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

Z = √(R² + (2πfL)²) (2)

Fig. 1,298.—Diagram of circuit for example II.

EXAMPLE II.—If an alternating current, having a frequency of 60, be impressed on a coil whose inductance is .05 henry and whose resistance is 6 ohms, what is the impedance?

Here R = 6; f = 60, and L = .05; substituting these values in (2)

Z = √(6² + (2π × 60 × .05)²) = √(393) = 19.8 ohms.

Fig. 1,299.—Diagram of circuit for example III.

EXAMPLE III.—If an alternating current, having a frequency of 60, be impressed on a circuit whose inductance is .05 henry, and whose capacity reactance is 10 ohms, what is the impedance?

X_i = 2πfL = 2 × 3.1416 × 60 × .05 = 18.85 ohms