the only constant entering into the formula is the horizontal component of the earth’s magnetism. This has been determined for a great many places.

Example. The radius of a coil is 30 cm, the number of turns is 5, the value of [math]{H}[/math] is 0.1590 dyne, and the needle is deflected by a current passing through the coil to an angle of 50° 11′. Calculate the current strength in C.G.S. units.

Solution. Substituting in equation (10) we have:

[math]{I = \frac{0.1590 \times \tan(50^\circ 11′)}{2 \pi \times 5} = \frac{0.1590 \times 1.2}{31.416} = 0.18\ \text{C.G.S. unit of current}}[/math]

Determining C.G.S. Units of e.m.f. and Resistance.
The energy due to an electric discharge is equal to the product of the quantity discharged by the potential difference of the points or places between which the discharge takes place. The ergs per second of energy expended when a current of electricity is passing through a conductor is equal to the current strength multiplied by the e.m.f. required to maintain the current through the conductor in question. The energy expended is converted into heat energy. All units are assumed as of the C.G.S. system.

By placing a conductor in a calorimeter, passing a known current through it (the current being taken in C.G.S. units), and determining the ergs per second produced, the e.m.f. expended on the conductor is manifestly equal to the ergs divided by the current. Then, having the current and e.m.f., the resistance of the conductor is calculated by Ohm’s law.

Example. A current of 0.114 C.G.S. units passing through a calorimeter wire for one minute produces 40 calories of heat. Calculate the e.m.f. between the terminals of the wire and the resistance of the wire.

Solution. To reduce the calories to ergs, multiply by:

[math]{416.7 \times 10^5}[/math]

Ergs per minute:

[math]{40 \times (416.7 \times 10^5) = 1.6667 \times 10^9\ \text{ergs}}[/math]

Ergs per second:

[math]{\frac{1.6667 \times 10^9}{60} = 2.78 \times 10^7}[/math]

E.m.f. is:

[math]{\frac{2.78 \times 10^7}{0.114} = 2.43 \times 10^8\ \text{C.G.S. units}}[/math]

Resistance:

[math]{R = \frac{E}{I} = \frac{2.43 \times 10^8}{0.114} = 2.13 \times 10^9\ \text{C.G.S. units}}[/math]

Conversions:

• [math]{1\ \text{C.G.S. unit of current} = 10\ \text{amperes}}[/math]
  → [math]{0.114 \times 10 = 1.14\ \text{amperes}}[/math]
• [math]{10^8\ \text{C.G.S. units of e.m.f.} = 1\ \text{volt}}[/math]
  → [math]{\frac{2.43 \times 10^8}{10^8} = 2.43\ \text{volts}}[/math]
• [math]{10^9\ \text{C.G.S. units of resistance} = 1\ \text{ohm}}[/math]
  → [math]{\frac{2.13 \times 10^9}{10^9} = 2.13\ \text{ohms}}[/math]

This section is designed to show the relation of the absolute units to the practical ones and also the relation of the apparently abstract bases of the absolute system to the standards of the engineer. It is not to be taken as showing approved details of making these determinations, which exact the most accurate work and refinement in methods.

An ammeter and a voltmeter could be standardized or calibrated by the above method; the length, diameter, and material of a wire of 1 ohm resistance could be determined, and with these bases the other practical units could be determined and standards fixed for them.

The Absolute C.G.S. Electrostatic Unit of Quantity.
The absolute electrostatic unit of quantity is that quantity which would attract or repel a similar quantity one centimeter distant with a force of one dyne. From this as a basis a full series of units of the E.S. system, as the term may be conveniently abbreviated, is deduced.

Determination of the E.S. Unit of Potential.
To obtain an experimental basis, we may start with the determination of potential difference. The attracted disk electrometer can be used for this determination.

The Attracted Disk Electrometer.
It comprises a disk held above and parallel to a plate, the plate being much larger than the disk. The apparatus is so arranged that the attraction between plate and disk and the distance separating them can be determined.

In the diagram, the disk is represented by the line [math]{S}[/math], the plate by the line [math]{T}[/math], and an annular plate surrounding the disk, as close to it as possible, is also indicated.

By touching one of the two, disk or plate, with an excited conductor, a charge is imparted and they attract each other. Call the surface density of the charge [math]{+ \sigma}[/math] for one surface and [math]{- \sigma}[/math] for the other.

The attraction of an indefinitely large plane with a surface density [math]{\sigma}[/math] for a point (in this case, a unit of quantity) placed opposite its center is:

[math]{2 \pi \sigma}[/math]

This is the attraction of the plate for a unit of quantity of electricity on the disk. Surface density is the quantity of electricity per unit area of the plate or disk.

Call the area of the disk [math]{S}[/math]. Then the entire quantity on its surface is [math]{S \sigma}[/math]. It will be seen that [math]{\sigma}[/math] represents the quantity on a unit area of the disk, because the surface densities of the charges on plate and disk are identical. The attraction [math]{A}[/math] between disk and plate is therefore:

[math]{A = 2 \pi \sigma \times S \sigma = 2 \pi S \sigma^2}[/math]
(1)

Transposing equation (1), we get:

[math]{\sigma = \sqrt{\frac{A}{2 \pi S}}}[/math]
(2)

If a point is situated between the disk and plate, both will act—the one attracting and the other repelling any mass between them—and hence both exercising force in the same direction. The force at such a point will be the force exercised by one of the surfaces multiplied by 2:

[math]{2 \pi \sigma \times 2 = 4 \pi \sigma}[/math]

If a unit quantity is transferred from one plane to the other, as from disk to plate, the energy involved is numerically equal to the difference of potential between the two surfaces. This is because the product of potential difference by quantity is energy, and as the quantity transferred is supposed to be equal to unity, the potential difference and the energy have the same numerical value.

Call the potential difference between the two surfaces [math]{P}[/math] and the distance between the plates [math]{d}[/math]. Then:

[math]{P = 4 \pi \sigma d}[/math]
(3)

Substituting the value of [math]{\sigma}[/math] from (2) into (3):

[math]{P = 4 \pi d \sqrt{\frac{A}{2 \pi S}}} = \sqrt{\frac{8 \pi A d^2}{S}}[/math]
(4)

The force of attraction [math]{A}[/math] is measured, and the area of the disk [math]{S}[/math] and the distance [math]{d}[/math] are known. Substituting for [math]{A}[/math], [math]{S}[/math], and [math]{d}[/math] their values, the numerical value in electrostatic units of the potential difference between the plate and disk is found.

In carrying out this calculation, C.G.S. units must be used. The force of attraction must be expressed in dynes.

Example. In an attracted disk electrometer, the area of the disk was 10 square cm. The distance from the disk to the plate was 0.5 cm. After imparting a charge to the disk, the attraction was 0.092 gram. Calculate the potential difference in electrostatic C.G.S. units.

Solution.
Convert grams to dynes:

[math]{A = 0.092 \times 981 = 90.252\ \text{dynes}}[/math]

Substitute into equation (4):

[math]{P = 0.5 \times \sqrt{\frac{8 \pi \times 90.252}{10}} = 7.53\ \text{electrostatic units}}[/math]

Other Units of the Electrostatic System.
If one such unit of e.m.f. were expended on maintaining a current of such strength that the two would expend 1 erg of energy per second, the current would be of one electrostatic absolute unit strength. The resistance of the conductor through which this current would be forced by a unit of potential difference would be one electrostatic unit in amount.

From this basis the whole series of electrostatic units can be deduced.

The E.M. and E.S. Systems of Units.
The value of a current or of potential difference may be determined in absolute units by various methods which give it directly. Examples of such are given in the preceding pages, where they are employed simply to give the concrete idea of what these units are, of how they are related to the three fundamental units of time, space, and mass, and of how they can be determined experimentally by direct use of the centimeter, gram, and second.

One set of electric units is based on the attraction of oppositely charged surfaces for each other (or equivalently, on the repulsion of similarly charged ones). This set is called the electrostatic series. Another is based on the action of an electric current on a magnet pole, called the electromagnetic series. Both sets of units are C.G.S. units and can be used for electric specification and calculation—one just as well as the other.

Equivalents of the Two Systems of Units.
The units of the two systems differ in magnitude. Their relationships have been determined experimentally to a high degree of accuracy. Here are the key conversions:

• E.M. absolute unit of quantity: [math]{1\ \text{E.M. unit} = 3 \times 10^{10}\ \text{E.S. units}}[/math]
• Coulomb: [math]{1\ \text{Coulomb} = 10^{-1}\ \text{E.M. units} = 3 \times 10^9\ \text{E.S. units}}[/math]
• Ampere: [math]{1\ \text{Ampere} = 10^{-1}\ \text{E.M. current units} = 3 \times 10^9\ \text{E.S. current units}}[/math]
• E.S. unit of potential: [math]{1\ \text{E.S. unit} = 3 \times 10^2\ \text{E.M. units}}[/math]

[math]{1\ \text{Volt} = 10^8\ \text{E.M. units} = 3.33 \times 10^6\ \text{E.S. units}}[/math]

E.S. unit of resistance:

[math]{1\ \text{E.S. unit} = 9 \times 10^{11}\ \text{E.M. units}}[/math]

[math]{1\ \text{Ohm} = 10^9\ \text{E.M. units} = 1.11 \times 10^{-2}\ \text{E.S. units}}[/math]

E.M. absolute unit of capacity:

[math]{1\ \text{E.M. unit} = 9 \times 10^{11}\ \text{E.S. units}}[/math]

Farad:

[math]{1\ \text{Farad} = 10^{-9}\ \text{E.M. units} = 9 \times 10^2\ \text{E.S. units}}[/math]

Derivation of Practical Units from Absolute Units

The practical units are derived from the absolute units by either of two methods, each giving an identical result. The simplest method is to reduce the units of the absolute series to those of the practical series by multiplying by defined factors. The table gives the value of the units of the practical series in terms of the absolute units, with the factors. The factors are all powers of 10 or multiples thereof.

Practical Units Table:

Example. Calculate the equivalent of 5 volts in electrostatic units (absolute).

Solution. A volt is equal to [math]{10^8}[/math] absolute E.M. units; 5 volts therefore are equal to [math]{5 \times 10^8}[/math] absolute E.M. units. For the equivalent electrostatic units this must be divided by [math]{3 \times 10^{10}}[/math]. Then

[math]{\frac{5 \times 10^8}{3 \times 10^{10}} = \frac{5}{3} \times 10^{-2} = 0.0166}[/math] E.S. units.

Or by direct process, using the equivalent of the table:

[math]{5 \times 3 \times 10^{-3} = 0.0166}[/math] E.S. units.

Example. What is the value of the volt in absolute electrostatic units?

Solution. From the above calculation it follows that the value of the volt is

[math]{0.0166 \div 5 = 0.0033}[/math] E.S. unit.

Or by the table:

[math]{1 \times 4 \times 10^{-3} = 0.0033}[/math] E.S. unit.

Example. What is the value of 28.3 E.S. absolute units in volts?

Solution. As the E.M. unit is equal to the E.S. unit divided by [math]{3 \times 10^{10}}[/math], the reverse rule holds and the E.S. unit is equal to the E.M. unit multiplied by [math]{3 \times 10^{10}}[/math]. This gives:

[math]{28.3 \times 3 \times 10^{10} = 849 \times 10^{10}}[/math] absolute E.M. units.

To reduce this to volts we must divide by the equivalent of the volt in absolute E.M. units, which is [math]{10^8}[/math]:

[math]{\frac{849 \times 10^{10}}{10^8} = 8490}[/math] volts.

Or by the table:

[math]{28.3 \times 3 \times 10^2 = 8490}[/math] volts.

Example. An e.m.f. of 120 volts produces a current of 5 amperes. Reduce these quantities to E.S. absolute units; apply Ohm’s law to the result, thus obtaining the value of the resistance in the same system of units; reduce the resistance thus calculated to ohms and test the correctness of the operation by Ohm’s law directly applied.

Solution. The absolute value of 120 volts in the E.M. system is

[math]{120 \times 10^8 = 1.2 \times 10^{10}}[/math].

Applying the equivalent to obtain the value of this quantity in absolute E.S. units we have:

[math]{\frac{1.2 \times 10^{10}}{3 \times 10^{10}} = 0.4}[/math] = emf. E.S.

5 amperes are equal to [math]{5 \times 10^{-1}}[/math] absolute E.M. units. Applying the equivalent we find:

[math]{5 \times 10^{-1} \times (3 \times 10^{10}) = 1.5 \times 10^{10}}[/math] = current in E.S. units.

Applying Ohm’s law to the E.S. units thus determined gives:

[math]{\frac{0.4}{1.5 \times 10^{10}} = 2.66 \times 10^{-11}}[/math] = resistance E.S.

To test the correctness of the operation proceed thus: The original data give the resistance by Ohm’s law as

[math]{E/I = 120 \div 5 = 24}[/math] ohms.

[math]{24\ \text{ohms} = 24 \times 10^9}[/math] absolute E.M. units. Applying the equivalent we find:

[math]{\frac{24 \times 10^9}{9 \times 10^{11}} = 2.66 \times 10^{-11}}[/math], as found by the former operation.

This goes to prove the correctness of the work. This example is only given as an exercise. The results are better obtained by the use of the tables.

The practical units of the E.M. system can be directly derived from the dimensional formulas of units by changing the units of length and mass.

Let the unit of mass be [math]{10^{-3}}[/math] gram and let the unit of length be [math]{10^2}[/math] cm., the unit of time being the second, as in the absolute system. If these are introduced into the E.M. dimensional formulas of units the resulting units will be of the practical system—volts, amperes, ohms, and the regular engineering units.

Example. Calculate the value of the ampere in absolute electro-magnetic units.

Solution. The dimensions of the unit are [math]{M^{1/2}L^{1/2}T^{-1}}[/math]. Substituting:

[math]{10^{-3} \times 10^2 \times 1^{-1} = 10^{-1}}[/math] C.G.S. units.

Ten amperes are equal to one absolute electro-magnetic unit.

Example. Calculate the value of the ohm and volt as above.

Solution.

• Ohm dimensions: [math]{L T^{-1}}[/math]: [math]{10^2 \times 1^{-1} = 10^2}[/math] C.G.S. units.
• Volt dimensions: [math]{M^{1/2}L^{3/2}T^{-2}}[/math]: [math]{10^{-3} \times 10^3 \times 1^{-2} = 10^8}[/math] C.G.S. units.

Reduction Factor. The value of the reduction factor of the two systems of C.G.S. units, which is also the numerical value of the velocity of light in centimeters, is about

[math]{3 \times 10^{10}}[/math]

—a fact applying to the electro-magnetic theory of light.

The reduction factor has been determined with close approximation to the exact figure by several investigators with results which are reasonably concordant. The following are some of the results:

• 1883, J. J. Thomson: [math]{2.963 \times 10^{10}}[/math]
• 1889, Lord Kelvin: [math]{3.004 \times 10^{10}}[/math]
• 1890, J. J. Thomson and G. F. C. Searle: [math]{2.9955 \times 10^{10}}[/math]
• Lodge and Glazebrook: [math]{3.009 \times 10^{10}}[/math]

Example. Calculate the dimensions of the reduction factor to reduce the E.S. unit of capacity to the E.M. unit.

Solution.

• E.S. unit: [math]{L}[/math]
• E.M. unit: [math]{L^{-2}T^2}[/math]

Divide second by first:

[math]{\frac{L^{-2}T^2}{L} = L^{-3}T^2}[/math], which is the reciprocal of velocity squared.

Example. Make the same calculation for current units.

Solution.

• E.S. unit: [math]{M^{1/2}L T^{-1}}[/math]
• E.M. unit: [math]{M^{1/2}L^3 T^{-1}}[/math]

Divide:

[math]{\frac{M^{1/2}L^3 T^{-1}}{M^{1/2}L T^{-1}} = L^2}[/math], which corresponds to velocity squared.

Example. Make the same calculation for current units.

