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.