Question

A series LCR circuit is connected to an ac voltage source. When L is removed from the circuit, the phase difference between current and voltage is $\frac {\pi}{3}$ . If instead C is removed from the circuit, the phase difference is again $\frac {\pi}{3}$ between current and voltage. The power factor of the circuit is :

• A. zero
• B. 0.5
• C. 1
• D. -1

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the behavior of a series LCR (Inductor-Capacitor-Resistor) circuit when components are removed one at a time.

Initially, when the inductor \( L \) is removed, the circuit becomes an RC (Resistor-Capacitor) circuit. In an RC circuit, the phase difference (\( \phi \)) between current and voltage is given by:

\(\phi = \tan^{-1} \left( \frac{-1}{\omega RC} \right)\)

It is given that the phase difference is \( \frac{\pi}{3} \), so:

\(\phi = -\frac{\pi}{3}\)

When the capacitor \( C \) is removed, the circuit becomes an RL (Resistor-Inductor) circuit. In an RL circuit, the phase difference is given by:

\(\phi = \tan^{-1} \left( \frac{\omega L}{R} \right)\)

Again, it is given that the phase difference is \( \frac{\pi}{3} \), so:

\(\phi = \frac{\pi}{3}\)

The key insight is that the same phase difference \( \frac{\pi}{3} \) is obtained regardless of whether \( L \) or \( C \) is removed. This symmetry indicates that the circuit is at resonance when all components are present. At resonance in an LCR circuit:

\(\omega L = \frac{1}{\omega C}\)

At resonance, the phase difference between voltage and current is 0, which implies current and voltage are in phase. Therefore, the power factor \( \text{cos} \phi \) is \( 1 \).

Thus, the power factor of the circuit when all components are present is 1.

Therefore, the correct answer is 1.