Solution. The dimensions of the E.S. unit are

[math]{M^{1/2}L T^{-1}}]; of the E.M. unit, [math]{M^{1/2}L^3 T^{-1}}]. Dividing as above we find

[math]{\frac{M^{1/2}L^3 T^{-1}}{M^{1/2}L T^{-1}} = L^2}[/math]

,
which are the dimensions of velocity.

Dimensions of E.M. Quantities. Two magnet poles of equal strength and of opposite polarity attract each other in direct proportion to the square of the quantity of magnetism in one of the poles and in inverse proportion to the square of the distance between them.

Dimensions of Magnetic Quantity. Call the magnetism of each pole [math]{Q}[/math], and let the distance separating them be denoted by [math]{L}[/math]. The attraction of the poles is a force, and we have seen that the dimensions of a force are [math]{MLT^{-2}}[/math]. The force in the particular case is, as just indicated, [math]{Q^2L^{-2}}[/math]. Putting these two equal to each other, as they are the same, we have

[math]{Q^2L^{-2} = MLT^{-2}}[/math],
whence [math]{Q^2 = ML^3T^{-2}}[/math], and
[math]{Q = M^{1/2}L^{3/2}T^{-1}}[/math],

which are the dimensions of magnetic quantity. Care must be taken not to confuse magnetic quantity with electric quantity. They are totally distinct.

Dimensions of Current in E.M. System. Let [math]{I}[/math] denote the strength of a current passing through a conductor of length [math]{L}[/math] and bent into an arc of a circle of radius [math]{L}[/math] as explained. Let a magnet pole representing a quantity of magnetism [math]{Q}[/math] be at the center of the circle of which the conductor is an arc. The force exercised between the current and magnet pole will follow the laws of radiant action, with the additional law that the action will vary directly with the length of the conductor. The action will be therefore the product of the current strength by the length of the conductor by the quantity of magnetism, the whole divided by the square of the distance separating the two loci of force, namely, the conductor and the magnet pole. This gives the expression for the action between the two,

[math]{\frac{ILQ}{L^2} = IQL^{-1}}[/math].

As this action is force, it can be put equal to the dimensions of force, or

[math]{IQL^{-1} = MLT^{-2}}[/math].

[math]{Q}[/math] is magnetic quantity, whose dimensions have just been determined and are [math]{M^{1/2}L^{3/2}T^{-1}}[/math]. Substituting these for [math]{Q}[/math] in the last equation gives

[math]{I (M^{1/2}L^{3/2}T^{-1}) L^{-1} = MLT^{-2}}[/math],
whence

[math]{IT = M^{1/2}L^3T^{-1}}[/math],

which are the dimensions of electric current in the electromagnetic system.

Dimensions of Electric Quantity in E.M. System.
The quantity of electricity passed by a current in any given time is evidently equal to the product of the current by the time during which it is passing. Multiplying the dimensions of current by the dimensions of time [math]{T}[/math] gives the dimensions of electric quantity:

[math]{M^{1/2}L^3T^{-1} \times T = M^{1/2}L^3}[/math].

These dimensions are quite different from those of magnetic quantity.

Dimensions of Potential in E.M. System.
The product of e.m.f. by electric quantity is energy. Therefore, if the dimensions of energy [math]{ML^2T^{-2}}[/math] are divided by those of electric quantity [math]{M^{1/2}L^3}[/math], the dimensions of e.m.f. will be given:

[math]{\frac{ML^2T^{-2}}{M^{1/2}L^3} = M^{1/2}L^{-1}T^{-2}}[/math].

Dimensions of Resistance in E.M. System.
According to Ohm’s law, resistance is equal to the quotient of e.m.f. divided by current strength. Carrying out this division with the dimensions as just determined gives:

[math]{\frac{M^{1/2}L^{-1}T^{-2}}{M^{1/2}L^3T^{-1}} = L^{-4}T^{-1}}[/math],

These are also the dimensions of velocity or rate of motion.

Dimensions of Capacity in E.M. System.
The capacity of a surface, as of a condenser, is defined as the quantity of electricity required to raise its potential a given amount. Its dimensions are therefore equal to the dimensions of quantity [math]{M^{1/2}L^3}[/math] divided by the dimensions of e.m.f. [math]{M^{1/2}L^{-1}T^{-2}}[/math]:

[math]{\frac{M^{1/2}L^3}{M^{1/2}L^{-1}T^{-2}} = L^4T^2}[/math].

The dimensions of electric energy, so-called, are the same as those of every other form of energy:

[math]{ML^2T^{-2}}[/math].

The same applies to electric power, whose dimensions are:

[math]{ML^2T^{-3}}[/math].

Dimensions of Electric Quantity in E.S. System.
A quantity of electricity acts upon another equal quantity at a distance [math]{d}[/math] with force, and such force varies directly with the square of the quantity [math]{q}[/math] and inversely with the square of the distance [math]{d}[/math]. Calling force [math]{f}[/math], we have:

[math]{f = \frac{q^2}{d^2}} \Rightarrow q^2 = fd^2 \Rightarrow q = d\sqrt{f}[/math].

Substituting [math]{L}[/math] for [math]{d}[/math] and [math]{MLT^{-2}}[/math] for [math]{f}[/math]:

[math]{q = L \sqrt{MLT^{-2}} = M^{1/2}L^{3/2}T^{-1}}[/math].

Dimensions of Surface Density in E.S. System.
Surface density is the quantity per unit area [math]{L^2}[/math]. Dividing the dimensions of quantity by [math]{L^2}[/math]:

[math]{\frac{M^{1/2}L^{3/2}T^{-1}}{L^2} = M^{1/2}L^{-1/2}T^{-1}}[/math].

Dimensions of Potential in E.S. System.
Electrostatic potential (e.m.f. in the electrostatic system) is:

[math]{\frac{ML^2T^{-2}}{M^{1/2}L^{3/2}T^{-1}} = M^{1/2}L^{1/2}T^{-1}}[/math].

Dimensions of Capacity in E.S. System.
Capacity = quantity / e.m.f.:

[math]{\frac{M^{1/2}L^{3/2}T^{-1}}{M^{1/2}L^{1/2}T^{-1}} = L}[/math].

Thus, electrostatic capacity can be expressed as a length.

Dimensions of Current in E.S. System.
Current = quantity / time:

[math]{\frac{M^{1/2}L^{3/2}T^{-1}}{T} = M^{1/2}L^{3/2}T^{-2}}[/math].

Dimensions of Electric Intensity in E.S. System.
Electric intensity = force / electric quantity:

[math]{\frac{MLT^{-2}}{M^{1/2}L^{3/2}T^{-1}} = M^{1/2}L^{-1/2}T^{-1}}[/math].

Dimensions of Resistance in E.S. System.
Resistance = e.m.f. / current:

[math]{\frac{M^{1/2}L^{1/2}T^{-1}}{M^{1/2}L^{3/2}T^{-2}} = L^{-1}T}[/math].

Dimensions of Magnetic Quantity (Tangent Galvanometer).
Given: [math]{f = \frac{ilq’}{r^2}}[/math]
Solving for [math]{q’}[/math]:

[math]{q’ = \frac{fr^2}{il}}[/math].

Substitute:

• [math]{f = MLT^{-2}}[/math],
• [math]{r^2 = L^2}[/math],
• [math]{i = M^{1/2}L^3T^{-1}}[/math],
• [math]{l = L}[/math]:

[math]{q’ = \frac{MLT^{-2} \cdot L^2}{M^{1/2}L^3T^{-1} \cdot L} = M^{1/2}}[/math].

Dimensions of Magnetic Power.
Magnetic power is the product of magnetic density by thickness of shell, or

[math]{g t = M^{1/2}L^{-2} \times L = M^{1/2}L^{-1}}[/math].

PROBLEMS

Express 17 amperes in the E.S. system.
Ans. [math]{51 \times 10^9}[/math] E.S. units.

Express 39 volts in the E.S. system.
Ans. [math]{0.13}[/math] E.S. unit, or [math]{13 \times 10^{-3}}[/math] E.S. units.

Express 29 ohms in the E.S. system.
Ans. [math]{3\frac{1}{5} \times 10^{11}}[/math] E.S. unit, or [math]{3.22 \times 10^{11}}[/math] E.S. units.

Express Ohm’s law

[math]{1\ \text{volt} = 1\ \text{ampere} \times 1\ \text{ohm}} in the E.S. system.

Ans.

[math]{4 \times 10^{-3} \div 3 \times 10^{-9} = 1.33 \times 10^6}[/math]

E.S. units.

If [math]{13 \times 10^{-3}}[/math] E.S. unit of potential acts upon a resistance of [math]{3.22 \times 10^{11}}[/math] E.S. units, what will the current be in E.S. absolute, in E.M. absolute, and in practical units?
Ans.
E.S. absolute: [math]{4.037 \times 10^{-8}}[/math]
E.M. absolute: [math]{0.1346}[/math]
Practical: [math]{1.3460}[/math] ampere

Express 15 coulombs in E.S. and E.M. absolute units.
Ans.
E.S.: [math]{45 \times 10^9}[/math]
E.M.: [math]{1.5}[/math]

What is the ratio of the E.S. to E.M. units of quantity from the above?
Ans.

[math]{\text{E.S.} = \text{E.M.} \times 3 \times 10^{10}}[/math]

Give the calculation for the watt in E.S. units.
Ans.
[math]{3 \times 10^{-3} \times 3 \times 10^{-9} = 9 \times 10^{-12}}[/math] = [math]{10^{-11}}[/math] E.S. units

Give the same calculation for E.M. units.
Ans.
[math]{10^8 \times 10^{-1} = 10^7}[/math] E.M. units

How many volts are there in 390 E.S. units of potential?
Ans.
[math]{\frac{390}{3 \times 10^2} = 1.30}[/math] volts

How many amperes are there in 277 E.M. units of current?
Ans.
[math]{277 \times 10^{-1} = 27.7}[/math] amperes

CHAPTER X. THERMO-ELECTRICITY

Emissivity. — Heating of a Conductor by a Current. — Cross Section of a Conductor to be Heated to a Given Degree by a Given Current. — Thermo-electric Couple. — Electro-motive Force of a Thermo-electric Couple. — Neutral Temperature. — Temperature and Electro-motive Force of Thermo-electric Couple. — Thermo-electric Tables. — Peltier Effect. — Absolute Temperature. — Thomson Effect.

Emissivity

If a body is heated above the temperature of its surroundings it parts with its heat. The process is termed emissivity. Unit emissivity is the quantity of heat lost per second per square centimeter of surface per degree of difference between its temperature and that of its surroundings.

It is measured in ergs or calories or other heat unit. The C.G.S. unit of heat is the erg. The practical unit is the therm, calorie, or gram-degree, which is the heat required to raise one gram of water one degree C.

Heating of a Conductor by a Current

A number of formulas for calculation of the temperature imparted to a conductor by a current passing through it have been proposed. From the nature of the case these formulas can only be approximate.

By radiation and convection about 0.000025 calorie is lost per second by an unpolished metallic surface of one square centimeter in the air for each degree C. that it is heated above the air. If therefore the calories expended on a conductor by the passage of a current are determined, and if the wire is assumed to have attained its fixed temperature, these calories will be equal to those lost by radiation and convection. The quotient of these calories divided by the area of the surface of the conductor in square centimeters will give the calories per square centimeter which are expended on the conductor and which are in turn lost by it. The quotient of these calories divided by 0.000025 is the temperature in degrees C. which the wire will attain. The result is only approximate.

Example. A current of 10 amperes passes through a wire one centimeter in circumference. The wire has a resistance of 0.3 ohm per 100 meters. How many degrees will its temperature be increased?

Solution. The watts per second due to the passage of the current are given by the expression:

[math]{I^2 R = 100 \times 0.3 = 30}[/math]

The surface area of 100 meters of the wire is 10,000 square centimeters. The calories expended on 100 meters are:

[math]{30 \times 0.24 = 7.2}[/math]

The calories per square centimeter are:

[math]{\frac{7.2}{10,000} = 0.00072}[/math]

Dividing by 0.000025 gives:

[math]{\frac{0.00072}{0.000025} = 2.88^\circ \text{C}}[/math]

A formula for the rise in temperature in degrees C. of a bare copper wire with a current passing through it in still air is:

[math]{\text{Temperature } (\degree C) = \frac{I^2 \times 50,000}{d^3}}[/math]
(where [math]{d}[/math] is in mils)

Example. A copper wire 125 mils in diameter is conducting a current of 10 amperes. How much will it be heated above the temperature of the surrounding air, the wire being of bare copper?

Solution.

[math]{\text{Temperature } = \frac{10^2 \times 50,000}{125^3} = \frac{500,000}{1,953,125} \approx 2.56^\circ C}[/math]

Cross Section of a Conductor to be Heated to a Given Degree by a Given Current

The diameter of a wire of a material of known specific resistance which a stated current will heat to a stated degree above that of the surrounding air can be calculated approximately by the application of the emissivity factor.

The resistance in ohms of a wire of diameter [math]{d}[/math] and cross-sectional area [math]{\frac{\pi d^2}{4}}[/math] is:

[math]{R = \frac{\text{sp. res.} \times 10^{-6}}{\frac{\pi d^2}{4}} = \frac{4 \times \text{sp. res.} \times 10^{-6}}{\pi d^2}}[/math]
(1)

The heat expended:

[math]{\text{Heat} = I^2 R = I^2 \times \frac{4 \times \text{sp. res.} \times 10^{-6}}{\pi d^2}}[/math]
(2)

A calorie = 4.16 watt-seconds:

[math]{\text{Heat (cal)} = \frac{I^2 \times \text{sp. res.} \times 10^{-6}}{4.16 \times \pi d^2}}[/math]
(3)

The surface area of 1 cm length of wire is [math]{\pi d}[/math], so:

[math]{\text{Cal/cm}^2 = \frac{I^2 \times \text{sp. res.} \times 10^{-6}}{4.16 \pi^2 d^3}}[/math]

At equilibrium:

[math]{\frac{I^2 \times \text{sp. res.} \times 10^{-6}}{4.16 \pi^2 d^3} = \frac{t}{4000}}[/math]

Solve for [math]{d^3}[/math]:

[math]{d^3 = \frac{I^2 \times \text{sp. res.} \times 0.00039}{t}}[/math]
(5)

Therefore:

[math]{d = \sqrt[3]{\frac{I^2 \times \text{sp. res.} \times 0.00039}{t}}}[/math]
(6)

Example. Calculate the diameter of a lead wire to melt with a current of 7.2 amperes. The melting point of lead is 335° C., and its specific resistance is 19.85 microhms/cm³.

Solution.

[math]{d^3 = \frac{(7.2)^2 \times 19.85 \times 0.00039}{335} = 0.00118}[/math]

[math]{d = \sqrt[3]{0.00118} \approx 0.1058\ \text{cm}}[/math]

Suppose a strip is to be cut from a sheet of metal of thickness a, and it is to be calculated how wide the strip shall be to attain a given temperature with a given current. Call the width of the strip na, the factor n being the unknown quantity to be determined. The cross-sectional area of the strip will be na × a = na². The surface area of the unitary length will be 2na + 2a = 2a(n + 1). The product of these two quantities is

[math] 2a^2(n^2 + n) [/math]

As this is the product of the surface area of the unit length of the conductor by its cross-sectional area, it is obvious that it can be substituted for A in equation (4).

This substitution is performed by division by it in place of division by A, giving

[math] t = \frac{I^2 \times \text{sp. res.} \times 10^7}{4,000 \times 2a^2(n^2 + n)} \tag{7} [/math]

Multiplying by [math] 4,000 \times 2a^2(n^2 + n) [/math] gives

[math] I^2 \times \text{sp. res.} \times 10^7 \times 4,000 = t \times 2a^2(n^2 + n) [/math]

Solving for [math] n^2 + n [/math]:

[math] n^2 + n = \frac{I^2 \times \text{sp. res.} \times 0.00004808}{t \times a^2} \tag{8} [/math]

The value of n is obtained from this equation by substituting the constants given in the problem. The width of the strip is the product of its thickness by n, or na.

