Question

The resistor '\(R\)' inductance '\(L\)' and capacitance '\(C\)' are connected in series with an a.c. source. When 'L' is removed from the circuit, the phase difference between voltage and current in the circuit is \(\frac{π}{3}\). If instead, C is removed from the circuit, the phase difference is again \(\frac{π}{3}\). The power factor of the circuit is

• A. \(\frac{\sqrt{3}}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{\sqrt{2}}\)
• D. \(1\)

Hint:

Both phase differences are \(\frac\pi3\), so \(X_L=X_C\).

Answer 1

The Correct Option is D

Step 1: Compare reactances
The equal magnitude of phase angles means $|X_C|=|X_L|$.

Step 2: Net effect
They cancel in the series circuit, so the impedance is purely resistive and the power factor is 1, option (D).

Final Answer:
The power factor is 1, option (D). \[ \boxed{1} \]