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- .root { background image: url(/_files/images/chemical engineering/common/root.gif); background position: top left; background repeat: no repeat; padding left: 10px; padding top: 1px; } 41 A, 410 W respectivelyB- .root { background image: url(/_files/images/chemical engineering/common/root.gif); background position: top left; background repeat: no repeat; padding left: 10px; padding top: 1px; } 35 A, 350 W respectivelyC-5 A, 250 W respectivelyD-11 A, 1210 W respectively-C-5 A, 250 W respectively

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The current i(t) through a 10 &Omega; resistor in series with an inductance is given by i(t) = 3 + 4 sin (100t + 45&deg;) + 4 sin (300t + 60&deg;) Amperes. The RMS value of the current and the power dissipated in the circuit are-A- .root { background image: url(/_files/images/chemical engineering/common/root.gif); background position: top left; background repeat: no repeat; padding left: 10px; padding top: 1px; } 41 A, 410 W respectivelyB- .root { background image: url(/_files/images/chemical engineering/common/root.gif); background position: top left; background repeat: no repeat; padding left: 10px; padding top: 1px; } 35 A, 350 W respectivelyC-5 A, 250 W respectivelyD-11 A, 1210 W respectively-C-5 A, 250 W respectively

APTITUDE CRACK · Accepted Answer

C 5 A, 250 W respectively  Power = I2R = 25 X 10 = 250 Watts.

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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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Exam Questions Papers - Engineering

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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