An ac source is rated at 220V, 50 Hz. The tim | vedclass

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(D) The frequency of the $ac$ source is $f = 50 	ext{ Hz}$.The time period $T$ of the $ac$ cycle is given by $T = 1/f = 1/50 	ext{ s} = 0.02 	ext{

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$5 	imes 10^{-3}$

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(D) The frequency of the $ac$ source is $f = 50 	ext{ Hz}$.The time period $T$ of the $ac$ cycle is given by $T = 1/f = 1/50 	ext{ s} = 0.02 	ext{ s}$.The voltage varies as $V = V_0 \sin(\omega t)$. The peak value occurs at $t = T/4$ and the voltage is zero at $t = T/2$ (or at $t=0$ for the start of the cycle).The time taken to go from the peak value $(V = V_0)$ to zero $(V = 0)$ corresponds to a quarter of the time period.Therefore,the required time $t = T/4 = 0.02 / 4 = 0.005 	ext{ s} = 5 	imes 10^{-3} 	ext{ s}$.

$(A) \ x_0 \sin \omega t$	$(i) \ x_0$
$(B) \ x_0 \sin \omega t \cos \omega t$	$(ii) \ \frac{x_0}{\sqrt{2}}$
$(C) \ x_0 \sin \omega t + x_0 \cos \omega t$	$(iii) \ \frac{x_0}{2 \sqrt{2}}$

An $ac$ source is rated at $220V, 50 Hz$. The time taken for voltage to change from its peak value to zero is.....$sec$.

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