Alternating Current - Result Question 11
11. 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
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Answer:
Correct Answer: 11. (b)
Solution:
- (b) When $L$ is removed,
Phase difference
$ \begin{equation*} \tan \phi=\frac{|X_C|}{R}=\tan \frac{\pi}{3}=\frac{X_C}{R} \tag{1} \end{equation*} $
When $C$ is removed,
Phase difference
$ \begin{equation*} \tan \phi=\frac{|X_L|}{R}=\tan \frac{\pi}{3}=\frac{X_L}{R} \tag{2} \end{equation*} $
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, $\cos \phi=\frac{R}{Z}=1$.
