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

Options

A. 0.75 s
B. 10.5 ms
C. 2.5 ms
D. 0.25 ms

Answer

To solve the problem of finding the time difference equivalent to a phase difference of \(\frac{\pi}{4}\) radians in an AC circuit with a frequency of 50 Hz, we can follow these steps:

### Step 1: Understand the relationship between phase difference and time difference
The phase difference \(\phi\) in radians can be related to time difference \(t\) using the angular frequency \(\omega\):
\[ \phi = \omega t \]
where \(\omega\) is the angular frequency in radians per second.

### Step 2: Calculate the angular frequency \(\omega\)
The angular frequency \(\omega\) is related to the frequency \(f\) (in Hz) by the formula:
\[ \omega = 2\pi f \]
Given that the frequency \(f = 50\) Hz, we can calculate \(\omega\):
\[ \omega = 2\pi \times 50 = 100\pi \text{ radians/second} \]

### Step 3: Substitute the values into the phase difference equation
Now, we can substitute the values of \(\phi\) and \(\omega\) into the equation:
\[ \frac{\pi}{4} = (100\pi) t \]

### Step 4: Solve for time difference \(t\)
To find \(t\), we can rearrange the equation:
\[ t = \frac{\frac{\pi}{4}}{100\pi} \]
This simplifies to:
\[ t = \frac{1}{400} \text{ seconds} \]

### Step 5: Convert the time difference into milliseconds
To express the time in milliseconds, we multiply by 1000:
\[ t = \frac{1}{400} \times 1000 = 2.5 \text{ milliseconds} \]

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

---