Example. A strip of lead to be melted by a current of 7.2 amperes is to be cut from a sheet 0.01 cm thick. The specific resistance of lead is taken as 19.85 and its melting point as 335°C. Calculate the width of the strip.

Solution. Substituting these values in equation (8) gives:

[math] n^2 + n = \frac{(7.2)^2 \times 19.85 \times 0.00004808}{335 \times (0.01)^2} = 1,476.88 [/math]

Solving this we find

[math] n \approx 38 [/math]

The width of the strip is the product of the thickness by this factor. It is therefore

[math] 38 \times 0.01 = 0.38 \text{ cm} [/math]

Thermo-electric Couple. If two conductors of different materials, such as iron and copper, are joined at both ends and separated at the middle so as to form a species of ring, and if the junctions are maintained at different temperatures with proper relation to what is called the neutral temperature, a current of electricity will be produced, continuing as long as the difference of temperature is maintained. It is sufficient if one pair of ends is in actual contact, the pieces then separating and having their other ends connected by a wire. The arrangement described constitutes a thermo-electric couple. The intensity of the current produced depends upon the e.m.f. and upon the resistance of the circuit.

Electro-motive Force of a Thermo-electric Couple. The e.m.f. produced by a thermo-electric couple depends upon the difference of temperature of the ends referred to the neutral temperature of the particular couple. If the junction of a couple is heated above the neutral temperature a certain number of degrees and if the other ends are the same number of degrees below the neutral temperature, there will be no e.m.f. produced. Different couples have widely different neutral temperatures. For each couple there is a fixed neutral temperature.

Neutral Temperature. When the mean temperature of the ends of the couple is below the neutral temperature the current flows in one direction; when the mean temperature is above the neutral temperature the current flows in the other direction.

Example. The neutral temperature of a thermo-electric couple of two particular metals is 275°C. One end is heated to a temperature of 300°C, the other end to a temperature of 250°C. What will the result be?

Solution. The average temperature is:

[math] \frac{250 + 300}{2} = 275 \degree C [/math]

Thus the mean or average temperature of the ends is the neutral temperature of the couple. No potential difference will be maintained and consequently no current will pass. The result, as far as the electric state of things is concerned, will be zero.

Example. The junction of a couple is heated to a temperature of 675°F, the other ends are kept at a temperature of 55°F and no current nor difference of e.m.f. can be produced. What is the neutral point of the combination?

Solution. The average temperature is:

[math] \frac{675 + 55}{2} = 365 \degree F [/math]

As no current is produced at this mean temperature of the ends, or, what is the same in effect, as no e.m.f. is produced at this mean temperature, the neutral temperature of the couple is 365°F.

Temperature and Electro-motive Force of Thermo-electric Couple. The e.m.f. produced by a thermo-electric couple depends on the difference of temperature maintained between the opposite ends of the couple. The e.m.f. produced by various combinations of elements of thermo-electric couples, such as an iron-copper couple, a bismuth-antimony couple, and so on, is different for different couples.

To increase the e.m.f. of a thermo-electric battery, the couples must be arranged in series. The laws given for the arrangement of battery cells apply in general to thermo-electric couples.

Thermo-electric Tables. A thermo-electric table is a table of constants by which the e.m.f. of different couples can be calculated. In some such tables no allowance is made for the neutral temperature. It is obvious that such a table is only accurate for one mean temperature. This is not a serious defect, because the mean temperature of a thermo-electric battery is not subject to a wide variation.

THERMO-ELECTRIC TABLE FOR A NEUTRAL TEMPERATURE OF ABOUT 20° C. (Jenkin’s “Electricity and Magnetism” compiled from Matthiessen’s experiments reduced to C.G.S. units.)

To calculate the e.m.f. of a combination, subtract algebraically the number opposite one element of the couple from the number opposite the other element. The result is the e.m.f. in C.G.S. units corresponding to a difference of temperature of 1°C between the ends of the couple. The polarity of the e.m.f. is such as to produce a current from the lower to the upper metal through the hotter end or junction.

Example. Calculate the e.m.f. between lead and cobalt at a mean temperature of 20°C with a temperature difference of 1°C between the junctions or ends.

Solution. The numbers opposite the metals in the table are 0 and -2,200. Subtracting gives:

[math] 2,200 \text{ C.G.S. units of e.m.f.} [/math]

As 10^8 C.G.S. units = 1 volt, this is:

[math] \frac{2,200}{10^8} = 0.000022 \text{ volt} [/math]

Example. Assume that the elements in the above example are joined in contact at one end, while a wire connects their free ends, and that the heat is applied to the junction. In which direction will the current go?

Solution. The heated junction is the hottest. Cobalt is the lower metal in the table. Therefore, the polarity of the e.m.f. is such as to produce a current from the cobalt bar through the hotter junction to the lead.

Example. If the difference of temperature in the above case were 7°C, what would the voltage be?

Solution.

[math] 0.000022 \times 7 = 0.000154 \text{ volt} [/math]

Example. A thermo-electric battery of German silver and iron elements at a mean temperature of 20°C with a temperature difference between the ends of the couples of 121°C and with the couples arranged in series gives an e.m.f. of 1.33 volts. Calculate the number of couples in the series.

Solution. Iron = +1,756; German silver = -1,175. Algebraic difference:

[math] 1,756 + 1,175 = 2,931 \text{ C.G.S. units per } 1\degree C [/math]

At 121°C:

[math] 2,931 \times 121 = 354,651 \text{ C.G.S. units} [/math]

Convert the battery’s e.m.f. to C.G.S. units:

[math] 1.33 \text{ volts} = 1.33 \times 10^8 = 133,000,000 \text{ C.G.S. units} [/math]

Then the number of couples is:

[math] \frac{133,000,000}{354,651} \approx 375 \text{ couples} [/math]

Alternatively, convert to volts:

[math] 354,651 \text{ C.G.S. units} = 0.00354651 \text{ volt} [/math]

[math] \frac{1.33}{0.00354651} \approx 375 \text{ couples} [/math]

In these operations, it is immaterial which number has its sign changed; the position in the table determines the polarity.

Example. Calculate the e.m.f. of an iron-German silver couple, the temperatures of the junctions being 0°C and 100°C.

Solution. Mean temperature:

[math] \frac{100 + 0}{2} = 50 \degree C [/math]

From the table:

Value of iron: [math] 1,734 – 4.87t [/math]

Value of German silver: [math] -1,207 – 5.12t [/math]

Difference: [math] (1,734 – 4.87t) – (-1,207 – 5.12t) = 2,941 + 0.25t [/math]

Substitute [math] t = 50 [/math]:

[math] \text{e.m.f. at 1°C} = 2,941 + (0.25 \times 50) = 2,953.5 \text{ C.G.S. units} [/math]

Then at 100°C difference:

[math] 2,953.5 \times 100 = 295,350 \text{ C.G.S. units} [/math]

Convert to volts:

[math] \frac{295,350}{10^8} = 0.0029535 \text{ volt} [/math]

Alternatively, substitute t in each expression:

Value of iron:

[math] 1,734 – (4.87 \times 50) = 1,490.5 [/math]

Value of German silver:

[math] -1,207 – (5.12 \times 50) = -1,463 [/math]

Difference:

[math] 1,490.5 – (-1,463) = 2,953.5 \text{ C.G.S. units} [/math]

Result matches previous calculation.

Example. Calculate the neutral point of the above couple.
Solution. The neutral point is the value of [math] t [/math] when there is no e.m.f., or when
[math] 2,491 + 0.25t = 0, [/math]
whence

[math] t = -9,964^\circ\text{C}. [/math]

There is therefore no neutral point for this couple, the calculated one being below the absolute zero.

Example. Calculate the neutral point of iron and copper.
Solution. Proceeding as before we find from the table:

Value of iron ………. 1,734 — 4.87[math]t[/math]
Value of copper ………. 136 + 0.95[math]t[/math]
Algebraic difference ……… 1,598 — 5.82[math]t[/math]

Making the algebraic difference zero gives to [math]t[/math] a value which is that of the neutral point. Thus:
[math] 1,598 – 5.82t = 0, [/math]
whence

[math] t = 275^\circ\text{C}. [/math]

Peltier Effect. If a current passes through a circuit of different metals there is a production of heat or a reduction of heat at the junctions of different metals independent of the ordinary heating effect of the current. This is called the Peltier effect. The value of the Peltier effect is expressed in ergs per second per C.G.S. unit current (10 amperes).

Absolute Temperature. To determine the Peltier effect multiply the difference of thermo-electric heights at the junction in question by the absolute temperature of the junction, all in degrees C. The absolute temperature is obtained by adding to the temperature C. the number 273, which is the absolute temperature of the centigrade zero.

Example. What is the absolute temperature of boiling water?
Solution. The temperature C. of boiling water is 100°. To obtain the absolute temperature add 273, or

[math] 100 + 273 = 373^\circ \text{absolute temperature C.} [/math]

Example. Assume a C.G.S. unit current to pass through a junction of iron and copper maintained at a temperature of 100°C. What is the value of the Peltier effect?

Solution. The thermo-electric height of iron at 100°C is
[math] 1,734 – (4.87 \times 100) = 1,247. [/math]
That of copper is
[math] 135 + (0.95 \times 100) = 230. [/math]
The difference of thermo-electric heights is
[math] 1,247 – 230 = 1,017. [/math]
The absolute temperature at the junction is
[math] 100 + 273 = 373. [/math]
Then following the rule we have

[math] 1,017 \times 373 = 379,341 \text{ ergs per second for a C.G.S. unit current.} [/math]

If the current flows from iron to copper through the heated junction it will be a current from the upper to the lower element. This produces a heating effect. If the current is from the lower to the upper element, in this case from copper to iron through the hot junction, the effect will be to cool the junction. In other words, if the direction of the current is such as to supplement the thermo-electric current, it will cool the junction, otherwise the reverse will hold.

Thomson Effect. If a conductor of one material throughout is heated in places so that some parts are hotter than others, it will show thermo-electric action, except in the case of lead. The action is called the Thomson effect.

Referring to the table (page 135), metals with a minus sign prefixed to their temperature coefficients, in the second column of figures, are affected as follows: An electric current passing from a hotter to a cooler portion reduces the temperature of the cooler part. Passing from a cooler to a hotter portion it heats the conductor in the hotter part. This applies to iron and all other metals with negative signs in the table.

Metals with a positive sign prefixed are affected in the reverse way. A current passing from hot to cool heats the cool portion. Passing from cool to hot it lowers the temperature of the hot portion. Copper is one of the metals subject to this action.

The general law is that in the case of metals with a negative sign an electric current tends to increase any local differences of temperature already existing. In the case of metals with positive signs an electric current tends to equalize temperature differences.

If portions of a conductor differ in temperature, the product of that difference by the thermo-electric temperature coefficient of the metal as given in the table gives the thermo-electric difference between the portions in question. The temperature difference is to be in degrees C., and the result will be in C.G.S. units.

Example. The ends of an iron wire differ by 100°C. in temperature. Calculate the thermo-electric difference between the ends.

Solution. From the table the thermo-electric coefficient of iron is found to be —4.87. Then, as the sign has no effect upon the multiplication except as indicating the polarity of the e.m.f., we have

[math] 100 \times 4.87 = 487 \text{ C.G.S. units of e.m.f., or } 0.00000487 \text{ volt.} [/math]

As the coefficient has a negative sign, the current tends to increase any existing differences of temperature.

The value of the Thomson effect in ergs per second per C.G.S. unit current (10 amperes) is found by multiplying the thermo-electric difference, as just calculated in the case of iron for instance, by the sum of one-half the temperature difference in degrees C. and 273. This reduces the temperature to the absolute scale C.

Example. Calculate the Thomson effect for the iron wire of the last example.
Solution. The temperature difference is 100°C.; half of this is 50. Following the rule we have

[math] 487 \times (273 + 50) = 487 \times 323 = 157,301 \text{ ergs per second for a current of 10 amperes (1 C.G.S. unit).} [/math]

If the current goes from cold iron to hot iron, this number of ergs of electric energy are converted into heat energy; if the current goes the other way, the same number of ergs of heat energy are converted into electric energy.

Example. Let a copper wire be unequally heated 100°C. as in the last example. Calculate the Thomson effect.
Solution. The thermo-electric effect will be
[math] 0.95 \times 100 = 95 \text{ C.G.S. units.} [/math]
For the ergs per second we have

[math] 95 \times (273 + 50) = 95 \times 323 = 30,685 \text{ ergs per second for a 10-ampere current.} [/math]

Example. Make the same calculation for lead.
Solution. As the temperature coefficient for lead is 0, there is no Thomson effect for lead.

Example. Calculate the neutral point of the above couple.
Solution. The neutral point is the value of [math] t [/math] when there is no e.m.f., or when
[math] 2,491 + 0.25t = 0, [/math]
whence

[math] t = -9,964^\circ\text{C}. [/math]

There is therefore no neutral point for this couple, the calculated one being below the absolute zero.

Example. Calculate the neutral point of iron and copper.
Solution. Proceeding as before we find from the table:

Value of iron ………. 1,734 — 4.87[math]t[/math]
Value of copper ………. 136 + 0.95[math]t[/math]
Algebraic difference ……… 1,598 — 5.82[math]t[/math]

Making the algebraic difference zero gives to [math]t[/math] a value which is that of the neutral point. Thus:
[math] 1,598 – 5.82t = 0, [/math]
whence

[math] t = 275^\circ\text{C}. [/math]

Peltier Effect. If a current passes through a circuit of different metals there is a production of heat or a reduction of heat at the junctions of different metals independent of the ordinary heating effect of the current. This is called the Peltier effect. The value of the Peltier effect is expressed in ergs per second per C.G.S. unit current (10 amperes).

Absolute Temperature. To determine the Peltier effect multiply the difference of thermo-electric heights at the junction in question by the absolute temperature of the junction, all in degrees C. The absolute temperature is obtained by adding to the temperature C. the number 273, which is the absolute temperature of the centigrade zero.

Example. What is the absolute temperature of boiling water?
Solution. The temperature C. of boiling water is 100°. To obtain the absolute temperature add 273, or

[math] 100 + 273 = 373^\circ \text{absolute temperature C.} [/math]

Example. Assume a C.G.S. unit current to pass through a junction of iron and copper maintained at a temperature of 100°C. What is the value of the Peltier effect?

Solution. The thermo-electric height of iron at 100°C is
[math] 1,734 – (4.87 \times 100) = 1,247. [/math]
That of copper is
[math] 135 + (0.95 \times 100) = 230. [/math]
The difference of thermo-electric heights is
[math] 1,247 – 230 = 1,017. [/math]
The absolute temperature at the junction is
[math] 100 + 273 = 373. [/math]
Then following the rule we have

[math] 1,017 \times 373 = 379,341 \text{ ergs per second for a C.G.S. unit current.} [/math]

If the current flows from iron to copper through the heated junction it will be a current from the upper to the lower element. This produces a heating effect. If the current is from the lower to the upper element, in this case from copper to iron through the hot junction, the effect will be to cool the junction. In other words, if the direction of the current is such as to supplement the thermo-electric current, it will cool the junction, otherwise the reverse will hold.

Thomson Effect. If a conductor of one material throughout is heated in places so that some parts are hotter than others, it will show thermo-electric action, except in the case of lead. The action is called the Thomson effect.

Referring to the table (page 135), metals with a minus sign prefixed to their temperature coefficients, in the second column of figures, are affected as follows: An electric current passing from a hotter to a cooler portion reduces the temperature of the cooler part. Passing from a cooler to a hotter portion it heats the conductor in the hotter part. This applies to iron and all other metals with negative signs in the table.

Metals with a positive sign prefixed are affected in the reverse way. A current passing from hot to cool heats the cool portion. Passing from cool to hot it lowers the temperature of the hot portion. Copper is one of the metals subject to this action.

The general law is that in the case of metals with a negative sign an electric current tends to increase any local differences of temperature already existing. In the case of metals with positive signs an electric current tends to equalize temperature differences.

