Question

In an LCR series circuit, a. c. voltage source is connected in series. When L is removed from the circuit, the phase difference between the voltage and the current is \(\pi /3\). If instead C is removed, the phase difference is again \(\pi /3\). The power factor of the circuit is \(\left(\tan \frac{\pi }{3}=\sqrt{3}\right)\)

• A. \(0.5\)
• B. \(\frac{1}{\sqrt{2}}\)
• C. \(1\)
• D. \(\frac{\sqrt{3}}{2}\)

Hint:

Equal phase angles with L removed and with C removed mean the two reactances are equal in magnitude.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
With L removed the circuit is R and C: \(\tan\phi=\frac{X_C}{R}\). With C removed it is R and L: \(\tan\phi=\frac{X_L}{R}\).

Step 2: Use the Given Angles
Both angles are \(\frac{\pi}{3}\), so \(\frac{X_C}{R}=\frac{X_L}{R}=\tan\frac{\pi}{3}=\sqrt{3}\).

Step 3: Conclude
So \(X_L=X_C\), which is the resonance condition for the full circuit.

Step 4: Power Factor
At resonance the impedance is \(Z=R\), so \(\cos\phi=\frac{R}{Z}=1\), option (C).

Final Answer:
The power factor of the circuit is 1. \[ \boxed{\text{(C) }1} \]