The frequency of an alternating current is `50 Hz`. The minimum time taken by it is reaching from zero to peak value is

Options

A. `5ms`
B. `10ms`
C. `20ms`
D. `50ms`

Explanation

To find the minimum time taken by an alternating current (AC) to reach from zero to its peak value when the frequency is given as 50 Hz, we can follow these steps:

### Step-by-step Solution:

1. **Understand the relationship between frequency and time period**:
The time period (T) of an AC signal is the reciprocal of the frequency (f). This can be expressed as:
\[ T = \frac{1}{f} \]

2. **Substitute the given frequency**:
Given that the frequency \( f = 50 \, \text{Hz} \), we can substitute this value into the equation for the time period:
\[ T = \frac{1}{50} \, \text{seconds} \]

3. **Calculate the time period**:
Performing the calculation gives:
\[ T = 0.02 \, \text{seconds} \quad (\text{or } 20 \, \text{ms}) \]

4. **Determine the time to reach from zero to peak value**:
The time taken to reach from zero to the peak value is one-fourth of the time period (T/4). This is because in a sinusoidal wave, the current reaches its peak value at a quarter of the time period.
\[ \text{Time to peak} = \frac{T}{4} \]

5. **Substitute the time period into the equation**:
Now substituting the value of T:
\[ \text{Time to peak} = \frac{0.02}{4} \, \text{seconds} \]

6. **Calculate the time to peak**:
Performing the calculation gives:
\[ \text{Time to peak} = 0.005 \, \text{seconds} \quad (\text{or } 5 \, \text{ms}) \]

7. **Final answer**:
Therefore, the minimum time taken by the current to reach from 0 to its peak value is:
\[ \text{Time} = 5 \, \text{ms} \]

### Final Answer:
The minimum time taken by the current to reach from 0 to its peak value is **5 milliseconds (ms)**.

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