Question

A resistance of \( 20 \Omega \) is connected to a source of an alternating potential \( V = 200 \sin(10\pi t) \). If \( t \) is the time taken by the current to change from the peak value to the rms value, then \( t \) is (in seconds):

• A. \( 25 \times 10^{-1} \)
• B. \( 2.5 \times 10^{-4} \)
• C. \( 25 \times 10^{-2} \)
• D. \( 2.5 \times 10^{-2} \)

Hint:

For an AC circuit with only resistance, current follows the same sinusoidal variation as voltage. Use the equation \( I = I_0 \sin(\omega t) \) and set \( I = I_{\text{rms}} \) to determine the required time interval.

Answer 1

The Correct Option is D

Step 1: Understanding the Given AC Voltage Expression The given alternating voltage is: \[ V = 200 \sin(10\pi t) \] From the standard AC voltage equation \( V = V_0 \sin(\omega t) \), we identify: \[ V_0 = 200, \quad \omega = 10\pi \] Since the circuit contains only resistance, the current follows the same sinusoidal form as voltage: \[ I = I_0 \sin(\omega t) \] where: \[ I_0 = \frac{V_0}{R} = \frac{200}{20} = 10 A \] Step 2: Finding the Time Interval The peak value of the current is \( I_0 \), and the rms value of the current is: \[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \] Since \( I = I_0 \sin(\omega t) \), setting \( I = I_{\text{rms}} \): \[ \frac{10}{\sqrt{2}} = 10 \sin(10\pi t) \] \[ \sin(10\pi t) = \frac{1}{\sqrt{2}} \] Step 3: Solving for \( t \) From trigonometry: \[ \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \] So, \[ 10\pi t = \frac{\pi}{4} \] \[ t = \frac{\pi}{4} \times \frac{1}{10\pi} \] \[ t = \frac{1}{40} = 2.5 \times 10^{-2} \text{ sec} \] Step 4: Conclusion Thus, the time taken by the current to change from the peak value to the rms value is \( 2.5 \times 10^{-2} \) sec.