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

Options

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

Explanation

To solve the problem step-by-step, we will first find the peak value of the alternating current (AC) and then calculate the time taken for the current to reach from zero to its maximum value.

### Step 1: Determine the Peak Value of the AC
Given that the RMS (Root Mean Square) value of the AC is 10 A, we can use the relationship between the RMS value and the peak value (I₀) of an AC current. The relationship is given by:

\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]

Rearranging this formula to find the peak value:

\[ I_0 = I_{rms} \times \sqrt{2} \]

Substituting the given RMS value:

\[ I_0 = 10 \, \text{A} \times \sqrt{2} \]

Calculating \(\sqrt{2}\):

\[ \sqrt{2} \approx 1.414 \]

Thus,

\[ I_0 \approx 10 \times 1.414 \approx 14.14 \, \text{A} \]

### Step 2: Calculate the Time Period (T)
The frequency (f) of the AC is given as 50 Hz. The time period (T) is the reciprocal of the frequency:

\[ T = \frac{1}{f} \]

Substituting the frequency:

\[ T = \frac{1}{50} = 0.02 \, \text{s} \]

### Step 3: Calculate the Time Taken to Reach from Zero to Peak Value
In a sinusoidal AC waveform, the time taken to go from zero to the peak value is one-quarter of the time period (T):

\[ \text{Time from 0 to } I_0 = \frac{T}{4} \]

Substituting the value of T:

\[ \text{Time from 0 to } I_0 = \frac{0.02}{4} = 0.005 \, \text{s} \]

This can also be expressed in milliseconds:

\[ 0.005 \, \text{s} = 5 \, \text{ms} \]

### Final Results
- The peak value (I₀) of the AC is approximately **14.14 A**.
- The time taken to reach from zero to the peak value is **0.005 s** or **5 ms**.