A current i=(3+acos100πt) A is flowing through a 10 Ω resistor. If the average power dissipated in the resistor is 170 watt, value of ‘a’, in ampere is
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a
1 amp
b
2 amp
c
3 amp
d
4 amp
answer is D.
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Detailed Solution
Irms=32+a22Average power =I2rms.R=9+a22 x 10∴9+a22 x 10=170⇒a=4 amp