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The potential difference V and the current i flowing through an instrument in an ac circuit of frequency f are given by V=5cosωt volts and I = 2 sin ωt amperes (where ω=2πf). The power dissipated in the instrument is

a
Zero
b
10 W
c
5 W
d
2.5 W

detailed solution

Correct option is A

V=5cosω t=5sinω t+π2  and i=2sinω tPower  =Vr.m.s.×ir.m.s.×cosφ= 0(Since φ=π2, therefore cosφ=cosπ2=0)

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