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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

    2. A.
    41 A, 410 W, respectively
    B.
    35 A, 350 W, respectively
    C.
    5 A, 250 W, respectively
    D.
    11 A, 1210 W, respectively

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    Answer : Option A

    Explanation :

    Irms = = 6.4

    Power P = I2R = 25 x 10 = 41 x 10 = 410 W.

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