In a series LCR ac circuit, if only L is removed, then the phase difference between the voltage and the current is \( \displaystyle\frac{\pi}{3} \). If only C is removed, the phase difference is again \( \displaystyle\frac{\pi}{3} \). The power factor of the circuit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1 \)
In a series circuit the phase difference between voltage and current is given by \( \tan\phi = (X_L - X_C)/R \). Use the two partial circuits to compare the reactances with R.
With L removed (an RC circuit):
\( \displaystyle \tan\frac{\pi}{3} = \frac{X_C}{R} \;\Rightarrow\; X_C = \sqrt{3}\,R \)
With C removed (an RL circuit):
\( \displaystyle \tan\frac{\pi}{3} = \frac{X_L}{R} \;\Rightarrow\; X_L = \sqrt{3}\,R \)
So in the full LCR circuit \( X_L = X_C \): the circuit is at resonance, the impedance is R and
\( \displaystyle \cos\phi = \frac{R}{Z} = \frac{R}{R} = 1 \)
Note: A similar question was also asked in CUET (UG) 2023 on 20-06-2023 (Slot 3).