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1. The current i(t), though a 10 Ω resistor in series with an inductance, is given by i(t) = 3 + 4 sin (10t + 45°) + 4 sin (300t + 60°) Amperes. The RMS value of the current and the power dissipated in the circuit are





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  • By: guest on 02 Jun 2017 12.56 am
    Irms = = 6.4 Power P = I2R = 25 x 10 = 41 x 10 = 410 W.
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