Alternating Current Ques 5
5. A series $LCR$ circuit is connected to an ac voltage source. When $L$ is removed from the circuit, the phase difference between current and voltage is $\frac{\pi}{3}$. If instead $C$ is removed from the circuit, the phase difference is again $\frac{\pi}{3}$ between current and voltage. The power factor of the circuit is :
[2020]
(a) $0.5$
(b) $1.0$
(c) $-1.0$
(d) zero
Show Answer
Answer:
Correct Answer: 5.(b)
Solution:
- (b) When $L$ is removed,
Phase difference
$\tan \phi=\frac{|X_C|}{R}=\tan \frac{\pi}{3}=\frac{X_C}{R} $ $\quad$ …….(1)
When $C$ is removed,
Phase difference
$\tan \phi=\frac{|X_L|}{R}=\tan \frac{\pi}{3}=\frac{X_L}{R}$ $\quad$ …….(2)
From eqs. (1) and (2), $X_L=X_C$
Since, $X_L=X_C$, the circuit is in resonance.
In this case, $Z=R$
$\therefore$ Power factor, $\cos \phi=\frac{R}{Z}=1$.
BETA
