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The voltage of an ac source varies with time according to the equation
$100\sin 100\pi t\cos 100\pi t$
where t is in seconds and V is in volts. Then

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The peak voltage of the source is 100 volts

The frequency of the source is 50 Hz

The peak voltage of the source is 50 volts

The peak voltage of the source is 100 $\sqrt 2$ volts

answer is B.

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Detailed Solution

$V = 50 \times 2\,\sin \,100\pi t\,\cos \,100\pi t = 50\sin 200\pi t$

$\Rightarrow \,\,{V_0} = 50\,Volts\,\,and\,\,v = 100\,Hz$

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