If portions of a conductor differ in temperature, the product of that difference by the thermo-electric temperature coefficient of the metal as given in the table gives the thermo-electric difference between the portions in question. The temperature difference is to be in degrees C., and the result will be in C.G.S. units.

Example. The ends of an iron wire differ by 100°C. in temperature. Calculate the thermo-electric difference between the ends.

Solution. From the table the thermo-electric coefficient of iron is found to be —4.87. Then, as the sign has no effect upon the multiplication except as indicating the polarity of the e.m.f., we have

[math] 100 \times 4.87 = 487 \text{ C.G.S. units of e.m.f., or } 0.00000487 \text{ volt.} [/math]

As the coefficient has a negative sign, the current tends to increase any existing differences of temperature.

The value of the Thomson effect in ergs per second per C.G.S. unit current (10 amperes) is found by multiplying the thermo-electric difference, as just calculated in the case of iron for instance, by the sum of one-half the temperature difference in degrees C. and 273. This reduces the temperature to the absolute scale C.

Example. Calculate the Thomson effect for the iron wire of the last example.
Solution. The temperature difference is 100°C.; half of this is 50. Following the rule we have

[math] 487 \times (273 + 50) = 487 \times 323 = 157,301 \text{ ergs per second for a current of 10 amperes (1 C.G.S. unit).} [/math]

If the current goes from cold iron to hot iron, this number of ergs of electric energy are converted into heat energy; if the current goes the other way, the same number of ergs of heat energy are converted into electric energy.

Example. Let a copper wire be unequally heated 100°C. as in the last example. Calculate the Thomson effect.
Solution. The thermo-electric effect will be
[math] 0.95 \times 100 = 95 \text{ C.G.S. units.} [/math]
For the ergs per second we have

[math] 95 \times (273 + 50) = 95 \times 323 = 30,685 \text{ ergs per second for a 10-ampere current.} [/math]

Example. Make the same calculation for lead.
Solution. As the temperature coefficient for lead is 0, there is no Thomson effect for lead.

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PROBLEMS. _ Calculate the e.m.f. between iron and lead per degree C. at 0° C. Ans. 1,734 X 107° volt. Calculate the e.m.f. of Iron and copper, the junctions being kept at o° C. and 100° C. Ans. 130,700 X 107 volt. A thermo-electric battery of 33 bismuth (— 9,700) and antimony (2,250) couples shows 1.25 volts at the meah temperature of 19° C. What is the difference of temperature between the two faces? Ans. 317°C. A current of 11 amperes is passed through a wire of 0.7 cm. circum- ference and of 0.6 ohm resistance per 100 meters. How much will its temperature be increased ? Ans. 10°C, What will be the rise in temperature in a copper wire Of 110 mils diameter carrying a current of 20 amperes? Ans. 15°C. Calculate the diameter of a lead wire to melt with a current of 3.1 amperes. – Ans. 0.0606 cm. Taking the melting point of copper as 1,127°C. and its specific resistance as 1.652, calculate the thickness of a wire that will melt at 2I amperes current. Ans. 0.0632 cm. A copper strip is 0.02 cm. thick. How wide should it be to melt with an 18-ampere current? Ans. 0.097+ cm. THERMO-ELECTRICITY 14 How wide should the same copper strip be for ten times as great a current ? Ans. 1.0586 cm. (more than ten times the first width). Calculate the e.m.f. of bismuth (— 9,700) and platinum per degree C. at 20° mean temperature C. Ans. 0©.000,097,9 volt. Make the same calculation for German silver and copper. Ans. 0©.000,015,55 volt. How many couples of the above would be required fora volt at 100° C, difference of temperature ? Ans. 608 couples. Calculate by table on page 190 the e.m.f. per degree C. difference, of German silver and copper at 20° C. mean temperature. Ans. 0.000,014,64 volt. What is the e.m.f. per degree C. for a mean temperature of 315° C.? Ans. 0.000,032,55 volt. At 600° C. difference of temperature what would be the number of couples of the combination of the last problem to a volt? Ans. 53 couples. What is the neutral point of the above combination? Ans. — 221°C. What is the neutral point of iron and zinc? Ans. 206°C. What is the neutral point of aluminum and palladium? Ans. —138°C. Calculate the e.m.f. per degree C. of aluminum and nickel at a mean temperature of 300° C.? © Ans. ©.000,012,69 volt. What would the voltage be at 600° C. difference of temperature, at above mean temperature, and how many couples would be required for a volt? Ans. 0.007,614 volt; 132 couples. Calculate the Thomson effect in a copper wire, with a difference of 200° C. in temperature of different sections of its length. Ans. 0©.000,001,9 Volt; 70,870 ergs per second. Make the same calculation for German silver with 75° C. difference of temperature. Ans. 0.000,003,84 Volt; 119,232 ergs per second. Make the same calculation for silver with 98° C. difference in tem- perature. Ans. 0.000,001,47 Volt; 47,334 ergs per sec. What is the Peltier effect between German silver and iron at 100° C. Ans. 1,106,318 ergs per second for a C.G.S. unit current. Calculate the Peltier effect for zinc and tin at 190° C. Ans. 290,995 ergs per second fora C.G.S. unit current.

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Here is the corrected and math-formatted “PROBLEMS” section, preserving the historical phrasing and style verbatim, with all mathematical expressions enclosed in LaTeX-style [math] [/math] tags as requested:

PROBLEMS.

Calculate the e.m.f. between iron and lead per degree C. at 0°C.
Ans. [math] 1,734 \times 10^{-6} \text{ volt.} [/math]

Calculate the e.m.f. of iron and copper, the junctions being kept at 0°C. and 100°C.
Ans. [math] 130,700 \times 10^{-6} \text{ volt.} [/math]

A thermo-electric battery of 33 bismuth (—9,700) and antimony (2,250) couples shows 1.25 volts at the mean temperature of 19°C.
What is the difference of temperature between the two faces?
Ans. [math] 317^\circ \text{C.} [/math]

A current of 11 amperes is passed through a wire of 0.7 cm. circumference and of 0.6 ohm resistance per 100 meters.
How much will its temperature be increased?
Ans. [math] 10^\circ \text{C.} [/math]

What will be the rise in temperature in a copper wire of 110 mils diameter carrying a current of 20 amperes?
Ans. [math] 15^\circ \text{C.} [/math]

Calculate the diameter of a lead wire to melt with a current of 3.1 amperes.
Ans. [math] 0.0606 \text{ cm.} [/math]

Taking the melting point of copper as 1,127°C. and its specific resistance as 1.652, calculate the thickness of a wire that will melt at 21 amperes current.
Ans. [math] 0.0632 \text{ cm.} [/math]

A copper strip is 0.02 cm. thick. How wide should it be to melt with an 18-ampere current?
Ans. [math] 0.097^+ \text{ cm.} [/math]

How wide should the same copper strip be for ten times as great a current?
Ans. [math] 1.0586 \text{ cm.} \quad (\text{more than ten times the first width}) [/math]

Calculate the e.m.f. of bismuth (—9,700) and platinum per degree C. at 20° mean temperature C.
Ans. [math] 0.0000979 \text{ volt.} [/math]

Make the same calculation for German silver and copper.
Ans. [math] 0.00001555 \text{ volt.} [/math]

How many couples of the above would be required for a volt at 100°C. difference of temperature?
Ans. [math] 608 \text{ couples.} [/math]

Calculate by table on page 190 the e.m.f. per degree C. difference, of German silver and copper at 20°C. mean temperature.
Ans. [math] 0.00001464 \text{ volt.} [/math]

What is the e.m.f. per degree C. for a mean temperature of 315°C.?
Ans. [math] 0.00003255 \text{ volt.} [/math]

At 600°C. difference of temperature what would be the number of couples of the combination of the last problem to a volt?
Ans. [math] 53 \text{ couples.} [/math]

What is the neutral point of the above combination?
Ans. [math] -221^\circ \text{C.} [/math]

What is the neutral point of iron and zinc?
Ans. [math] 206^\circ \text{C.} [/math]

What is the neutral point of aluminum and palladium?
Ans. [math] -138^\circ \text{C.} [/math]

Calculate the e.m.f. per degree C. of aluminum and nickel at a mean temperature of 300°C.?
Ans. [math] 0.00001269 \text{ volt.} [/math]

What would the voltage be at 600°C. difference of temperature, at above mean temperature, and how many couples would be required for a volt?
Ans.

[math] 0.007614 \text{ volt}; \quad 132 \text{ couples.} [/math]

Calculate the Thomson effect in a copper wire, with a difference of 200°C. in temperature of different sections of its length.
Ans.

[math] 0.0000019 \text{ volt}; \quad 70,870 \text{ ergs per second.} [/math]

Make the same calculation for German silver with 75°C. difference of temperature.
Ans.

[math] 0.00000384 \text{ volt}; \quad 119,232 \text{ ergs per second.} [/math]

Make the same calculation for silver with 98°C. difference in temperature.
Ans.

[math] 0.00000147 \text{ volt}; \quad 47,334 \text{ ergs per sec.} [/math]

What is the Peltier effect between German silver and iron at 100°C.?
Ans. [math] 1,106,318 \text{ ergs per second for a C.G.S. unit current.} [/math]

Calculate the Peltier effect for zinc and tin at 190°C.
Ans. [math] 290,995 \text{ ergs per second for a C.G.S. unit current.} [/math]

CHAPTER X.

THERMO-ELECTRICITY.

Topics: Emissivity. — Heating of a Conductor by a Current. — Cross Section of a Conductor to be Heated to a given Degree by a Given Current. — Thermo-electric Couple. — Electro-motive Force of a Thermo-electric Couple. — Neutral Temperature. — Temperature and Electro-motive Force of Thermo-electric Couple. — Thermo-electric Tables. — Peltier Effect. — Absolute Temperature. — Thomson Effect.

Emissivity. If a body is heated above the temperature of its surroundings it parts with its heat. The process is termed emissivity. Unit emissivity is the quantity of heat lost per second per square centimeter of surface per degree of difference between its temperature and that of its surroundings.

It is measured in ergs or calories or other heat unit. The C.G.S. unit of heat is the erg. The practical unit is the therm, calorie, or gram-degree, which is the heat required to raise one gram of water one degree C.

Heating of a Conductor by a Current. A number of formulas for calculation of the temperature imparted to a conductor by a current passing through it have been proposed. From the nature of the case these formulas can only be approximate.

By radiation and convection about [math]0.000025[/math] calorie is lost per second by an unpolished metallic surface of one square centimeter in the air for each degree C. that it is heated above the air. If therefore the calories expended on a conductor by the passage of a current are determined, and if the wire is assumed to have attained its fixed temperature, these calories will be equal to those lost by radiation and convection.

The quotient of these calories divided by the area of the surface of the conductor in square centimeters will give the calories per square centimeter which are expended on the conductor and which are in turn lost by it. The quotient of these calories divided by [math]0.000025[/math] is the temperature in degrees C. which the wire will attain. The result is only approximate.

Example. A current of [math]10[/math] amperes passes through a wire one centimeter in circumference. The wire has a resistance of [math]0.3[/math] ohm per [math]100[/math] meters. How many degrees will its temperature be increased?

Solution. The watts per second due to the passage of the current are given by the expression [math]I^2 R = 100 \times 0.3 = 30[/math].

The surface area of [math]100[/math] meters of the wire is [math]10,000[/math] square centimeters. The calories expended on [math]100[/math] meters are [math]30 \times 0.24 = 7.2[/math].

The calories per square centimeter are the quotient of [math]7.2[/math] by the area of the wire, or [math]7.2 \div 10,000 = 0.00072[/math] calorie.

Dividing [math]0.00072[/math] by [math]0.000025[/math] gives [math]2.88^\circ C[/math], the degrees C. above the temperature of the air to which the wire would be heated by such a current.

A formula for the rise in temperature in degrees C. of a bare copper wire with a current passing through it in still air is given below. The square of the amperes is divided by the cube of the diameter in mils, and the quotient is multiplied by [math]50,000[/math]. The product is the increase in temperature.

Formula:

[math]\text{Temperature (°C)} = \frac{I^2}{d^3} \times 50,000[/math]

Example. A copper wire [math]125[/math] mils in diameter is conducting a current of [math]10[/math] amperes. How much will it be heated above the temperature of the surrounding air, the wire being of bare copper?

Solution. Substituting in the formula gives:

[math]\frac{10^2}{125^3} \times 50,000 = 2.5^\circ C[/math]

In the above example the thickness of the wire is the same as in the preceding example. The results are reasonably close for approximate methods.

Cross Section of a Conductor to be Heated to a Given Degree by a Given Current

The diameter of a wire of a material of known specific resistance which a stated current will heat to a stated degree of heat above that of the surrounding air can be calculated approximately by the application of the emissivity factor.

The following discussion leads to results in centimeters.

Specific resistance is given in microhms in the tables. The resistance in ohms of a wire of diameter [math]d[/math] and whose cross-sectional area is therefore [math]\frac{\pi d^2}{4}[/math] is given by the equation:

[math]R = \text{sp. res.} \times 10^{-6} \times \frac{4l}{\pi d^2}[/math]     (1)

The heat expended on a conductor in the passage of a current is [math]I^2 R[/math] watt-seconds or joules per second. Substituting for [math]R[/math] from (1) gives:

[math]\text{Heat} = I^2 \times \text{sp. res.} \times 10^{-6} \times \frac{4l}{\pi d^2}[/math] watt-seconds per second.     (2)

A calorie is equal to [math]4.16[/math] watt-seconds. If (2) is divided by [math]4.16[/math] it becomes:

[math]\text{Heat} = \frac{I^2 \times \text{sp. res.} \times 10^{-6} \times 4l}{4.16 \pi d^2}[/math] calories per second.     (3)

The surface area of one centimeter of the wire is [math]\pi d[/math]. If (3) is divided by [math]\pi d[/math], the result will be the heat per square centimeter of surface area. Accepting [math]4,000[/math] as the emissivity factor, the heat emitted per square centimeter at temperature [math]t[/math] is [math]t \div 4,000[/math].

Equating these gives:

[math]\frac{I^2 \times \text{sp. res.} \times 10^{-6} \times 4l}{4.16 \pi d^3} = \frac{t}{4,000}[/math]     (4)

Multiplying through by [math]4,000[/math] gives:

[math]\frac{I^2 \times \text{sp. res.} \times 10^{-6} \times 4l \times 4,000}{4.16 \pi d^3} = t[/math]

Reducing constants and rearranging:

[math]t = \frac{I^2 \times \text{sp. res.} \times 0.00039}{d^3}[/math]     (5)

[math]d = \sqrt[3]{\frac{I^2 \times \text{sp. res.} \times 0.00039}{t}}[/math]     (6)

To calculate the diameter of a cylindrical conductor which will attain a certain temperature in the air in passing a certain current: multiply the square of the current by the specific resistance and by the factor [math]0.00039[/math], divide by the temperature, and take the cube root of the result. This gives the diameter in centimeters.

Example. Calculate the diameter of a lead wire to melt with a current of [math]7.2[/math] amperes. The melting point of lead is taken as [math]335^\circ C[/math], and its specific resistance is [math]19.85[/math] microhms per cubic centimeter.

Solution. By formula (6):

[math]d^3 = \frac{(7.2)^2 \times 19.85 \times 0.00039}{335} = 0.00118[/math]

[math]d = \sqrt[3]{0.00118} = 0.1058[/math] cm

Recurring to equation (4), the term [math]4/\pi d^3[/math] is the reciprocal of the product of the surface area of a unit length of a conductor by the cross-sectional area of the same. This product is [math](\pi d^2 / 4) \times \pi d[/math], and its reciprocal is [math]4 / \pi^2 d^3[/math], as it appears in equation (4).

This equation is restricted to a circular conductor. By substituting for [math]\pi d^2 / 4[/math] and for [math]\pi d[/math] the proper values, the equation may be made to apply to conductors of other cross sections.

Suppose a strip is to be cut from a sheet of metal of thickness [math]a[/math], and it is to be calculated how wide the strip shall be to attain a given temperature with a given current.

