Question

A coil has resistance of 20 W and inductance of 0.35 H. Compute its reactance and its impedance to an alternating current of 25 cycles/s.

Answer 1

Answer:

• Reactance of the coil is 55 W

• Impedance of the coil is 59 W

Explanation:

Given;

Resistance of the coil, R = 20 W

Inductance of the coil, L = 0.35 H

Frequency of the alternating current, F = 25 cycle/s

Reactance of the coil is calculated as;

$X_L=$ 2πFL

Substitute in the given values and calculate the reactance $(X_L)$

$X_L =$ 2π(25)(0.35)

$X_L$ = 55 W

Impedance of the coil is calculated as;

$Z = \sqrt{R^2 + X_L^2} \\\\Z = \sqrt{20^2 + 55^2} \\\\Z = 59 \ W$

Therefore, the reactance of the coil is 55 W and Impedance of the coil is 59 W