The voltage of an AC source varies with time | vedclass

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(B) Given 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 rewrit

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

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(B) Given 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 $\omega = 2\pi 
u$ is the angular frequency:$V_0 = 50 \ V$.$\omega = 200\pi \Rightarrow 2\pi 
u = 200\pi \Rightarrow 
u = 100 \ Hz$.Thus,the peak voltage is $50 \ V$ and the frequency is $100 \ Hz$.

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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