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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
0.5
B
1
C
-1
D
zero
Detailed Solution
When L is removed: $\tan\frac{\pi}{3} = \frac{X_C}{R}$ ...(i)
When C is removed: $\tan\frac{\pi}{3} = \frac{X_L}{R}$ ...(ii)
From (i) and (ii), $X_L = X_C$, so the circuit is at resonance and Z = R.
Power factor $= \cos\phi = \frac{R}{Z} = 1$