Question

An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 47.2 H inductor, determine the inductive reactance and the rms current through the inductor.

Answer 1

The inductance is 17784.96 ohm and rms current is 4.77 mA.

Step-by-step explanation:

Voltage, V = 120 V

frequency, f = 60 Hz

Inductance, L = 47.2 H

The rms voltage is

`V_(rms)=(V_o)/(\sqrt 2)\\\\V_(rms)=(120)/(\sqrt 2)\\\\V_(rms) = 84.87 V`

The reactance is given by

`X_L = 2\pi f L\\\\X_L = 2* 3.14* 60* 47.2 \\\\X_L = 17784.96 ohm`

The rms current is

`I_(rms) =(V_(rms))/(X_L)\\\\I_(rms)=(84.87)/(17784.96)\\\\I_(rms) = 4.77* 10^(-3) A = 4.77 mA`