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 $2 \sin 	heta \cos 	heta = \sin(2 	heta)$,we can

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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 $2 \sin 	heta \cos 	heta = \sin(2 	heta)$,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 $\omega = 200 \pi \, rad/s$.Frequency $f = \frac{\omega}{2 \pi} = \frac{200 \pi}{2 \pi} = 100 \, Hz$.Thus,the peak voltage is $50 \, V$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

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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