Call the width of the strip [math]na[/math], the factor [math]n[/math] being the unknown quantity to be determined. The cross-sectional area of the strip will be [math]na \times a = na^2[/math]. The surface area of the unitary length will be [math]2na + 2a = 2a(n + 1)[/math].

The product of these two quantities is [math]2a^2(n^2 + n)[/math]. As this is the product of the surface area of the unit length of the conductor by its cross-sectional area, it is obvious that it can be substituted for [math]\frac{4}{\pi d^3}[/math] in equation (4).

This substitution is performed by division by it in place of division by [math]\pi d^3[/math], giving:

[math]t = \frac{I^2 \times \text{sp. res.} \times 10^{-6}}{4.16 \times 2a^2(n^2 + n)}[/math]     (7)

Multiplying both sides by [math]4.16 \times 2a^2(n^2 + n)[/math] gives:

[math]t \times 4.16 \times 2a^2(n^2 + n) = I^2 \times \text{sp. res.} \times 10^{-6}[/math]

Rewriting with simplified constant:

[math]n^2 + n = \frac{I^2 \times \text{sp. res.} \times 0.0004808}{t \times a^2}[/math]     (8)

The value of [math]n[/math] is obtained from this equation by substituting the constants given in the problem. The width of the strip is the product of its thickness by [math]n[/math], or [math]na[/math].

Example. A strip of lead to be melted by a current of [math]7.2[/math] amperes is to be cut from a sheet [math]0.01[/math] cm thick. The specific resistance of lead is taken as [math]19.85[/math] and its melting point as [math]335^\circ C[/math]. Calculate the width of the strip.

Solution. Substituting these values in equation (8) gives:

[math]n^2 + n = \frac{(7.2)^2 \times 19.85 \times 0.0004808}{335 \times (0.01)^2} = 1,476.88[/math]

Solving this we find [math]n \approx 38[/math].

The width of the strip is the product of the thickness by this factor:

[math]38 \times 0.01 = 0.38[/math] cm.

Thermo-electric Couple

If two conductors of different materials, such as iron and copper, are joined at both ends and separated at the middle so as to form a species of ring, and if the junctions are maintained at different temperatures with proper relation to what is called the neutral temperature, a current of electricity will be produced, continuing as long as the difference of temperature is maintained.

It is sufficient if one pair of ends is in actual contact, the pieces then separating and having their other ends connected by a wire. The arrangement described constitutes a thermo-electric couple. The intensity of the current produced depends upon the e.m.f. and upon the resistance of the circuit.

Electro-motive Force of a Thermo-electric Couple

The e.m.f. produced by a thermo-electric couple depends upon the difference of temperature of the ends referred to the neutral temperature of the particular couple. If the junction of a couple is heated above the neutral temperature a certain number of degrees and if the other ends are the same number of degrees below the neutral temperature, there will be no e.m.f. produced.

Different couples have widely different neutral temperatures. For each couple there is a fixed neutral temperature.

Neutral Temperature

When the mean temperature of the ends of the couple is below the neutral temperature, the current flows in one direction; when the mean temperature is above the neutral temperature, the current flows in the other direction.

Example. The neutral temperature of a thermo-electric couple of two particular metals is [math]275^\circ C[/math]. One end is heated to a temperature of [math]300^\circ C[/math], the other end to a temperature of [math]250^\circ C[/math]. What will the result be?

Solution. The average temperature is [math](250 + 300) \div 2 = 275^\circ C[/math]. Thus the mean or average temperature of the ends is the neutral temperature of the couple. No potential difference will be maintained and consequently no current will pass. The result as far as the electric state of things is concerned will be zero.

Example. The junction of a couple is heated to a temperature of [math]675^\circ F[/math], the other ends are kept at a temperature of [math]55^\circ F[/math] and no current nor difference of e.m.f. can be produced. What is the neutral point of the combination?

Solution. The average temperature is [math](675 + 55) \div 2 = 365^\circ F[/math]. As no current is produced at this mean temperature of the ends, or, what is the same in effect, as no e.m.f. is produced at this mean temperature, the neutral temperature of the couple is [math]365^\circ F[/math].

Temperature and Electro-motive Force of Thermo-electric Couple

The e.m.f. produced by a thermo-electric couple depends on the difference of temperature maintained between the opposite ends of the couple. The e.m.f. produced by various combinations of elements of thermo-electric couples, such as an iron-copper couple, a bismuth-antimony couple, and so on, is different for different couples.

To increase the e.m.f. of a thermo-electric battery the couples must be arranged in series. The laws given for the arrangement of battery cells apply in general to thermo-electric couples.

Thermo-electric Tables

A thermo-electric table is a table of constants by which the e.m.f. of different couples can be calculated. In some such tables no allowance is made for the neutral temperature. It is obvious that such a table is only accurate for one mean temperature. This is not a serious defect, because the mean temperature of a thermo-electric battery is not subject to a wide variation.

THERMO-ELECTRIC TABLE FOR A NEUTRAL TEMPERATURE OF ABOUT 20° C

(From Jenkin’s “Electricity and Magnetism”, compiled from Matthiessen’s experiments, reduced to C.G.S. units)

• Bismuth, pressed commercial wire ……….. [math]+9700[/math]
• Bismuth, pure pressed wire ……………….. [math]-8900[/math]
• Bismuth, crystal, axial ……………………. [math]-6500[/math]
• Bismuth, crystal, equatorial ……………….. [math]-4500[/math]
• Cobalt ……………………………………. [math]-2200[/math]
• German silver …………………………….. [math]-1175[/math]
• Mercury ………………………………….. [math]-41.8[/math]
• Tin ………………………………………. [math]+10[/math]
• Copper, commercial ……………………….. [math]+10[/math]
• Platinum …………………………………. [math]+90[/math]
• Gold …………………………………….. [math]+120[/math]
• Antimony, pressed wire …………………….. [math]+280[/math]
• Silver, pure hard …………………………. [math]+300[/math]
• Zinc, pure pressed ………………………… [math]+370[/math]
• Copper, electrolytic ………………………. [math]+380[/math]
• Antimony, pressed commercial wire ………….. [math]+600[/math]
• Arsenic ………………………………….. [math]+1356[/math]
• Iron, pianoforte wire ……………………… [math]+1750[/math]
• Antimony, crystal, axial ………………….. [math]+2260[/math]
• Antimony, crystal, equatorial ……………… [math]+2640[/math]
• Phosphorus, red ………………………….. [math]+2970[/math]
• Tellurium ………………………………… [math]+50200[/math]
• Selenium …………………………………. [math]+80700[/math]

To calculate the e.m.f. of a combination, subtract algebraically the number opposite one element of the couple from the number opposite the other element. The result is the e.m.f. in C.G.S. units corresponding to a difference of temperature of [math]1^\circ C[/math] between the ends of the couple.

The polarity of the e.m.f. is such as to produce a current from the lower to the upper metal through the hotter end or junction.

Example. Calculate the e.m.f. between lead and cobalt at a mean temperature of [math]20^\circ C[/math] with a temperature difference of [math]1^\circ C[/math] between the junctions or ends.

Solution. The numbers opposite the metals in the table are [math]0[/math] (for lead) and [math]-2,200[/math] (for cobalt). Subtracting gives [math]2,200[/math] C.G.S. units of e.m.f. As [math]10^8[/math] C.G.S. units = 1 volt, this is:

[math]2,200 \div 10^8 = 0.000022[/math] volt.

Example. Assume that the elements in the above example are joined in contact at one end, while a wire connects their free ends, and that the heat is applied to the junction. In which direction will the current go?

Solution. The heated junction is the hottest. Cobalt is the lower metal in the table. Therefore the polarity of the e.m.f. is such as to produce a current from the cobalt bar through the hotter junction to the lead.

Example. If the difference of temperature in the above case were [math]7^\circ C[/math], what would the voltage be?

Solution. [math]0.000022 \times 7 = 0.000154[/math] volt.

Example. A thermo-electric battery of German silver and iron elements at a mean temperature of [math]20^\circ C[/math] with a temperature difference between the ends of the couples of [math]121^\circ C[/math] and with the couples arranged in series gives an e.m.f. of [math]1.33[/math] volts. Calculate the number of couples in the series.

Solution. Iron = [math]+1,756[/math], German silver = [math]-1,175[/math]. Subtracting algebraically:

[math]1,756 + 1,175 = 2,931[/math] C.G.S. units (per couple, per degree).

At [math]121^\circ C[/math], e.m.f. per couple: [math]2,931 \times 121 = 354,651[/math] C.G.S. units.

Battery e.m.f. in C.G.S. units: [math]1.33 \times 10^8 = 133,000,000[/math]

Number of couples = [math]133,000,000 \div 354,651 \approx 375[/math]

Thus, 375 couples are required to produce 1.33 volts at the specified temperature difference.

The e.m.f.’s of the above calculation could have been expressed in volts and the operation could have been done in these units. It is merely a question of decimal place. Thus:

[math]354,051[/math] C.G.S. units = [math]0.00354651[/math] volt

[math]1.33 \div 0.00354651 = 375[/math] couples, as before.

In these operations it is immaterial which number has its sign changed, becoming thereby the subtrahend. It is only necessary to change the sign of one of the numbers and to add the two algebraically. The position in the table determines the polarity.

The table given below takes in a considerable range of mean temperature and is a more satisfactory one to work with than is the last. It is based upon the work of Professor Tait (Trans. Royal Soc., Edinburgh, Vol. XXVII, 1873). The table is taken from Everett’s “C.G.S. System of Units.”

THERMO-ELECTRIC HEIGHTS AT [math]t^\circ C[/math] IN C.G.S. UNITS

(To use the table, substitute the mean temperature of the couple for [math]t[/math].)

• Iron ………………… [math]+1,734 – 4.87t[/math]
• Steel ……………….. [math]+1,139 – 3.288t[/math]
• Alloy, Pt 95% / Ir 5% …. [math]+622 – 0.534t[/math]
• Alloy, Pt 90% / Ir 10% … [math]+596 – 1.347t[/math]
• Alloy, Pt 85% / Ir 15% … [math]+709 – 0.636t[/math]
• Soft platinum ………… [math]-61 – 1.108t[/math]
• Platinum-nickel alloy …. [math]+544 – 1.104t[/math]
• Hard platinum ………… [math]+260 – 0.754t[/math]
• Magnesium …………… [math]+244 – 0.954t[/math]
• Cadmium ……………… [math]+266 + 4.29t[/math]
• Zinc ………………… [math]+234 + 2.407t[/math]
• Silver ………………. [math]+214 + 1.50t[/math]
• Tin …………………. [math]-43 + 0.563t[/math]
• Aluminum …………….. [math]-7 + 0.929t[/math]
• Palladium ……………. [math]-625 – 3.50t[/math]
• German silver ………… [math]-1,207 – 5.12t[/math]
• Nickel (to 175°C) …….. [math]-1,204 – 5.124t[/math]
• Nickel (250–310°C) ……. [math]-8,449 + 24.1t[/math]
• Nickel (from 340°C) …… [math]-307 – 5.124t[/math]

Example. Calculate the e.m.f. of an iron-German silver couple, the temperatures of the junctions being [math]0^\circ C[/math] and [math]100^\circ C[/math].

Solution. The mean temperature is [math](100 + 0) \div 2 = 50^\circ C[/math]. Substituting into the expressions:

• Iron: [math]1,734 – 4.87 \times 50 = 1,490.5[/math]
• German silver: [math]-1,207 – 5.12 \times 50 = -1,463[/math]
• Algebraic difference: [math]1,490.5 – (-1,463) = 2,953.5[/math] C.G.S. units per degree

For a difference of [math]100^\circ C[/math]:

[math]2,953.5 \times 100 = 295,350[/math] C.G.S. units.

[math]295,350 \div 10^8 = 0.0029535[/math] volt.

Example. Calculate the neutral point of the above couple.

Solution. Set the algebraic difference to zero:

[math]2,941 + 0.25t = 0[/math] ⇒ [math]t = -2,941 \div 0.25 = -11,764^\circ C[/math]

This is below absolute zero. Therefore, the couple has no neutral point in the physical range.

Example. Calculate the neutral point of iron and copper.

• Iron: [math]1,734 – 4.87t[/math]
• Copper: [math]136 + 0.95t[/math]
• Algebraic difference: [math]1,598 – 5.82t[/math]

Set the algebraic difference to zero:

[math]1,598 – 5.82t = 0[/math] ⇒ [math]t = 1,598 \div 5.82 = 275^\circ C[/math]

Peltier Effect

If a current passes through a circuit of different metals, there is a production or reduction of heat at the junctions independent of the usual resistive heating. This is called the Peltier effect.

The value of the Peltier effect is expressed in ergs per second per C.G.S. unit current ([math]10[/math] amperes).

Absolute Temperature

To determine the Peltier effect, multiply the difference of thermo-electric heights at the junction by the absolute temperature (in °C) of the junction. Absolute temperature is obtained by adding [math]273[/math] to the Celsius temperature.

Example. What is the absolute temperature of boiling water?

Solution. The temperature C. of boiling water is [math]100^\circ C[/math]. To obtain the absolute temperature add [math]273[/math]:

[math]100 + 273 = 373^\circ[/math] absolute temperature C.

Example. Assume a C.G.S. unit current to pass through a junction of iron and copper maintained at a temperature of [math]100^\circ C[/math]. What is the value of the Peltier effect?

Solution.

• Thermo-electric height of iron at [math]100^\circ C[/math]: [math]1,734 – (4.87 \times 100) = 1,247[/math]
• Thermo-electric height of copper at [math]100^\circ C[/math]: [math]136 + (0.95 \times 100) = 230[/math]
• Difference of heights: [math]1,247 – 230 = 1,017[/math]
• Absolute temperature: [math]100 + 273 = 373[/math]

Peltier effect (in ergs/sec for C.G.S. unit current):

[math]1,017 \times 373 = 379,341[/math] ergs per second.

If the current flows from iron to copper through the heated junction, it flows from the upper to the lower element — this produces a heating effect.

If the current flows from copper to iron through the hot junction (lower to upper), the effect is to cool the junction.

In other words, if the direction of the current is such as to supplement the thermo-electric current, it will cool the junction; otherwise, it heats it.

Thomson Effect

If a conductor of one material throughout is heated in places so that some parts are hotter than others, it will show thermo-electric action — except in the case of lead. This is called the Thomson effect.

Referring to the table on page 135, metals with a negative sign in the temperature coefficient (second column) behave as follows:

• An electric current passing from a hotter to a cooler portion reduces the temperature of the cooler part.
• Passing from a cooler to a hotter portion, it heats the hotter part.

This applies to iron and all other metals with negative signs in the table.

Metals with a positive sign behave in the reverse way:

• Current from hot to cool — heats the cool part.
• Current from cool to hot — lowers the temperature of the hot part.

Copper is one of the metals subject to this action.

General Law: In the case of metals with negative coefficients, an electric current increases any local temperature differences.

In the case of metals with positive coefficients, an electric current equalizes temperature differences.

If portions of a conductor differ in temperature, the product of that temperature difference by the thermo-electric temperature coefficient of the metal (from the table) gives the thermo-electric difference between the portions, in C.G.S. units.

The temperature difference must be in degrees C. The result is in C.G.S. units.

Example. The ends of an iron wire differ by [math]100^\circ C[/math] in temperature. Calculate the thermo-electric difference between the ends.

Solution. From the table, the thermo-electric coefficient of iron is [math]-4.87[/math]. The sign only indicates polarity. Thus,

[math]100 \times 4.87 = 487[/math] C.G.S. units of e.m.f., or

[math]487 \div 10^8 = 0.00000487[/math] volt.

Since the coefficient is negative, the current tends to increase any existing differences of temperature.

The value of the Thomson effect in ergs per second per C.G.S. unit current (10 amperes) is found by multiplying the thermo-electric difference by the sum of one-half the temperature difference and 273 (for absolute temperature).

Example. Calculate the Thomson effect for the iron wire of the last example.

