The voltage of an ac source varies with time | vedclass

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(B) Given the equation $V = 100 \sin(100 \pi t) \cos(100 \pi t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we can r

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The peak voltage of the source is $50 \, V$.

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(B) Given the equation $V = 100 \sin(100 \pi t) \cos(100 \pi t)$.Using the trigonometric identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we can rewrite the equation as:$V = 50 	imes [2 \sin(100 \pi t) \cos(100 \pi t)]$$V = 50 \sin(200 \pi t)$.Comparing this with the standard form $V = V_0 \sin(\omega t)$,we get:Peak voltage $V_0 = 50 \, V$.Angular frequency $\omega = 200 \pi \, rad/s$.Since $\omega = 2 \pi f$,we have $200 \pi = 2 \pi f$,which gives $f = 100 \, Hz$.Therefore,the peak voltage is $50 \, V$.

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

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