current is flowing, the quantity of electricity is multiplied by the difference of potential of the two terminals of the thing in which the work is done.

Obviously this method of computing work applies only to a part of a circuit, not to a complete one (a little thought will show one that the term difference of potential has no definite meaning when applied to a complete circuit).

In computing the electrical energy which has been converted into other forms of energy in a complete circuit, the term electromotive force, or E.M.F., is used.

The electromotive force of a circuit is numerically equal to the work* done when a unit quantity of electricity is carried around the complete circuit.

If the work done in carrying 1 unit of electricity is equal to the electromotive force, then the work done in carrying Q units is Work = Q × E.M.F.

(24)

The electromotive force is measured in volts; hence the work in a complete circuit may be indicated by the equation

or

Joules = coulombs × volts

= amperes X seconds X volts.

If both sides of equation (24) are divided by the time t,

(25)

(26)

(27)

Equation (23), in practical units, applies:

Watts = amperes X volts.

In any circuit there must always be some source of energy, usually located in a cell or in a dynamo; and since the electromotive force is used in computing the total energy supplied, it is customary to associate the electromotive force with this source of energy. Hence the electromotive force is said to be in the cell or in the dynamo. For example, the electromotive force of a dry cell is given as 1.5 volts. What is meant by this statement is that the work this cell can do in sending a unit charge (1 coulomb) around the entire circuit is equal to 1.5 joules. If the cell produces a current, I, which flows for t seconds, the quantity, or charge, is It, and the work done by the cell is equal to the electromotive force times the quantity, or 1.5 It.

Electromotive force has a misleading name, for it is not a force but a work factor; when multiplied by quantity, it gives work. If it were a force, it would have to be multiplied by a distance to give work. The name had its origin in the analogy of an electric current to the flow of a liquid. To keep water in motion a mechanical force is necessary, and in the case of electricity the electromotive force was taken to be the moving force. While the name is not accurate, the beginner often finds the point of view helpful. 423. Summary. Current of electricity is the rate at which quantity of electricity passes a given place.

If the current is uniform, the quantity Q passing in a time t is, by equation (10),

Q = It;

if the flow is not uniform, this equation is true if I is the average current. A summary of the facts regarding potential and potential differences is given in section 415. The difference of potential between any two points is equal to the work done in transferring unit quantity of electricity from the point of lower potential to that of higher potential. The work, W, done when a quantity, Q, is transferred from one place to another, through a difference of potential of V2 - V1, is, by equation (18), W = Q (V2-V1). In practical units

2

Joules = coulombs × volts.

2

When a current flows through conductors of any shape and made of any kind of material, there is always a loss, or fall, of potential. The resistance of the conductor is so defined that, by equation (12),

Fall of potential = current × resistance.

In practical units, by equation (13),

Loss in volts = amperes × ohms.

This equation is one form of Ohm's law.

Experiments show that the resistances of conductors made of the com

mon metals are independent of the size of the current.

The electromotive force of a circuit is equal to the energy supplied to the circuit when unit quantity of electricity flows around the complete circuit. When Q units flow, the work done is, by equation (24),

where V is the difference of potential between the terminals of that part of the circuit.

In a complete circuit, by equation (27),

Power = 1 x E. M. F.

and

Work = 1 x E. M. F. xt.

In general, by equation (23),

Watts = amperes × volts.

Joules = watt-seconds = amperes × volts X seconds

Watt-hours = watts × hours

(28)

(29)

4. The resistance of an electric toaster is 20 ohms. What will be the

current when it is connected to two wires having a difference of potential of 106 volts?

2. An incandescent lamp has a current of 0.5 ampere when the voltage across it is 100 volts. What is its resistance?

3. A copper trolley wire has a resistance of 0.5 ohm per mile. A street car 2 mi. from the power house is using 80 amperes. What voltage is lost on the line ?

4. 60 street lamps are connected in series. Each lamp requires 6 amperes and has a resistance of 3 ohms. (a) How much voltage is required for each lamp? (6) How much for all the lamps?

5. A current of 6 amperes flows for 4 min. through a resistance of 12 ohms. Compute (a) the applied voltage and (b) the total energy supplied in joules.

6. A house is connected to a power circuit in the alley by a line having a resistance of 0.5 ohms. If the voltage in the alley is 108 volts, what will it be in the house when a current of 10 amperes is flowing into the house?

7. A house is connected to a power circuit by 400 ft. of wire which has a resistance of 1.6 ohms per 1000 ft. What is the voltage drop in the wires (a) if the current is 5 amperes? (b) if the current is 10 amperes?

8. A headlight lamp on an automobile has 3 amperes at 6 volts. How many watts does it use?

9. (a) Find the resistance and the normal current for a lamp labeled $110 volts, 25 watts"; (b) for one labeled "60 volts, 25 watts"; (c) for one labeled "110 volts, 100 watts."

10. An electric toaster has a resistance of 20 ohms. How much power does it use when connected to a 100-volt circuit?

11. An incandescent lamp at 100 volts takes 0.5 ampere. Find the cost per hour of running this if the rate is 12 & per kilowatt-hour.

12. Find the cost per hour of running an electric stove which is labeled "8 amperes, 110 volts" if the charge is 10¢ per kilowatt-hour.

13. An electric fan running on a 100-volt circuit uses 40 watts of power. At 10 per kilowatt-hour, how much does it cost to run the fan 8 hr.?

14. A dynamo, E. M. F. of 125 volts, is connected to a distant motor by wires having a resistance of 0.7 ohm. If the current is 40 amperes, (a) how much is the voltage drop along the wires? (b) what is the total power developed? (c) what is the power wasted on the wires?

CHAPTER XXIX

HEATING EFFECTS; JOULE'S LAW

The electric heater, 424. Electric fuses, 425. The incandescent lamp, 426. Joule's law, 427. Determination of the mechanical equivalent of heat by the use of Joule's law, 428. A theoretical proof of Joule's law, 429. A special case of electric heating; constant voltage, 430. Losses along transmission lines, 431. Why electrical energy is transmitted at high potentials, 432. Summary, 433.

424. The electric heater. The heating effects of an electric current are well known. In the ordinary incandescent lamp the current raises the temperature of the filament to "white" heat. Electric irons, toasters, and other heating devices are now common household appliances. While it is desirable in many cases for a current to develop heat, yet there are instances where this heating effect is a source of trouble and great danger. For example, the overheating of electric motors sometimes results in the destruction of the insulation of the wires, thus rendering the whole machine worthless. The principles that will be explained in this chapter apply to all cases: to those where heating is desirable, as well as to those where it is to be eliminated as far as possible.

Coils of wire are used as conductors in some types of electric heaters, and in others long, flat, metallic ribbons. As these have to be heated by the current to a bright red, they must have a rather high resistance; that is, they must be poor conductors, for good conductors are not so readily heated. In addition, the metal which is heated by the current must be one that is capable of being heated to a high temperature without oxidizing, or suffering some other form of deterioration. There are several kinds of alloys on the market that are excellent for heating-coils; for example, the nickel-chromium alloys are well suited for coils which are to be exposed to the air. Different alloys