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(D) Given: $R = 20 \Omega$,$V = 200 \sin (10 \pi t)$.Using Ohm's law,$I = \frac{V}{R} = \frac{200}{20} \sin (10 \pi t) = 10 \sin (10 \pi t)$.The peak

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$2.5 	imes 10^{-2}$

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(D) Given: $R = 20 \Omega$,$V = 200 \sin (10 \pi t)$.Using Ohm's law,$I = \frac{V}{R} = \frac{200}{20} \sin (10 \pi t) = 10 \sin (10 \pi t)$.The peak current is $I_0 = 10 	ext{ A}$.The rms current is $I_{	ext{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} 	ext{ A}$.At peak value,$10 = 10 \sin (10 \pi t_1) \Rightarrow \sin (10 \pi t_1) = 1 \Rightarrow 10 \pi t_1 = \frac{\pi}{2} \Rightarrow t_1 = \frac{1}{20} 	ext{ s}$.At rms value,$\frac{10}{\sqrt{2}} = 10 \sin (10 \pi t_2) \Rightarrow \sin (10 \pi t_2) = \frac{1}{\sqrt{2}} \Rightarrow 10 \pi t_2 = \frac{\pi}{4} \Rightarrow t_2 = \frac{1}{40} 	ext{ s}$.The time taken to change from peak to rms is $\Delta t = t_1 - t_2 = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} = 0.025 	ext{ s} = 2.5 	imes 10^{-2} 	ext{ s}$.

$A$ 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 rms value,then $t$ is (in seconds).

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