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

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Question

(b)$E = {E_0}\cos \omega t = {E_0}\cos \frac{{2\pi t}}{T}$
$ = 10\cos \frac{{2\pi 	imes 50 	imes 1}}{{600}} = 10\;\cos \frac{\pi }{6}$$ = 5\sqrt 3

Vedclass · Accepted Answer

$5\sqrt 3 \,V$

Vedclass · Answer

(b)$E = {E_0}\cos \omega t = {E_0}\cos \frac{{2\pi t}}{T}$
$ = 10\cos \frac{{2\pi 	imes 50 	imes 1}}{{600}} = 10\;\cos \frac{\pi }{6}$$ = 5\sqrt 3 \;volt.$

The peak value of an alternating e.m.f. E is given by $E = {E_0}\cos \omega \,t$ is $10\, volts$ and its frequency is $50\; Hz$. At time$t = \frac{1}{{600}}\;sec$, the instantaneous e.m.f. is

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