Discussion :: Exam Questions Paper
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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
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A.
41 A, 410 W respectively
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B.
35 A, 350 W respectively
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C.
5 A, 250 W respectively
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D.
11 A, 1210 W respectively
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Answer : Option C
Explanation :
Power = I2R = 25 X 10 = 250 Watts.
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