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1. The current i(t) through a 10 Ω resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are
(A): 11 A, 1210 W respectively
(B): 35 A, 350 W respectively
(C): 41 A, 410 W respectively
(D): 5 A, 250 W respectively
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By: guest on 02 Jun 2017 12.54 am
Power = I2R = 25 X 10 = 250 Watts.
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