Question

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 :

• A. zero
• B. 0.5
• C. 1.0
• D. -1.0

Answer 1

Answer: (C)

When L is removed,
$tan\,\phi=\frac{\left|X_{C}\right|}{R}$
$\Rightarrow tan \frac{\pi}{3}=\frac{X_{c}}{R} \ldots\left(i\right)$
When C is removed,
$tan \,\phi=\frac{\left|X_{L}\right|}{R}$
$\Rightarrow tan \frac{\pi}{3}=\frac{X_{L}}{R} \ldots\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
Since $X_{L}=X_{C}$, the circuit is in resonance
$Z=R$
Power factor $=cos\,\phi=\frac{R}{Z}=1$