Question

The phase difference between the voltage and the current in an a.c. circuit is \( \frac{\pi}{4} \). If the frequency is 50 Hz, then the phase difference is equivalent to a time of:

• A. \( 4.5 \times 10^{-3} \, \text{s} \)
• B. \( 1.5 \times 10^{-3} \, \text{s} \)
• C. \( 3.5 \times 10^{-3} \, \text{s} \)
• D. \( 2.5 \times 10^{-3} \, \text{s} \)

Hint:

To convert phase difference into time, divide the phase difference by \( 2\pi f \), where \( f \) is the frequency of the oscillating quantity.

Answer 1

The Correct Option is D

Step 1: Relation between Phase Difference and Time.
The phase difference \( \Delta \phi \) is related to the time difference \( \Delta t \) by the formula: \[ \Delta t = \frac{\Delta \phi}{2\pi f} \] where \( f \) is the frequency and \( \Delta \phi = \frac{\pi}{4} \). Substituting the values: \[ \Delta t = \frac{\pi/4}{2\pi \times 50} = \frac{1}{4 \times 50} = 2.5 \times 10^{-3} \, \text{s} \] Step 2: Final Answer.
Thus, the phase difference is equivalent to \( 2.5 \times 10^{-3} \, \text{s} \).