The phase difference between the voltage and the current in an AC circuit is `pi//4`. If the frequency is `50Hz` then this phase difference will be equivalent to a time of

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Determine the Time Period (T)**:
The time period \(T\) of an AC signal is given by the formula:
\[ T = \frac{1}{f} \]
where \(f\) is the frequency. Given that \(f = 50 \, \text{Hz}\):
\[ T = \frac{1}{50} = 0.02 \, \text{s} = 20 \, \text{ms} \]

2. **Relate Phase Difference to Time**:
In one complete cycle (which corresponds to \(360^\circ\) or \(2\pi\) radians), the time taken is equal to the time period \(T\). Therefore:
\[ 360^\circ \text{ corresponds to } 20 \, \text{ms} \]

3. **Convert Phase Difference to Degrees**:
The given phase difference is:
\[ \frac{\pi}{4} \text{ radians} = 45^\circ \]

4. **Calculate the Time for the Phase Difference**:
To find the time equivalent of \(45^\circ\), we can set up a proportion based on the relationship established in step 2:
\[ \frac{45^\circ}{360^\circ} = \frac{t}{T} \]
where \(t\) is the time corresponding to the phase difference \(45^\circ\). Rearranging gives:
\[ t = T \cdot \frac{45^\circ}{360^\circ} \]

5. **Substituting the Values**:
Now substituting \(T = 20 \, \text{ms}\):
\[ t = 20 \, \text{ms} \cdot \frac{45}{360} \]

6. **Simplifying the Calculation**:
Simplifying the fraction:
\[ t = 20 \, \text{ms} \cdot \frac{1}{8} = 2.5 \, \text{ms} \]

### Final Answer:
The time equivalent to the phase difference of \(\frac{\pi}{4}\) radians is \(2.5 \, \text{ms}\).

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