14. Referring again to water flowing in a pipe, though the quantity of water which passes is the same at any cross-section of the pipe, the pressure per square inch is not the same. Even in the case of a horizontal pipe of the same diameter throughout, the water when flowing suffers a loss of head or pressure. It is this difference of pressure that causes the water to flow between two points against the friction of the pipe.

This is precisely similar to a current of electricity flowing through a conductor. Though the quantity of electricity that flows is equal at all cross-sections, the electromotive force is by no means the same at all points along the conductor. It suffers a loss or drop of electrical potential in the direction in which the current is flowing, and it is this difference of electrical potential that causes the electricity to flow against the resistance of the conductor. Ohm's law not only gives the strength of the current in a closed circuit, but also the difference of potential in volts along that circuit. The difference of potential (e) in volts between any two points along a circuit is equal to the product of the strength of the current (C) in amperes and the resistance (R) in ohms of that part of the circuit between those two points, or e = CR; e also represents the loss or drop of potential in volts between the two points. If any two of these quantities are known, the third can be readily found; for, by transposing,

C = e and R = e .

R C

Example

Fig. 4 represents part of a circuit in which a current of 3 amperes is flowing. The resistance from a to b is 1.5 ohms; from b to c is 2.3 ohms; and from c to d is 3.6 ohms. Find the difference of potential between a and b, b and c, c and d, and between a and d.

Solution

We apply the formula e = CR derived from Ohm's law.

The difference of potential between a and b = 3x1.5 = 4.5 volts.

The difference of potential between b and c = 3x2.3 = 6.9 volts.

The difference of potential between c and d = 3 X 3.6 = 10.8 volts.

The difference of potential between a and d = 4.5 + 6.9 + 10.8 = 22.2 volts; or, in other words, the loss or drop of potential caused by a current of 3 amperes flowing between a and d is 22.2 volts.

15. In a great many cases, it is desirable to have the current flow from the source a long distance to feed lamps or motors, and return without causing an excessive drop or loss of potential in the conductors leading to and from the two places. In such circuits, the greater part of the total generated electromotive force is expended in those lamps or motors, and only a small fraction of it is lost in the rest of the circuit. Under these conditions, it is customary to decide upon a certain drop or loss of potential beforehand; and from that and the current, calculate the resistance of the two conductors. For this purpose the formula

Solution 441

Fig. 4.

R = e (Art. 14)

C may be used.

Examfle

It is desired to transmit a current of 5 amperes to a motor situated 500 feet from the source; the total generated E. M. F. is 120 volts, and only 1/10 of this potential is to be lost in the conductors leading to and from the motor, (a) Find the resistance of the two conductors, and (b) the resistance per foot of the conductors, assuming each to be 500 feet long.

Solution

(a) 1/10 of 120 volts = 12 volts, which represent the drop or loss of potential on the two conductors. Let e = 12 volts; C = 5 amperes: and R = the total resistance of the two conductors. Then.

R = e = 12 = 2.4 ohms. Ans.

(b) The resistance per foot of any conductor is found by dividing the total resistance of a conductor by its length in feet; hence, since 2.4 ohms is the resistance of two conductors, each 500 feet long, then the resistance per foot of the conductors =

2.4 = .0024 ohm. Ans.

2 x 500