Question

A lamp consumes only 25% of the peak power in an AC circuit. The phase difference between the applied voltage and the current is

• A. \(\frac{\pi}{6}\)
• B. \(\frac{\pi}{3}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{\pi}{2}\)
• E. \(\pi\)

Hint:

Power factor \(=\cos\phi\). Smaller power → larger phase difference.

Answer 1

The Correct Option is B

Concept: In AC circuits, average power is related to peak power by: \[ P_{\text{avg}} = P_{\text{max}} \cos\phi \] where \(\phi\) is the phase difference between voltage and current.

Step 1: Understand the given condition.
The lamp consumes 25% of peak power: \[ P_{\text{avg}} = 0.25 P_{\text{max}} \]

Step 2: Use power relation. \[ P_{\text{avg}} = P_{\text{max}} \cos\phi \]

Step 3: Equate both expressions. \[ P_{\text{max}} \cos\phi = 0.25 P_{\text{max}} \]

Step 4: Cancel common term. \[ \cos\phi = 0.25 \]

Step 5: Solve for \(\phi\). \[ \phi = \cos^{-1}(0.25) \]

Step 6: Approximate value. \[ \cos^{-1}(0.25) \approx 60^\circ = \frac{\pi}{3} \]

Step 7: Final conclusion. \[ \boxed{\frac{\pi}{3}} \]