Question

A pure inductive coil having inductance 0.1henry is connected across 200V,50Hz supply. Write the equation for voltage and current

Answer 1

Answer:

The frequency of AC source =f=50Hz

Thus the impedance of the inductor, X

L

=ωL=(2πf)L=20πΩ

Resistance of the resistor =20Ω

Thus, the net impedance of the circuit =

X

L

2

+R

2

=65.94Ω

Thus, the current in the circuit =

X

V

=

65.94Ω

220V

=3.33A