The voltage of an AC source varies with time | vedclass

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(C) 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

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The peak voltage of the source is $50 	ext{ V}$.

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(C) 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 $V = V_0 \sin(\omega t)$,where $V_0$ is the peak voltage and $\omega = 2 \pi f$:Peak voltage $V_0 = 50 	ext{ V}$.Angular frequency $\omega = 200 \pi 	ext{ rad/s}$.Since $\omega = 2 \pi f$,we have $200 \pi = 2 \pi f$,which gives $f = 100 	ext{ Hz}$.Thus,the peak voltage is $50 	ext{ V}$ and the frequency is $100 	ext{ Hz}$. Therefore,option $C$ is correct.

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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