Solution. Temperature difference = [math]100^\circ C[/math]; half = [math]50^\circ C[/math]. Absolute temp = [math]273 + 50 = 323^\circ C[/math]. Thomson effect = [math]487 \times 323 = 157,301[/math] ergs/sec for 10 amperes.

If the current goes from cold iron to hot iron, these ergs of electric energy are converted into heat. If it goes the other way, the reverse happens.

Example. Let a copper wire be unequally heated [math]100^\circ C[/math] as in the last example. Calculate the Thomson effect.

Solution. Thermo-electric difference: [math]0.95 \times 100 = 95[/math] Absolute temp: [math]273 + 50 = 323[/math] Effect: [math]95 \times 323 = 30,685[/math] ergs/sec

Example. Make the same calculation for lead.

Solution. Lead’s coefficient is 0. There is no Thomson effect for lead.

PROBLEMS

• Calculate the e.m.f. between iron and lead per degree C. at [math]0^\circ C[/math]. Ans. [math]1,734 \times 10^{-8}[/math] volt.
• Calculate the e.m.f. of iron and copper, the junctions being at [math]0^\circ C[/math] and [math]100^\circ C[/math]. Ans. [math]130,700 \times 10^{-8} = 0.001307[/math] volt.
• A thermo-electric battery of 33 bismuth ([math]-9,700[/math]) and antimony ([math]+2,250[/math]) couples shows [math]1.25[/math] volts at [math]19^\circ C[/math]> mean temperature. Ans. [math]317^\circ C[/math] difference of temperature.
• Current = 11 A; wire circumference = 0.7 cm; resistance = 0.6 ohm per 100m. Ans. [math]10^\circ C[/math] rise in temperature.
• Current = 20 A; copper wire diameter = 110 mils. Ans. [math]15^\circ C[/math] rise.
• Lead wire melts at 3.1 A. Ans. [math]0.0606[/math] cm diameter.
• Copper melting at [math]1,127^\circ C[/math], resistance = 1.652. Ans. [math]0.0632[/math] cm diameter at 21 A.
• Copper strip 0.02 cm thick, 18 A current. Ans. [math]0.097+[/math] cm width.
• Same copper strip for 180 A. Ans. [math]1.0586[/math] cm width.
• Bismuth and platinum at [math]20^\circ C[/math] mean temp. Ans. [math]0.0000979[/math] volt per °C.
• German silver and copper at [math]20^\circ C[/math]. Ans. [math]0.00001555[/math] volt per °C.
• For 1 volt at [math]100^\circ C[/math] with above couple: Ans. [math]608[/math] couples.
• German silver and copper from page 190 table, [math]20^\circ C[/math]> mean: Ans. [math]0.00001464[/math] volt per °C.
• At [math]315^\circ C[/math] mean: Ans. [math]0.00003255[/math] volt per °C.
• For 1 volt at [math]600^\circ C[/math] difference: Ans. [math]53[/math] couples.
• Neutral point of German silver + copper: Ans. [math]-221^\circ C[/math]
• Neutral point of iron + zinc: Ans. [math]206^\circ C[/math]
• Neutral point of aluminum + palladium: Ans. [math]-138^\circ C[/math]
• e.m.f. of aluminum + nickel at [math]300^\circ C[/math]: Ans. [math]0.00001269[/math] volt/°C
• Voltage at [math]600^\circ C[/math] diff, same mean temp: Ans. [math]0.007614[/math] volt; [math]132[/math] couples
• Thomson effect in copper for [math]200^\circ C[/math] temp diff: Ans. [math]0.0000019[/math] volt; [math]70,870[/math] ergs/sec
• Same for German silver, [math]75^\circ C[/math] temp diff: Ans. [math]0.00000384[/math] volt; [math]119,232[/math] ergs/sec
• Same for silver, [math]98^\circ C[/math]: Ans. [math]0.00000147[/math] volt; [math]47,334[/math] ergs/sec
• Peltier effect between German silver and iron at [math]100^\circ C[/math]: Ans. [math]1,106,318[/math] ergs/sec for C.G.S. unit current
• Peltier effect for zinc + tin at [math]190^\circ C[/math]: Ans. [math]290,995[/math] ergs/sec for C.G.S. unit current

Example. The formula of sulphuric acid is H₂SO₄. Calculate the proportions of the constituent elements.

Solution. Multiply the atomic weight of each element by the number of atoms in the molecule:

• [math]H = 2 \times 1 = 2[/math]
• [math]S = 1 \times 32 = 32[/math]
• [math]O = 4 \times 16 = 64[/math]

This gives the ratio: [math]H : S : O :: 2 : 32 : 64[/math]

Example. The formula of sodium sulphate is Na₂SO₄. Calculate the proportions of the constituent elements.

Solution. Sodium : Sulphur : Oxygen :: [math]46 : 32 : 64[/math]

Example. Calculate the proportions of the constituent elements of silver nitrate, AgNO₃ ([math]N = 14[/math]).

Solution. Silver : Nitrogen : Oxygen :: [math]107.1 : 14 : 48[/math]

Chemical Saturation and Valency

When two elements combine to satisfy their chemical affinities, they are said to saturate each other.

If a single atom of one element saturates a single atom of another, their valencies are equal.

Example. The formula of copper oxide is CuO. It is saturated; thus both copper and oxygen have the same valency — they are dyads.

Example. The formula of water is H₂O. Two hydrogen atoms (monads) saturate one oxygen atom. Thus, oxygen is a dyad.

Example. Sulphur trioxide has the formula SO₃. Oxygen is a dyad, and there are 3 oxygen atoms, so sulphur must be a hexad:

[math]3 \times 2 = 6[/math]

Chemical Equivalents and Atomic Weights

The chemical equivalent of an element is:

[math]\text{Chemical Equivalent} = \frac{\text{Atomic Weight}}{\text{Valency}}[/math]

Example. Oxygen has an atomic weight of [math]16[/math] and is a dyad. Chemical equivalent = [math]16 \div 2 = 8[/math]

Example. Hydrogen has an atomic weight of [math]1[/math] and is a monad. Chemical equivalent = [math]1 \div 1 = 1[/math]

Example. Water molecule H₂O → Substitute chemical equivalents: [math]1 : 8[/math] (hydrogen : oxygen)

Example. Sulphur in different compounds:

• Hexad → [math]32 \div 6 = 5.3[/math]
• Tetrad → [math]32 \div 4 = 8[/math]
• Dyad (in H₂S) → [math]32 \div 2 = 16[/math]

Electric Decomposition (Electrolysis)

A definite amount of electricity decomposes a definite amount of substance, if that substance is electrolyzable.

The weight of an element separated is said to “correspond to” the quantity of electricity that causes its deposition.

Similarly, one can refer to the quantity of electricity required to separate a known amount of an element or compound as “corresponding to” that weight.

Relation of Chemical Equivalents to Electrolysis

Chemical equivalents express the relative weights of elements that correspond to any fixed quantity of electricity.

Thus, if a certain current deposits [math]1[/math] gram of hydrogen, it will deposit [math]8[/math] grams of oxygen, [math]107[/math] grams of silver, etc., all proportional to their chemical equivalents.

Note: Some molecules involve shared saturation (e.g., carbon atoms partially saturating each other), so chemical equivalents may not apply directly.

In the molecule C₂H₂, carbon is effectively a triad. In C₂H₆, carbon’s virtual valency is 4.

CHAPTER XI.

ELECTRO-CHEMISTRY.

Chemical Composition. — Chemical Saturation. — Chemical Equivalents and Atomic Weights. — Electric Decomposition or Electrolysis. — Relation of Chemical Equivalents to Electrolysis. — Electro-chemical Equivalents. — Thermo-electro Chemistry. — Calculation of Electro-motive Force of a Voltaic Couple. — Problems.

Chemical Composition.

The atomic weights of the elements indicate the proportions in which the elements combine, subject to Avogadro’s law. Thus, the atomic weight of hydrogen being [math]1[/math] and that of oxygen being [math]16[/math], it follows that in their combination with each other the ratio of [math]1:16[/math] or some multiple thereof must obtain. The water molecule contains two atoms of hydrogen and one atom of oxygen. The ratio follows from the atomic weights; it is [math]2:16[/math]. The atomic weight of the hydrogen is multiplied by [math]2[/math]. This example illustrates a fundamental law of chemistry, the law of multiple proportions.

The determination of the atomic weights is one of the most difficult problems of the analytic branch of chemistry, and atomic weights are subject to constant revision as new determinations are made.

Example. The formula of sulphuric acid is H₂SO₄. Calculate the proportions of the constituent elements.

Solution. Multiplying the atomic weight of each element by the number of the atoms in the molecule:

• [math]H = 2 \times 1 = 2[/math]
• [math]S = 1 \times 32 = 32[/math]
• [math]O = 4 \times 16 = 64[/math]

This gives the ratio: [math]H : S : O :: 2 : 32 : 64[/math]

Example. The formula of sodium sulphate is Na₂SO₄. Calculate the proportions of the constituent elements.

Solution. Sodium : Sulphur : Oxygen :: [math]46 : 32 : 64[/math]

Example. Calculate the proportions of the constituent elements of silver nitrate, AgNO₃ ([math]N = 14[/math]).

Solution. Silver : Nitrogen : Oxygen :: [math]107.1 : 14 : 48[/math]

Chemical Saturation.

When two elements combine so as to satisfy their chemical affinities, they are said to saturate each other. If a single atom of one element saturates a single atom of another, the two are said to have the same valency. Thus, the formula of copper oxide is CuO. It is completely saturated; therefore both copper and oxygen are of the same valency, or atomicity, as it is also called.

Valency.

If other ratios than the unitary obtain in a simple saturated molecule, it indicates that the valencies of the constituent elements differ one from the other. The formula of the water molecule is H₂O; therefore oxygen has a valency twice as great as that of hydrogen.

Elements of a valency of one are called monads; those of a valency of two, dyads; of three, triads; of four, tetrads; of five, pentads; of six, hexads; of seven, heptads; of eight, octads.

Example. The valency of hydrogen is [math]1[/math]. What is the valency of oxygen, the water molecule being a saturated one and its formula being H₂O?

Solution. As it requires two atoms of the monad hydrogen to saturate the one atom of oxygen, oxygen must be a dyad.

Example. The formula of sulphur trioxide is SO₃. What is the valency of sulphur?

Solution. As one atom of sulphur saturates three atoms of the dyad oxygen, the valency of sulphur is [math]3 \times 2 = 6[/math]; sulphur is a hexad.

The statements given above are rather illustrative than exhaustive, as other considerations apply in many cases in chemistry. It is also to be noted that in practical work the less important decimals are omitted from the atomic weights in calculations.

Chemical Equivalents and Atomic Weights.

The chemical equivalent of an element is the quotient of its atomic weight divided by its valency. It expresses the simplest ratio of the constituent elements of any simple saturated molecule of two elements, a binary molecule as it is termed.

The atomic weight of oxygen is [math]16[/math]; it is a dyad; its chemical equivalent is therefore [math]\frac{16}{2} = 8[/math]. The atomic weight of hydrogen is [math]1[/math]; it is a monad; therefore its chemical equivalent is [math]1[/math]. The molecule of water has the formula H₂O. Substituting for the constituent symbols their chemical equivalents gives [math]1 : 8[/math] as the ratio of the oxygen and hydrogen in water.

Example. Calculate the chemical equivalents of sulphur.

Solution. Sulphur in some compounds is a hexad. Its atomic weight is [math]32[/math]; its chemical equivalent is [math]\frac{32}{6} = 5.3[/math]. In other compounds, in SO₂ for example, it is a tetrad. The chemical equivalent of tetrad sulphur is [math]\frac{32}{4} = 8[/math]. Sometimes, as in H₂S, it is a dyad, when its chemical equivalent is [math]\frac{32}{2} = 16[/math].

Electric Decomposition or Electrolysis.

When a current of electricity is passed through a compound so as to electrolyze it, a definite amount of the compound is decomposed by a definite amount of electricity. For various reasons some compounds cannot be electrolyzed. The statement of the necessary requirements for electrolysis is outside the field of this book.

Assuming the possibility of the separation of any element from its compounds by electrolysis, the quantity of an element actually or potentially separable by a definite quantity of electricity will be referred to as the weight or quantity “corresponding to” the quantity of electricity in question. In the same way the quantities of electricity “corresponding to” given weights of the elements and of their compounds will be referred to.

Relation of Chemical Equivalents to Electrolysis.

Chemical equivalents are the relative weights of elements corresponding to any definite quantity of electricity.

In some molecules atoms of the same element partly saturate each other. Chemical equivalents do not apply to such. In the molecule C₂H₂ the tetrad carbon elements partly saturate each other, so that the carbon is virtually a triad. In the molecule C₂H₄, it has a virtual valency of 2.

The quantities of elements which a definite quantity of electricity will precipitate are proportional to their chemical equivalents.

Example. A certain quantity of electricity would precipitate 119 grams of silver. What weight of copper would the same quantity precipitate?

Solution. The chemical equivalents of silver and copper are respectively [math]107[/math] and [math]31.6[/math]. We then have the proportion:

[math]107 : 31.6 :: 119 : x[/math]

Solving, [math]x = 35.1[/math] grams.

Example. A current of electricity is passed through two solutions, one a solution of silver, the other a solution of an unknown salt. From the first solution 129 grams of silver are precipitated; from the other 35.1 grams of an unknown metal. What was the metal?

Solution. The ratio of the weights precipitated gives the ratio of the chemical equivalents:

[math]129 : 35.1 :: 107.1 : x[/math], which gives [math]x = 29.1[/math].

Referring to the table, this is the chemical equivalent of nickel, which therefore is the metal of the unknown solution.

Electro-chemical Equivalents.

The electro-chemical equivalent of an element is the weight in grams corresponding to, or which would be precipitated by, one C.G.S. unit of electricity. A table could be made for any desired unit.

The electro-chemical equivalents of the elements are proportional to their chemical equivalents.

Example. The electro-chemical equivalent of hydrogen is [math]0.00010438[/math]. What is the electro-chemical equivalent of sodium?

Solution. The chemical equivalent of sodium is [math]23[/math] and that of hydrogen is [math]1[/math]; therefore:

[math]0.00010438 \times 23 = 0.002401[/math]

Example. The electro-chemical equivalent of silver is [math]0.01118[/math]. Calculate the electro-chemical equivalent of zinc by direct proportion.

Solution. The chemical equivalents of silver and zinc are [math]107.1[/math] and [math]32.45[/math] respectively. Then:

[math]107.1 : 32.45 :: 0.01118 : x[/math], solving gives [math]x = 0.003387[/math].

Example. Calculate the electro-chemical equivalent of a dyad element whose atomic weight is [math]118.1[/math].

Solution. The chemical equivalent is [math]118.1 \div 2 = 59.05[/math].

Then the electro-chemical equivalent is:

[math]0.00010438 \times 59.05 = 0.006164[/math], which is the electro-chemical equivalent of stannous tin.

The reciprocal of the electro-chemical equivalent gives the number of units of electricity required to precipitate one gram of substance. For hydrogen:

[math]1 \div 0.00010438 = 9580.4[/math] C.G.S. units of electricity.

To reduce this to coulombs, multiply by [math]10[/math]. Thus, to precipitate one gram of hydrogen, [math]9,580.4[/math] coulombs are required.

To obtain the electro-chemical equivalent of any element, the electro-chemical equivalent of hydrogen may be multiplied by the chemical equivalent of the element. To obtain the reciprocal of any element, divide the reciprocal of hydrogen, [math]9,580.4[/math], by the chemical equivalent of the element.

Example. Calculate the electro-chemical equivalent and its reciprocal for the metal lead.

Solution. The electro-chemical equivalent of hydrogen is [math]0.00010438[/math]. Multiplying this by the chemical equivalent of lead, [math]102.67[/math], gives: [math]0.00010438 \times 102.67 = 0.010717[/math], the electro-chemical equivalent of lead. Dividing [math]9,580.4[/math] by [math]102.67[/math] gives: [math]9,580.4 \div 102.67 = 93.31[/math], the reciprocal for lead.

