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

Fig. 1,318.—Diagram for obtaining ohmic drop in the circuit fig. 1,316 when impressed pressure and reactance drop are given. Lay off BC to scale = reactance drop; draw AB at right angle and of indefinite length; with C as center and radius of length = impressed pressure, describe an arc cutting ohmic drop line at A, then AB = ohmic drop = 5 volts by measurement.

Fig. 1,319.—Graphical method of finding angle of lag when the ohmic drop and reactance drop are given. The angle of lag φ, is that angle included between the impressed pressure and the ohmic drop lines, that is, between AC and AB.

EXAMPLE.—An alternating current of 10 amperes having a frequency of 60, is impressed on a circuit containing a resistance of 5 ohms and an inductance of 15 milli-henrys. What is the impressed pressure?

The active pressure or ohmic drop is 5 × 10 = 50 volts.

Fig. 1,320.—Diagram of circuit containing 5 ohms resistance, 15 milli-henrys inductance, with 3 ampere 60 frequency current.

The inductance reactance or X_i = is 2 × 3.1416 × 60 × .015 = 5.66 ohms. Substituting this and the current value 10 amperes in the formula for inductance pressure or reactance drop (equation 2 on page [1,077]) gives E_i = 5.65 × 10 = 56.5 volts.