# A Textbook of Electric Lighting and Railways Volume 1

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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1901 edition. Excerpt: ...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 dzference 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', which is an example of the use of formula 8. 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, I I by transposing, C: % and R' = %, as already given in formulas 6 and 7. ExAMPLE.--Fig. 50 represents part 0f'a circuit in which a current of a, b 0 d 3 amperes is flowing. The re"'."," ""__ sistance from a to b is 1.5 ohms, FIG. 50. ' ' from b to c is 2.3 ohms, and from c to d is 36 ohms. Find the difference of potential between a and b, b and c, c and d, and a and d. SoLu'rIoN.--Since, by formula 8, E' = C R', then, 65. In a great many cases it is desirable to have the current flow from the source a long distance to some electrical receptive device and...