A current of 1A flows through a coil when it is connected across a DC . . .

Question : A current of 1A flows through a coil when it is connected across a DC battery of 100V. If DC battery is replaced by an AC source of 100 V and angular frequency 100 rad/s, the current reduces to 0.5 A. Find 
(i) impedance of the circuit
(ii) self-inductance of the coil.
(iii) phase difference between the voltage and the current

CBSE 2023 SET-2

Solution : 

In DC Circuit

I=1A

V=100V

Resistance of the inductor(R) = V/I = 100/1=100Ω

In AC Circuit

V=100V

ω=100 rad/s

I=0.5 A

(i) Impedance of the circuit (Z) = V/I = 100/0.5 = 200Ω

(ii) We know 

z²=X_L²+R²

X_L²=z²-R²

X_L²=(200)²-(100)²

X_L²=40000-10000

X_L²=30000

X_L=√30000

X_L=100√3

ωL=100√3

100L=100√3

L=√3 H

(iii) 

Phase Difference (φ) 

tanφ=X_L/R=ωL/R=100√3/100 = √3

φ=60°