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In an electrical circuit R, L, C and an a.c. voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\frac{\pi}{3}$. If instead, C is removed from the circuit, the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit is
A
$\frac{\sqrt3}{2}$
B
$\frac{1}{2}$
C
$\frac{1}{\sqrt2}$
D
1
Detailed Solution
With L removed (RC circuit): $\tan\frac{\pi}{3} = \frac{X_C}{R}$
With C removed (RL circuit): $\tan\frac{\pi}{3} = \frac{X_L}{R}$
So $X_L = X_C$: the full LCR circuit is at resonance.
$\tan\phi = \frac{X_L - X_C}{R} = 0 \Rightarrow \phi = 0$
Power factor $= \cos\phi = 1$
With C removed (RL circuit): $\tan\frac{\pi}{3} = \frac{X_L}{R}$
So $X_L = X_C$: the full LCR circuit is at resonance.
$\tan\phi = \frac{X_L - X_C}{R} = 0 \Rightarrow \phi = 0$
Power factor $= \cos\phi = 1$