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The peak value of an alternating e.m.f. $E = E_{0} cosωt$ 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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10 V

$5 \sqrt{3} V$

5 V

1 V

answer is B.

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Detailed Solution

$E = E_{0} cosωt = E_{0} \cos \frac{2 πt}{T} = E_{0} \cos 2 πvt$

$= 10 \cos \frac{2 π \times 50 \times 1}{600} = 10 \cos \frac{π}{6} = 5 \sqrt{3} volt$

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