Example. In one hour a current of electricity precipitates 3.740 grams of silver. What is the strength of the current?

Solution. There are [math]3,600[/math] seconds in an hour. [math]3.740 \div 3,600 = 0.001039[/math] grams per second. One ampere precipitates [math]0.001118[/math] gram of silver per second. Then [math]0.001039 \div 0.001118 = 0.93[/math] amperes.

Example. How many grams of silver are deposited in an hour by one ampere?

Solution. One ampere-hour = [math]3,600[/math] coulombs. One coulomb deposits [math]0.001118[/math] gram. So: [math]0.001118 \times 3,600 = 4.0248[/math] grams.

Example. For how long must a current of 7 amperes be maintained to precipitate 13 grams of silver?

Solution. 1 gram of silver requires [math]89.45[/math] C.G.S. units = [math]894.5[/math] coulombs. So: [math]13 \times 894.5 = 11,628.5[/math] coulombs. Current = [math]7[/math] amperes = [math]7[/math] coulombs/sec. Time = [math]11,628.5 \div 7 = 1,661.2[/math] seconds = 27 minutes 41 seconds.

Thermo-electric Chemistry

Let [math]H[/math] be the heat in calories due to the combination of one gram of any element. Let [math]Q[/math] be the number of coulombs, and [math]z[/math] the electro-chemical equivalent. Then the mass involved is [math]Qz[/math] grams, and the heat is:

[math]QzH[/math] calories. Since one calorie = [math]0.424[/math] kilogram-meters, the mechanical energy is:

[math]0.424 QzH[/math]   (1)

But the electrical energy is [math]QE[/math] joules, and since [math]1[/math] joule = [math]1 \div 9.81[/math] kilogram-meters, this becomes:

[math]\frac{QE}{9.81}[/math]   (2)

Equating (1) and (2):

[math]\frac{QE}{9.81} = 0.424 QzH[/math]

Solving for [math]E[/math]:

[math]E = \frac{9.81 \times 0.424 \times z \times H}{1} = 4.168 zH[/math]   (4)

Calculation of Electro-motive Force of a Voltaic Couple

This formula gives the value of the e.m.f. of a reaction in terms of the electro-chemical equivalent and the calories of heat per gram. The heat data [math]H[/math] for many reactions has been experimentally determined, and so this formula enables calculation of theoretical e.m.f. for a given voltaic reaction.

Example. What e.m.f. is required to decompose water?

Solution. Take the value [math]34,450[/math] calories as the heat due to the oxidation of one gram of hydrogen, producing water. The electro-chemical equivalent of hydrogen is, for coulomb notation, [math]0.000010438[/math]. Introducing these values in equation (4) gives:

[math]E = 4.16 \times 0.000010438 \times 34,450 = 1.495[/math] volts.

These calculations are not in exact accord with direct experiment, as the heat coefficient is not always exact, and subsidiary reactions may exist which are not taken into account in the calculation.

In the action of a battery, chemical decomposition, operating to reduce the e.m.f., sometimes has a place. Then in calculating the e.m.f. of such a battery the e.m.f. of decomposition must be subtracted from the e.m.f. of combination to get the e.m.f. of the battery.

Example. Calculate the e.m.f. of the Daniell battery.

Solution. In the Daniell couple zinc is dissolved in sulphuric acid. In the solution of one gram of zinc in sulphuric acid, [math]1,670[/math] calories are set free. The electro-chemical equivalent of zinc is, for coulomb notation, [math]0.0003387[/math]. Substituting in equation (4) we have:

[math]E = 4.16 \times 0.0003387 \times 1,670 = 2.353[/math] volts.

This is the voltage due to the chemical combination of the couple. There is also a decomposition, that of the copper sulphate, from which copper is deposited. In the solution of copper in sulphuric acid [math]881[/math] calories are set free or are absorbed in its separation, the two actions being strictly reciprocal. This reciprocal relation obtains in all chemical reactions. The electro-chemical equivalent of copper is [math]0.0003293[/math]. Substituting in equation (4) we have:

[math]E = 4.16 \times 0.0003293 \times 881 = 1.207[/math] volts.

As copper is separated in this reaction heat is absorbed and the voltage is reduced. The net voltage of the couple is therefore:

[math]2.353 – 1.207 = 1.146[/math] volts.

By actual test the voltage of the Daniell couple is found to be [math]1.079[/math] volts, a discrepancy of [math]0.067[/math] volts.

In e.m.f. or polarization calculations such as those just illustrated, the calculation may be based on any of the constituents of the reaction or on the final product. The result of the calculation will be the same. To prove this we may use the reaction of hydrogen and oxygen in the formation of water.

In the reaction in question 1 part of hydrogen unites with 8 parts of oxygen to form 9 parts of water. The calories due to the reaction are [math]34,450[/math] for 1 gram of hydrogen, as we have seen. This then is the heat due to the combination of 8 grams of oxygen with hydrogen, or due to the formation of 9 grams of water. Therefore, for 1 gram of oxygen the heat is:

[math]34,450 \div 8 = 4,306[/math] calories,

and for 1 gram of water it is:

[math]34,450 \div 9 = 3,828[/math] calories.

In applying the formula, these are the quantities which have to be multiplied by the electro-chemical equivalents of oxygen or water, as the case may be.

But the electro-chemical equivalent of oxygen is equal to that of hydrogen multiplied by [math]8[/math], and that of water is equal to that of hydrogen multiplied by [math]9[/math]. The result of the multiplication of the formula is therefore the same in all the three cases. The electro-chemical equivalent of oxygen is [math]0.0000829[/math]; that of water is [math]0.0000933[/math]. The formulas for the oxygen and water basis are obtained by substituting the respective factors in (4), giving:

For oxygen, [math]E = 4.16 \times 0.000083 \times 4,306 = 1.487[/math] volts,

For water, [math]E = 4.16 \times 0.00009334 \times 3,828 = 1.487[/math] volts, which are the same as those obtained on the hydrogen basis.

The quantity of electricity required for depositing the number of grams of an element equal to its chemical equivalent is the same for all elements. It is [math]9,580.4[/math] C.G.S. units, or [math]95,804[/math] coulombs. Thus this quantity of electricity will deposit [math]107.1[/math] grams of silver, [math]29.15[/math] grams of nickel, [math]65.23[/math] grams of gold, [math]29.52[/math] grams of tin, and so on.

The e.m.f. of a couple can be determined from this factor. The energy of a cell is expressible in two ways. It can be expressed in energy units, such as ergs, or in compound electric units, each one the product of a unit of e.m.f. by a unit of quantity. These two must be equal to each other. If the quantity of electricity is known, it is obvious that the quotient of the energy unit divided by the quantity unit will give the e.m.f.

Example. Calculate the e.m.f. of the Daniell couple, using the above factor.

Solution. [math]9,580.4[/math] C.G.S. units of electricity will precipitate [math]32.45[/math] grams of zinc. The calories of heat due to the solution of 1 gram of zinc we have taken as [math]1,670[/math]. The calories due to [math]32.45 \times 1,670 = 54,191.5[/math] calories.

The calories for the corresponding figure for copper are given by [math]31.55 \times 881 = 27,796[/math] calories.

The total calories of the couple are equal to [math]54,191.5 – 27,796 = 26,395.5[/math] calories.

To reduce calories to C.G.S. units or ergs they must be multiplied by [math]4.2 \times 10^7[/math].

[math]26,395.5 \times 4.2 \times 10^7 = 11,086 \times 10^8[/math] ergs.

These ergs correspond to a definite quantity of electric energy. Of this quantity one constituent is known. This constituent is [math]9,580.4[/math] C.G.S. units of quantity. If therefore the ergs are divided by this figure, the quotient will be the other constituent or factor. This other factor is e.m.f. expressed in C.G.S. units.

[math](11,086 \times 10^8) \div 9,580.4 = 1.16 \times 10^8[/math] C.G.S. units = [math]1.16[/math] volts.

Example. Calculate the e.m.f. of the decomposition of water.

Solution. The chemical equivalent of hydrogen being [math]1[/math], the calories per equivalent are the same as those per gram, namely, [math]34,450[/math].

[math]34,450 \times 4.2 \times 10^7 = 14,469 \times 10^8[/math] ergs.

[math](14,469 \times 10^8) \div 9,580.4 = 1.51 \times 10^8[/math] C.G.S. units = [math]1.51[/math] volts.

PROBLEMS.

Express the operation of calculating the electro-chemical equivalent of cupric copper. Ans. [math]31.55 \times 0.00010438 = 0.003293[/math].

With what factor must the electro-chemical equivalent of any element be multiplied to give grams per hour? Ans. As [math]1[/math] ampere = [math]10[/math] C.G.S. unit, the factor is [math]3,600 \div 10 = 360[/math].

If [math]0.0397[/math] gram of silver is precipitated by a current, what is the quantity of electricity in C.G.S. units and in coulombs? Ans. [math]3.55[/math] C.G.S. units; [math]35.5[/math] coulombs.

Using the reciprocal of the electro-chemical equivalent of gold, [math]146.87[/math], express the operation of determining the quantity of electricity required to precipitate [math]0.0781[/math] gram of gold. Ans. [math]0.0781 \times 146.87 = 11.47[/math] C.G.S. units; [math]114.7[/math] coulombs.

How long will it take [math]3.4[/math] amperes to precipitate [math]1.751[/math] grams of cupric copper? Ans. 26 minutes 4 seconds.

A current precipitates from a solution of copper sulphate [math]0.115[/math] gram of copper; from a second solution in series with the first the same current precipitates [math]0.23[/math] gram of copper. What is the second solution? Ans. A solution of cuprous copper.

The same current precipitates from one solution [math]0.071[/math] gram of silver and from another solution in series with the first [math]0.0658[/math] gram of another metal. What is the other metal? Ans. Mercuric mercury.

What weight of gold will [math]3.75[/math] amperes deposit in [math]5[/math] hours [math]21[/math] minutes? Ans. [math]49.18[/math] grams; [math]758.94[/math] grains.

What proportion will give the relative weights of cupric copper and gold precipitated by the same current, taking gold as unity?
Ans. [math]652 : 316 :: 1 : x = 0.484[/math]. For each grain of gold, [math]0.484[/math] grain of copper will be deposited.

What weight of lead will be deposited by [math]3.9[/math] amperes in [math]10[/math] seconds?
Ans. [math]0.042[/math] gram; [math]0.648[/math] grain.

A current deposits [math]1.75[/math] grams of silver in 1 hour. Calculate its strength.
Ans. [math]0.4348[/math] ampere.

The atomic weight of a pentad element is [math]14.01[/math]. What is its chemical equivalent?
Ans. [math]14.01 \div 5 = 2.802[/math].

A current passing through a solution deposits [math]0.00231[/math] gram of cupric copper each second. The e.m.f. expended is [math]1.22[/math] volts. Calculate the power or activity.
Ans. [math]\text{Current} = 0.00231 \div 0.001159 = 1.993[/math] amperes. Power = [math]1.993 \times 1.22 = 2.43[/math] watts. (Note: Book’s answer was [math]8.55[/math] watts — a sign it may use different e.c.e. or rounding.)

A current passing through an electrolytic voltameter and then through a calorimeter deposits [math]16.1[/math] grams of silver per hour and develops [math]6,912[/math] calories. What e.m.f. was expended in the calorimeter?
Ans. One ampere-hour deposits [math]4.025[/math] grams. [math]16.1 \div 4.025 = 4[/math] amperes. [math]6,912 \div 4 = 1,728[/math] calories per ampere. [math]1,728 \div 860 = 2[/math] volts (since [math]1[/math] volt-ampere-second = [math]0.24[/math] calorie).

The current in the above case is increased so as to precipitate [math]18[/math] grams of silver per hour. What will be the effect on the calorimeter?
Ans. [math]18 \div 4.025 = 4.47[/math] amperes. [math]4.47 \times 1,728 = 7,728[/math] calories.

A current of electricity is passed through two solutions, one of silver and one of cupric copper. After a time it deposits [math]5[/math] grams of silver. How much copper will be deposited?
Ans. [math]107.1 : 31.6 :: 5 : x = 1.473[/math] grams.

CHAPTER XII.

FIELDS OF FORCE.

Fields of Force. — The Unit Field. — Intensity of Field. — Polarity. — Quantity of Field. — Lines of Force per Square Centimeter and per Square Inch. — Kapp’s Unit. — Fields of Uniform and of Varying Strength. — Radiant Fields of Force. — Reciprocal Action in Fields of Force. — Induction of E.M.F. and Current.

Fields of Force. A field of force exists in any region in which, without physical contact, force is exercised upon a mass or other physical quantity. Thus when a magnet exercises force upon a piece of iron, attracting it without any contact between the two, a magnetic field of force is shown to exist in the space between them. The force exerted in such a field may also be repulsive, as when like poles repel each other.

Every mass attracts every other mass without physical contact. This is a field of gravitational force. In electrical studies, we are mainly concerned with electro-magnetic and electrostatic fields, especially the former.

Like other physical quantities, a field has two measurable aspects — its total value and its intensity. These are typically described in terms of lines of force. Intensity refers to lines of force per unit area, such as per square inch or square centimeter.

The Unit Field. Intensity of Field. A unit field of force is one which exerts unit force on unit quantity. In the electro-magnetic case, this is one dyne on a unit magnetic pole. A field of unit intensity has [math]1[/math] line of force per square centimeter.

Another name for this unit of magnetic field strength is the gauss. So, a field of [math]1[/math] gauss = [math]1[/math] line per square centimeter.

Example. A magnet pole of strength [math]1.339[/math] units is acted on by a field with [math]39.7[/math] dynes of force. What is the field strength?

Solution. The field strength = [math]39.7 \div 1.339 = 29.65[/math] dynes per unit pole = [math]29.65[/math] gausses.

Example. What force acts on a pole of strength [math]5.5[/math] units in a [math]12[/math]-gauss field?

Solution. [math]12 \times 5.5 = 66[/math] dynes.

Polarity. A field has direction, which is shown by the direction of the force it exerts. A pivoted magnet needle aligns itself with the polarity of the field, i.e., with its lines of force. Similarly, a weight falling toward Earth indicates the direction of Earth’s gravitational field.

Quantity of Field. Multiply the intensity of a field by its cross-sectional area to get the total number of lines of force, or quantity of field. For example, a field of strength [math]5[/math] over [math]3[/math] square centimeters has [math]5 \times 3 = 15[/math] lines of force. A line of force is sometimes called a weber.

Example. A field of force of [math]19[/math] lines to the square centimeter acts upon a unit magnet pole. What is the force in dynes exerted upon the pole?

Solution. As a line of force is equal to 1 dyne acting upon unit quantity, and as the magnet pole is a unit pole, the answer is [math]19[/math] dynes.

Example. An electro-magnet attracts a magnet pole of [math]31[/math] units strength with a force of [math]51[/math] grams. What is the strength of field?

Solution. [math]51[/math] grams = [math]51 \times 981 = 50,031[/math] dynes. This is the force on a pole of [math]31[/math] units strength, so the field strength is [math]50,031 \div 31 = 1,614[/math] lines of force per square centimeter.

Example. Expressing the strength of Earth’s gravity in gravitational lines of force, how many lines of force per square centimeter would it contain?

Solution. Earth’s gravity acts on a 1-gram mass with [math]981[/math] dynes of force. So the gravitational field has [math]981[/math] lines of force per square centimeter.

Lines of Force per Square Centimeter and per Square Inch. A square inch is equal to [math]6.45[/math] square centimeters. Therefore:

• To convert lines per square centimeter to lines per square inch: multiply by [math]6.45[/math]
• To convert lines per square inch to lines per square centimeter: multiply by [math]0.155[/math]

Example. A dynamo field has [math]5,000[/math] lines per square centimeter. What is the value per square inch?

Solution. [math]5,000 \times 6.45 = 32,250[/math] lines per square inch.

Example. Reduce [math]121,300[/math] lines per square inch to lines per square centimeter.

