A resistance of $20 Ω$ is connected to a source of an alternating potential $V=220 \sin (100\, πt)$. The time taken by the current to change from the peak value to rms value, is

$2.5×10^{-3}s$

The correct answer is Option (4) → $2.5×10^{-3}s$

$ I = \frac{V}{R} = 11 \sin 100\pi t$

$I_{peak} = 11$

$ 11 = 11 \sin 100\pi t_1 \Rightarrow \sin 100\pi t_1 = \sin \frac{\pi}{2}$

$ t_1 = \frac{1}{200}$

$ I_{rms} = \frac{11}{\sqrt 2} = 11 \sin 100\pi t_2 \Rightarrow \sin 100\pi t_2 = \frac{\pi}{4}$

$\Rightarrow t_2 = \frac{1}{400}$

$ \Delta t = t_1 - t_2 = \frac{1}{200} - \frac{1}{400} = \frac{1}{400} = 2.5\times 10^{-3}$