The phase difference between the current and the electromotive force in an ac circuit is `pi//4` radian. If the frequency is 50 Hz, then the time difference corresponding to this phase difference, will be –

Options

A. 0.25 s
B. 0.02 s
C. 2.5 ms
D. 25 ms

Answer

To solve the problem, we need to find the time difference corresponding to a phase difference of \(\frac{\pi}{4}\) radians in an AC circuit with a frequency of 50 Hz.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Phase difference, \(\Delta \phi = \frac{\pi}{4}\) radians
- Frequency, \(f = 50\) Hz

2. **Calculate the Time Period (T)**:
The time period \(T\) of an AC circuit is given by the formula:
\[ T = \frac{1}{f} \]
Substituting the given frequency:
\[ T = \frac{1}{50} = 0.02 \text{ seconds} \]

3. **Relate Phase Difference to Time**:
The phase difference in radians can be related to time using the formula:
\[ \Delta t = \frac{\Delta \phi}{2\pi} \times T \]
Here, \(\Delta t\) is the time difference corresponding to the phase difference \(\Delta \phi\).

4. **Substitute the Values**:
Now substituting the values we have:
\[ \Delta t = \frac{\frac{\pi}{4}}{2\pi} \times 0.02 \]

5. **Simplify the Expression**:
Simplifying the fraction:
\[ \Delta t = \frac{1}{8} \times 0.02 \]

6. **Calculate \(\Delta t\)**:
\[ \Delta t = 0.0025 \text{ seconds} = 2.5 \text{ milliseconds} \]

### Final Answer:
The time difference corresponding to a phase difference of \(\frac{\pi}{4}\) radians is **2.5 milliseconds**.