The root-mean-square value of an alternating current of 50 Hz frequency is 10 ampere. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be

Options

A. `2 xx 10^(-2) sec and 14.14 amp`
B. `1xx 10^(-2) sec and 7.07 amp`
C. `5 xx 10^(-3) sec and 7.07 amp`
D. `5 xx 10^(-3)sec and 14.14 amp`

Answer

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Root Mean Square (RMS) value of the current, \( I_{rms} = 10 \, \text{A} \)
- Frequency of the alternating current, \( f = 50 \, \text{Hz} \)

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

3. **Determine the Time to Reach Maximum Value (t'):**
- The time taken to go from zero to maximum value is one-fourth of the time period:
\[ t' = \frac{T}{4} = \frac{0.02}{4} = 0.005 \, \text{s} = 5 \times 10^{-3} \, \text{s} \]

4. **Calculate the Peak Value (I₀):**
- The peak value \( I_0 \) can be calculated from the RMS value using the formula:
\[ I_0 = \sqrt{2} \times I_{rms} \]
- Substituting the RMS value:
\[ I_0 = \sqrt{2} \times 10 \approx 1.414 \times 10 \approx 14.14 \, \text{A} \]

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

### Summary:
- Time taken from zero to maximum value: \( 5 \times 10^{-3} \, \text{s} \)
- Peak value of current: \( 14.14 \, \text{A} \)