Hawkins Electrical Guide v. 05 (of 10) / Questions, Answers, & Illustrations, A progressive course of study for engineers, electricians, students and those desiring to acquire a working knowledge of electricity and its applications

by N. Hawkins

X_i = 2πfL = 2 × 3.1416 × 60 × .5 = 188.5 ohms

The quantity 2πfL or reactance being of the same nature as a resistance is used in the same way as a resistance. Accordingly, since, by Ohm's law

E = RI (1)

an expression may be obtained for the volts necessary to overcome reactance by substituting in equation (1) the value of reactance given above, thus

E = 2πfLI (2)

Fig. 1,295.—Diagram of circuit for example II. As in example I, resistance is disregarded.

EXAMPLE II.—How many volts are necessary to force a current of 3 amperes with frequency 60 through a coil whose inductance is .5 henry? Substituting in equation (2) the values here given

E = 2πfLI = 2π × 60 × .5 × 3 = 565 volts.

The foregoing example may serve to illustrate the difference in behaviour of direct and alternating currents. As calculated, it requires 565 volts to pass only 3 amperes of alternating current through the coil on account of the considerable spurious resistance. The ohmic resistance of a coil is very small, as compared with the spurious resistance, say 2 ohms. Then by Ohm's law I = E ÷ R = 565 ÷ 2 = 282.5 amperes.