The peak value of an alternating emf e given | vedclass

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(D) The given equation for alternating emf is $e = e_0 \cos \omega t$.Given peak value $e_0 = 10 \ V$ and frequency $f = 50 \ Hz$.The angular frequenc

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$5 \sqrt{3} \ V$

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(D) The given equation for alternating emf is $e = e_0 \cos \omega t$.Given peak value $e_0 = 10 \ V$ and frequency $f = 50 \ Hz$.The angular frequency is $\omega = 2 \pi f = 2 \pi 	imes 50 = 100 \pi \ rad/s$.Substituting the values into the equation at $t = \frac{1}{600} \ s$:$e = 10 \cos(100 \pi 	imes \frac{1}{600})$$e = 10 \cos(\frac{\pi}{6})$Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we get:$e = 10 	imes \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ V$.

The peak value of an alternating emf $e$ given by $e = e_0 \cos \omega t$ is $10 \ V$ and its frequency is $50 \ Hz$. At time $t = \frac{1}{600} \ s$,the instantaneous e.m.f is:

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