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(D) The peak value of current is $i_0$ and the r.m.s. value is $i_{rms} = \frac{i_0}{\sqrt{2}}$.Given the alternating current equation $i = i_0 \sin(1

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$2.5 	imes 10^{-3} \, 	ext{s}$

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(D) The peak value of current is $i_0$ and the r.m.s. value is $i_{rms} = \frac{i_0}{\sqrt{2}}$.Given the alternating current equation $i = i_0 \sin(100 \pi t)$.At peak value,the phase $\omega t_1 = \frac{\pi}{2}$,so $t_1 = \frac{\pi}{2 	imes 100 \pi} = \frac{1}{200} \, 	ext{s}$.At r.m.s. value,$i = \frac{i_0}{\sqrt{2}}$,so $\sin(100 \pi t_2) = \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4})$.Thus,$100 \pi t_2 = \frac{\pi}{4} \Rightarrow t_2 = \frac{1}{400} \, 	ext{s}$.The time taken is $\Delta t = t_1 - t_2 = \frac{1}{200} - \frac{1}{400} = \frac{2-1}{400} = \frac{1}{400} \, 	ext{s}$.$\Delta t = 0.0025 \, 	ext{s} = 2.5 	imes 10^{-3} \, 	ext{s}$.

$A$ resistance of $20 \, \Omega$ is connected to a source of an alternating potential $V = 220 \sin(100 \pi t)$. The time taken by the current to change from its peak value to its r.m.s. value is:

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