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The peak value of an alternating e.m.f which is given by $E = E_{0} \cos w t$ is 10volts and its frequency is $50 H z .$ At time $t = \frac{1}{600} s,$ the instantaneous emf is

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$10 V$

$5 \sqrt{3} V$

$1 V$

$5 V$

answer is B.

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Detailed Solution

$E = E_{0} \cos ω t E_{0} = 10 V, t = \frac{1}{600} \sec, f = 50 H z E = 10 \cos (2 π \times 50 \times \frac{1}{600}) E = 10 \cos \frac{π}{6} = 5 \sqrt{3} v$

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