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Q.

The peak value of an alternating e.m.f. E is given by E=E0cosω t is 10 volts and its frequency is 50 Hz. At time t=1600sec, the instantaneous e.m.f. is

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a

10 V

b

53 V

c

5 V

d

1 V

answer is B.

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Detailed Solution

E=E0cosωt=E0cos2πtT=10cos2π×50×1600=10 cosπ6=53 volt.
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