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Q.

The peak value of an alternating emf E given by $E = E_{0} \cos ω t$ is 10 V and frequency is 50 HZ. At time t = (1/600) s, the instantaneous value of emf is

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a

$5 \sqrt{3} V$

b

10 V

c

1 V

d

5 V

answer is B.

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Detailed Solution

Given that $E_{0} = 10 V, t = \frac{1}{600} s$

$∴ E = E_{0} \cos 2 π f t$

$= 10 \cos [2 π \times 50 \times \frac{1}{600}]$

$= 10 \cos (π / 6) = 10 (\sqrt{3} / 2) = 5 \sqrt{3} V$

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