The rms value of an ac of 50Hz is 10A. The time taken be an alternating current in reaching from zero to maximum value and the peak value will be

Options

A. `2xx10^(-2) s and 14.14 A`
B. `1xx10^(-2) s and 7.07 A`
C. `5xx10^(-3) s and 7.07 A`
D. `5xx10^(-3) s and 14.14 A`

Explanation

To solve the problem, we need to find two things: the time taken for the alternating current to reach its maximum value from zero and the peak value of the current given the RMS value.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Frequency (f) = 50 Hz
- RMS value of current (I_RMS) = 10 A

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

3. **Determine the Time to Reach Maximum Value:**
- The time taken to reach the maximum value (peak value) from zero is \( \frac{T}{4} \) because the AC waveform is sinusoidal and takes a quarter of the time period to go from zero to the peak.
- Thus:
\[ \text{Time to reach peak} = \frac{T}{4} = \frac{0.02 \, \text{s}}{4} = 0.005 \, \text{s} = 5 \times 10^{-3} \, \text{s} \]

4. **Calculate the Peak Value (I_peak):**
- The relationship between the RMS value and the peak value is given by:
\[ I_{\text{peak}} = I_{\text{RMS}} \times \sqrt{2} \]
- Substituting the given RMS value:
\[ I_{\text{peak}} = 10 \, \text{A} \times \sqrt{2} \approx 10 \, \text{A} \times 1.414 \approx 14.14 \, \text{A} \]

### Final Results:
- **Time taken to reach maximum value:** \( 5 \times 10^{-3} \, \text{s} \)
- **Peak value of current:** \( 14.14 \, \text{A} \)