Question

An alternating current of frequency 50 Hz has the peak value as 14.14 A. The time taken by the alternating current in reaching from zero to maximum value and r.m.s. value of current will be respectively

• A. 0.025 s, 5 A
• B. 0.005 s, 5 A
• C. 0.005 s, 10 A
• D. 0.025 s, 10 A

Hint:

For AC circuits, remember that the current reaches its maximum value at one–fourth of the time period, and the r.m.s. value is always \( \frac{1}{\sqrt{2}} \) times the peak value.

Answer 1

The Correct Option is C

Step 1: Time taken to reach maximum current.
For an alternating current, the time period is given by \[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \, \text{s}. \] The current reaches its maximum value in one–fourth of the time period. Therefore, \[ t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \, \text{s}. \]
Step 2: Calculating the r.m.s. value of current.
The r.m.s. value of alternating current is given by \[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}}, \] where \( I_0 = 14.14 \, \text{A} \). Hence, \[ I_{\text{rms}} = \frac{14.14}{\sqrt{2}} = 10 \, \text{A}. \]
Step 3: Conclusion.
The time taken is \(0.005 \, \text{s}\) and the r.m.s. value of current is \(10 \, \text{A}\).