Question

A resistance of \(20\Omega\) is connected to a source of alternating potential \(V=220\,\sin (100\pi t)\). The time taken by current to change from its peak value to r.m.s. value is \(\left(\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right)\)

• A. \(0.2\) s
• B. \(0.25\) s
• C. \(25\times 10^{-3}\) s
• D. \(2.5\times 10^{-3}\) s

Hint:

In a resistor the current is in phase with the voltage, so find the phase gap between the peak and the r.m.s. value and divide it by the angular frequency.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The circuit is purely resistive, so the current is in phase with the voltage and \(I=I_0\sin(100\pi t)\).
The angular frequency is \(\omega=100\pi\) rad/s.

Step 2: Phase at each value
At the peak value \(\sin\phi=1\), so \(\phi_1=\dfrac{\pi}{2}\).
At the r.m.s. value \(I=\dfrac{I_0}{\sqrt{2}}\), so \(\sin\phi=\dfrac{1}{\sqrt{2}}\) and \(\phi_2=\dfrac{\pi}{4}\).

Step 3: Calculation
The phase change is \(\Delta\phi=\dfrac{\pi}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{4}\).
\(t=\dfrac{\Delta\phi}{\omega}=\dfrac{\pi/4}{100\pi}=\dfrac{1}{400}\text{ s}=2.5\times10^{-3}\text{ s}\).

Step 4: Compare the options
The value \(2.5\times10^{-3}\) s matches option (D).

Final Answer:
The current takes 2.5 milliseconds to fall from its peak value to its r.m.s. value. \[ \boxed{\text{(D) }2.5\times10^{-3}\text{ s}} \]