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

To solve for the time \( t \) taken by the current to change from its peak value to its RMS (root mean square) value in an AC circuit, we can start with the given alternating potential \( V = 200 \sin(10\pi t) \). Here, the peak voltage \( V_0 \) is 200 V.

The RMS value \( V_{\text{rms}} \) of an AC voltage is given by:

\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]

Substituting the peak voltage, we have:

\[ V_{\text{rms}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \text{ V} \]

The current \( I \) in the circuit at any time is given by Ohm's law:

\[ I(t) = \frac{V(t)}{R} \]

Where \( R = 20 \Omega \). Thus:

\[ I(t) = \frac{200 \sin(10\pi t)}{20} = 10 \sin(10\pi t) \]

The peak current \( I_0 \) is 10 A, and the RMS current \( I_{\text{rms}} \) is:

\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \text{ A} \]

To find the time \( t \) when the current changes from \( I_0 \) to \( I_{\text{rms}} \), we equate:

\[ 10 \sin(10\pi t) = 5\sqrt{2} \]

Simplifying gives:

\[ \sin(10\pi t) = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} \]

The value \(\frac{\sqrt{2}}{2}\) corresponds to \(\sin(45^\circ) = \sin\left(\frac{\pi}{4}\right)\).

Thus, we have:

\[ 10\pi t = \frac{\pi}{4} \]

Solving for \( t \):

\[ t = \frac{\pi}{4 \times 10\pi} = \frac{1}{40} = 2.5 \times 10^{-2} \text{ seconds} \]

Therefore, the correct answer is \( 2.5 \times 10^{-2} \) seconds.

Answer 2

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.