In a series LCR circuit an ac voltage source is connected. When inductor is removed from the circuit, the phase difference between the voltage and the current in the circuit is \( \displaystyle\frac{\pi}{3} \). If capacitor is removed from the circuit instead of inductor, the phase difference is again \( \displaystyle\frac{\pi}{3} \). What is the power factor of the circuit?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1
In a series circuit the phase difference between voltage and current is given by \( \tan\phi = X/R \), and the power factor is \( \cos\phi = R/Z \).
With the inductor removed (RC circuit):
\( \displaystyle \tan\frac{\pi}{3} = \frac{X_C}{R} \;\Rightarrow\; X_C = \sqrt{3}\,R \)
With the capacitor removed (RL circuit):
\( \displaystyle \tan\frac{\pi}{3} = \frac{X_L}{R} \;\Rightarrow\; X_L = \sqrt{3}\,R \)
So \( X_L = X_C \) in the full LCR circuit: the circuit is at resonance, and the reactances cancel.
\( \displaystyle Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = R \;\Rightarrow\; \cos\phi = \frac{R}{Z} = 1 \)
Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).