An `AC` source is rated at `220V, 50 Hz.` The time taken for voltage to change from its peak value to zero is

Options

A. `50 sec`
B. `0.02 sec`
C. `5 sec`
D. `5xx10^(-3)sec`

Explanation

To solve the problem of finding the time taken for the voltage of an AC source to change from its peak value to zero, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the AC Voltage Function**:
The voltage of an AC source can be represented as a sinusoidal function. The peak voltage (V_peak) is the maximum voltage, and in this case, it is given as 220V. The AC voltage can be expressed as:
\[ V(t) = V_{peak} \sin(\omega t) \]
where \(\omega\) is the angular frequency.

2. **Determine the Frequency**:
The frequency (f) of the AC source is given as 50 Hz. The angular frequency (\(\omega\)) can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of f:
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

3. **Identify the Time Period (T)**:
The time period (T) of the AC source is the reciprocal of the frequency:
\[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \, \text{s} = 20 \, \text{ms} \]

4. **Determine the Time to Change from Peak to Zero**:
The voltage changes from its peak value to zero during the first quarter of the cycle. Since the full cycle corresponds to the time period T, the time taken to go from peak to zero is:
\[ t = \frac{T}{4} \]
Substituting the value of T:
\[ t = \frac{20 \, \text{ms}}{4} = 5 \, \text{ms} \]

5. **Final Result**:
Therefore, the time taken for the voltage to change from its peak value to zero is:
\[ t = 5 \, \text{ms} = 5 \times 10^{-3} \, \text{s} \]

### Conclusion:
The answer is \(5 \, \text{ms}\).

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