The peak value of an alternating e.m.f. E giv | vedclass

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(B) Given: Peak value $E_0 = 10 \ V$,frequency $f = 50 \ Hz$,and time $t = \frac{1}{600} \ s$.The instantaneous e.m.f. is given by $E = E_0 \cos(\omeg

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$5\sqrt{3} \ V$

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(B) Given: Peak value $E_0 = 10 \ V$,frequency $f = 50 \ Hz$,and time $t = \frac{1}{600} \ s$.The instantaneous e.m.f. is given by $E = E_0 \cos(\omega t)$.Since $\omega = 2\pi f$,we have $\omega = 2 	imes \pi 	imes 50 = 100\pi \ rad/s$.Substituting the values into the equation:$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 e.m.f. $E$ 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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