A resistance of \( 20\,\Omega \) is connected to a source of alternating voltage \( V=220\sin(100\pi t) \) (V in volts, t in seconds). The time taken by the current to change from its peak value to its rms value is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2.5\times10^{-3}\,\mathrm{s} \)
In a purely resistive circuit the current is in phase with the voltage, \( i=i_m\sin\omega t \), with \( \omega=100\pi\,\mathrm{rad\,s^{-1}} \). The resistance fixes only the amplitude, \( i_m=V_m/R \). The timing depends on \( \omega \) alone. The current is at its peak when \( \sin\omega t=1 \), and it next equals its rms value \( i_m/\sqrt{2} \) when \( \sin\omega t=1/\sqrt{2} \).
The phase angles at the two instants are:
\( \displaystyle \omega t_1=\frac{\pi}{2} \quad \text{(peak)} \)
\( \displaystyle \omega t_2=\frac{3\pi}{4} \quad \text{(rms value, just after the peak)} \)
The time taken is:
\( \begin{aligned} t_2-t_1 &= \frac{3\pi/4-\pi/2}{\omega} \\ &= \frac{\pi/4}{100\pi\,\mathrm{s^{-1}}} \\ &= \frac{1}{400}\,\mathrm{s} \\ &= 2.5\times10^{-3}\,\mathrm{s} \end{aligned} \)