The phase difference between current and voltage in an AC circuit is `(pi)/(4)` radian, If the frequency of AC is 50 Hz, then the phase difference is equivalent to the time difference:-

Options

A. 0.78 s
B. 15.7 ms
C. `2.5 s `
D. 2.5 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. **Use the formula for time difference:**
The time difference \(\Delta t\) can be calculated using the formula:
\[ \Delta t = \frac{\Delta \phi}{2\pi f} \]

3. **Substitute the values into the formula:**
\[ \Delta t = \frac{\frac{\pi}{4}}{2\pi \times 50} \]

4. **Simplify the expression:**
- First, simplify the denominator:
\[ 2\pi \times 50 = 100\pi \]
- Now substitute this back into the equation:
\[ \Delta t = \frac{\frac{\pi}{4}}{100\pi} \]
- The \(\pi\) in the numerator and denominator cancels out:
\[ \Delta t = \frac{1}{4 \times 100} = \frac{1}{400} \]

5. **Calculate the time difference:**
\[ \Delta t = 0.0025 \text{ seconds} \]

6. **Convert seconds to milliseconds:**
\[ \Delta t = 0.0025 \times 1000 = 2.5 \text{ milliseconds} \]

### Final Answer:
The time difference corresponding to the phase difference of \(\frac{\pi}{4}\) radians at a frequency of 50 Hz is **2.5 milliseconds**.

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