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 the peak value of the alternating current and the time taken to reach from zero to the maximum value. Let's break it down step by step.

### Step 1: Understand the relationship between RMS and peak value
The root mean square (RMS) value of an alternating current (AC) is related to its peak value (I₀) by the formula:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
Given that the RMS value is 10 A, we can rearrange the formula to find the peak value.

### Step 2: Calculate the peak value
Using the formula:
\[ I_0 = I_{rms} \times \sqrt{2} \]
Substituting the given RMS value:
\[ I_0 = 10 \, \text{A} \times \sqrt{2} \approx 10 \times 1.414 \approx 14.14 \, \text{A} \]

### Step 3: Determine the time period of the AC signal
The time period (T) of the AC signal can be calculated using the frequency (f):
\[ T = \frac{1}{f} \]
Given that the frequency is 50 Hz:
\[ T = \frac{1}{50} \, \text{s} = 0.02 \, \text{s} \]

### Step 4: Calculate the time taken to reach from zero to maximum value
The time taken for the current to go from zero to its maximum value is one-fourth of the time period:
\[ t = \frac{T}{4} \]
Substituting the value of T:
\[ t = \frac{0.02}{4} = 0.005 \, \text{s} = 5 \, \text{ms} \]

### Final Results
- The peak value of the current is approximately **14.14 A**.
- The time taken to reach from zero to maximum value is **5 ms**.