Solution. [math]121,300 \times 0.155 = 18,801[/math] lines per square centimeter.

Kapp’s Unit. One Kapp line = [math]6,000[/math] lines per square inch = [math]930[/math] lines per square centimeter. This unit is rarely used.

Example. How many Kapp lines are there in a field of [math]10[/math] square inches with [math]3,906[/math] lines per square inch?

Solution. Total lines = [math]3,906 \times 10 = 39,060[/math]. Divide by [math]6,000[/math] gives [math]6.51[/math] Kapp lines.

Example. Calculate the Kapp lines in a field of [math]27[/math] square centimeters with [math]2,750[/math] lines per cm².

Solution. Total lines = [math]2,750 \times 27 = 74,250[/math]. Divide by [math]930[/math] gives [math]79.8[/math] Kapp lines.

Fields of Uniform and Varying Strength. A uniform field has the same intensity everywhere. Its lines of force are parallel. Earth’s gravitational field is essentially uniform over small terrestrial distances.

Example. Compare the attraction of Earth on a mass at [math]1[/math] cm and at [math]100[/math] cm height.

Solution. The field is uniform at small distances, so the force of attraction (i.e., weight) is the same at both points.

Radiant Fields of Force. A field of force established by a point source, such as a very small magnet pole, varies in strength according to the inverse square law of central forces. If two such points act upon each other, the force varies inversely with the square of the distance between them and directly with the product of their strengths.

The formula expressing this relationship is:

[math]F = \frac{Nm m’}{r^2}[/math]

where:

• [math]F[/math] is the force in dynes,
• [math]N[/math] is a constant,
• [math]m[/math] and [math]m'[/math] are the interacting quantities (masses or pole strengths),
• [math]r[/math] is the distance separating them in cm.

In gravitation, [math]N = 6.6576 \times 10^{-8}[/math] (in C.G.S. units). In magnetism, [math]N = 1[/math].

Example. A mass of 7 grams and a mass of 3 grams are 11 cm apart. What gravitational force acts between them?

Solution. Substitute into the formula:

[math]F = 6.6576 \times 10^{-8} \times \frac{7 \times 3}{11^2} = 6.6576 \times 10^{-8} \times \frac{21}{121} = 1.15 \times 10^{-8}[/math] dyne.

Reciprocal Action in Fields of Force. The force is mutual: both bodies exert the same force on each other. If either [math]m[/math] or [math]m'[/math] is zero, the force is zero.

Example. Two equal magnet poles act on each other with a force of 3.4 dynes, at 2.5 cm apart. What is the strength of each?

Solution. Since [math]m = m'[/math], the formula becomes:

[math]F = \frac{m^2}{r^2}[/math] ⇒ [math]m^2 = F \cdot r^2 = 3.4 \cdot (2.5)^2 = 21.25[/math] ⇒ [math]m = 4.6[/math]

Each pole has a strength of [math]4.6[/math] unit magnet poles.

Example. Two equally charged disks repel each other with 10.4 dynes of force. The distance between them is 3.1 cm. What is the charge on each?

Solution. Using [math]F = \frac{m^2}{r^2}[/math],

[math]m^2 = F \cdot r^2 = 10.4 \cdot (3.1)^2 = 99.944[/math] ⇒ [math]m = 10[/math]

Each disk is charged with [math]10[/math] electrostatic units.

Example. One of the above disks (with 10 units of charge) is placed 3.4 cm from another charged disk. It repels the second disk with a force of 2.1 dynes. What is the charge on the second disk?

Solution. Use [math]F = \frac{m m’}{r^2}[/math]. Let [math]m = 10[/math], [math]r = 3.4[/math], [math]F = 2.1[/math]:

[math]m’ = \frac{F \cdot r^2}{m} = \frac{2.1 \cdot (3.4)^2}{10} = \frac{2.1 \cdot 11.56}{10} = 2.43[/math]

The second disk has a charge of [math]2.43[/math] electrostatic units.

Example. A magnet pole attracts another at 1 cm distance with a force of 3 dynes. One pole is a unit pole. What is the field strength at 1 cm from the other pole?

Solution. Since field strength is force per unit pole, and a unit pole is acted on with 3 dynes, the field strength is [math]3[/math] lines of force per square cm.

Induction of E.M.F. and Current. If a unit length of conductor is moved at unit speed across a unit field, a unit potential difference (or e.m.f.) will be generated between its ends. The conductor must move at right angles to the field’s polarity.

Example. An electromagnetic field has [math]21[/math] lines of force per cm². A wire [math]121[/math] cm long is moved across it at [math]32[/math] cm/sec. What is the induced e.m.f.?

Solution. In a unit field, the e.m.f. would be:

[math]121 \times 32 = 3,872[/math] C.G.S. units.

Since the field has [math]21[/math] lines/cm²:

[math]3,872 \times 21 = 81,312[/math] C.G.S. units

This is equal to: [math]0.000813[/math] volt.

Induced current results when a circuit is closed and field quantity changes. The induced e.m.f. is proportional to the rate of change of flux (lines of force) within the circuit.

Example. A wire loop forms a circle 1 meter in diameter. A field of [math]257[/math] lines/cm² is reduced to [math]61[/math] in [math]9[/math] seconds. What is the e.m.f.?

Solution. Area of the circle is:

[math]A = \pi r^2 = \pi (50)^2 \approx 7,854[/math] cm²

Rate of change in field: [math]\frac{257 – 61}{9} = 21.78[/math] lines/cm²/sec

e.m.f. = [math]7,854 \times 21.78 = 171,060[/math] C.G.S. units = [math]0.00171[/math] volt

Note: 1 e.m. unit = 1 line of force/sec. 1 volt = [math]10^8[/math] lines/sec

Example. A wire 11 cm long moves at 17 cm/sec across a field of 39 lines/cm². What is the e.m.f.?

Solution.

[math]E = 11 \times 17 \times 39 = 7,293[/math] C.G.S. units = [math]0.00007293[/math] volt

Example. A train moves at 30 mi/hr on rails 4 ft 8 in apart. The vertical component of Earth’s magnetic field is [math]0.438[/math] lines/cm². What is the e.m.f. induced in the axle?

Solution.

Convert speed: 30 mi/hr = [math]1,341[/math] cm/sec

Gauge = [math]56[/math] in = [math]142.24[/math] cm

Area swept/sec = [math]1,341 \times 142.24 = 190,743[/math] cm²

Lines cut/sec = [math]190,743 \times 0.438 = 83,545[/math] C.G.S. units

Voltage: [math]E = \frac{83,545}{10^8} = 0.000835[/math] volt

Dynamo and Motor Working Fields

A dynamo’s or motor’s working field is the region swept by the armature conductors. In a two-pole machine with a drum armature, each conductor cuts the field twice per revolution, but due to the parallel winding configuration, the effective e.m.f. is based on one pass per revolution.

The e.m.f. in such machines is given by:

[math]\text{e.m.f. (volts)} = \frac{nSN}{10^8}[/math]

• [math]n[/math]: revolutions per second
• [math]S[/math]: number of conductors in series
• [math]N[/math]: lines of force cut per revolution

For a multipolar machine (with fields taken per pole pair), use:

[math]\text{e.m.f. (volts)} = \frac{pnSN}{10^8}[/math]

• [math]p[/math]: number of pole pairs

Example 1

A two-pole dynamo has [math]7.17 \times 10^6[/math] lines of force, [math]S = 120[/math] conductors, and rotates at [math]780[/math] RPM ([math]13[/math] rev/sec). Find the e.m.f.:

Solution:

[math]\text{e.m.f.} = \frac{13 \times 120 \times 7.17 \times 10^6}{10^8} = 111.7[/math] volts

Example 2

An 8-pole machine with [math]9 \times 10^6[/math] lines per pole, [math]90[/math] conductors in series, and running at [math]20[/math] rev/sec. Find the e.m.f.:

Solution: There are [math]p = 4[/math] pole pairs.

[math]\text{e.m.f.} = \frac{4 \times 20 \times 90 \times 9 \times 10^6}{10^8} = 648[/math] volts

Problems

• A field of area [math]363[/math] cm² contains [math]76,321[/math] lines. Find its intensity:
  Ans: [math]\frac{76,321}{363} = 210[/math] gausses
• A field has strength [math]1,100[/math] gausses and area [math]36[/math] cm². Find its total flux:
  Ans: [math]1,100 \times 36 = 39,600[/math] lines
• A field has strength [math]2,700[/math] gausses and total flux [math]297,000[/math] webers. Find area:
  Ans: [math]\frac{297,000}{2,700} = 110[/math] cm²
• Convert [math]4,000[/math] lines/cm² to lines/in²:
  Ans: [math]4,000 \times 6.45 = 25,800[/math] lines/in²
• A field has [math]2,000[/math] lines over [math]5[/math] cm². What is its force on a unit pole?
  Ans: [math]\frac{2,000}{5} = 400[/math] dynes
  = [math]\frac{400}{981} \approx 0.408[/math] gram-force
  = [math]0.408 \times 15.43 = 6.30[/math] grains

Worked Problems: Magnetic Field and Electromagnetic Induction

• An electro-magnet attracts a magnet pole of [math]29[/math] units strength with a force of [math]5[/math] grams. What is the strength of the field?
  Ans: [math]\frac{5 \times 981}{29} = 169[/math] lines/cm²; [math]169 \times 6.45 = 1,090[/math] lines/in²
• Two equal magnet poles attract each other with a force of [math]0.005[/math] gram at [math]0.5[/math] cm apart. What is the strength of each?
  Ans: [math]F = \frac{m^2}{r^2} \Rightarrow m = \sqrt{0.005 \times 981 \times 0.5^2} = 1.107[/math] unit poles
• Two disks attract each other with [math]0.003[/math] gram-force at [math]0.7[/math] cm. What is the static charge on each?
  Ans: [math]m = \sqrt{0.003 \times 981 \times 0.7^2} = 1.208[/math] electrostatic units
• A magnet pole of [math]6[/math] units strength attracts another with [math]0.0414[/math] gram-force at [math]2[/math] cm. What is the strength of the second magnet?
  Ans: [math]m = \frac{F \times r^2}{6} = \frac{0.0414 \times 981 \times 2^2}{6} = 27.07[/math] units
• A magnet of [math]7.0[/math] unit poles is attracted at [math]1.5[/math] cm by another with a force of [math]0.31[/math] gram. What is the field at [math]1[/math] cm?
  Ans: [math]F = \frac{m_1 m_2}{r^2} \Rightarrow \text{Field} = \frac{0.31 \times 981}{7} = 97.75[/math] lines/cm²; [math]97.75 \times 6.45 = 630.5[/math] lines/in²

Electromagnetic Induction in Dynamos

• A two-pole dynamo has [math]9,000,000[/math] lines of force and [math]140[/math] conductors. At [math]1,200[/math] RPM, what is the e.m.f.?
  Ans: [math]n = 20[/math] rev/sec; [math]e = \frac{20 \times 140 \times 9 \times 10^6}{10^8} = 252[/math] volts
• A two-pole dynamo has pole faces of [math]36[/math] in² and induction of [math]65,000[/math] lines/in²; it has [math]200[/math] turns and spins at [math]1,500[/math] RPM. Find the e.m.f.
  Ans: [math]N = 36 \times 65,000 = 2.34 \times 10^6[/math]; [math]n = 25[/math] rev/sec
  [math]e = \frac{25 \times 200 \times 2.34 \times 10^6}{10^8} = 117[/math] volts
• An 8-pole dynamo has [math]120,000[/math] lines/in² and pole area of [math]48[/math] in². There are [math]2,400[/math] conductors in [math]8[/math] parallel groups. The speed is [math]10[/math] rev/sec. Find the e.m.f.
  Ans: [math]N = 120,000 \times 48 = 5.76 \times 10^6[/math] per pole
  [math]S = \frac{2,400}{8} = 300[/math]; [math]p = 4[/math]
  [math]e = \frac{4 \times 10 \times 300 \times 5.76 \times 10^6}{10^8} = 691.2[/math] volts
• A two-pole machine has pole faces of [math]9[/math] in² each and induction of [math]100,000[/math] lines/in². At [math]25[/math] rev/sec, find required conductors for [math]200[/math] volts.
  Ans: [math]N = 9 \times 100,000 = 9 \times 10^5[/math];
  [math]200 = \frac{25 \times S \times 9 \times 10^5}{10^8} \Rightarrow S = \frac{2 \times 10^8}{25 \times 9 \times 10^5} = 889[/math] conductors

Chapter XIII. Magnetism

• Moments
• Lever Arm of a Force
• The Couple
• Unit of Moment
• The Magnetic Filament
• Lines of Force in a Filament
• Magnetic Quantity and Strength of Pole
• Measure of Magnetic Quantity
• Lines of Force produced by Unit Quantity of Magnetism
• Moment of a Magnet
• Intensity of Magnetization
• Turning Moment of a Magnet
• Magnetic Traction
• Problems

Moments

If a bar or lever is pivoted so it can rotate about a point, and a force is applied at a distance from that point, the turning effect of that force is called a moment.

Definition: The moment of a force is the product of the force and the perpendicular distance from the pivot (center of rotation) to the line of action of the force. This perpendicular distance is called the lever arm.

[math]\text{Moment} = \text{Force} \times \text{Lever Arm}[/math]

Lever Arm of a Force

The lever arm is the perpendicular from the pivot to the line (or extension) along which the force acts. This determines how effectively the force causes rotation.

The Couple

A couple consists of two equal and opposite forces whose lines of action do not coincide but act to produce rotation. The distance between their lines of action is the effective lever arm.

In magnetism, the two opposite poles of a magnet form a couple when the magnet is mounted to turn freely about its center. The forces due to each pole act in opposite directions, creating a net torque that tends to align the magnet with the external magnetic field.

Moment of a couple:
[math]\text{Moment} = \text{Force} \times \text{Distance between forces}[/math]

Unit of Moment

The unit of moment is defined as the moment produced by a force of 1 dyne acting with a lever arm of 1 centimeter. Any combination of force and lever arm whose product equals 1 dyne-centimeter represents 1 unit of moment.

For example:

• 1 dyne × 10 cm = 10 units
• 10 dynes × 1 cm = 10 units

The Magnetic Filament

A magnetic filament is an idealized concept: a slender, perfectly magnetized strand of material where all magnetic lines are confined within, with no leakage from the sides. The poles are so distant from one another that their mutual influence is negligible.

This approximates the behavior of the central region of long, slender magnets, where the magnetization is effectively linear and confined.

Lines of Force in a Filament

If an isolated magnet pole were possible, it would radiate magnetic lines of force in all directions uniformly, forming a spherical field. The number of lines through the surface of a unit sphere gives the total field of the pole.

Since the surface area of a sphere of radius 1 cm is [math]4\pi = 12.5664[/math] cm², and the field intensity is defined as the number of lines per cm² at 1 cm distance, the total lines of force is:

[math]\text{Total lines} = 4\pi \times \text{field at 1 cm}[/math]

Example 1

A pole establishes a field of 10 lines/cm² at 1 cm. How many lines emerge?

Solution: [math]10 \times 12.5664 = 125.66[/math] lines

Example 2

A pole produces 9 lines/cm² at 3 cm. Find the total lines emerging from it.

Solution:

• At unit distance: [math]9 \times 3^2 = 81[/math] lines/cm²
• Total lines: [math]81 \times 12.5664 = 1,017.88[/math]

|  4   0  -2 |
| -2  -2   6 |  = -64
|  0   4  -2 |

|  2   0  -2 |
|  0  -2   6 |  = -32
| -2   4  -2 |

|  2  -1   0  -1   0 |
| -1   5  -1  -1   0 |
|  0  -1   3   0  -1 |
| -1  -1   0   3  -1 |
|  0   0  -1  -1   3 |

|  5  -1  -1   0 |
| -1   3   0  -1 |
| -1   0   3  -1 |
|  0  -1  -1   3 |

| g   p     -q |
| -(g+r+s)  r  -s |
| s   b   q+s |

| 0   b   -q |
| r   r   -s |
| s   b   q+s |