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A circuit has an AC voltage source in series with a 50 ohm resistor and a 113 mH inductor. The frequency is 100 cycles/sec, and the peak voltage is 140 volts. Find the average power dissipated by the resistor. Solve using imaginary numbers.

Respuesta :

Answer:

Explanation:

The rms voltage = 140/√2 = 140/1.414 = 99 V.

Reactance of inductor  = wL = 2 X 3.14 X 100 X 113 X 10⁻³ =70.96 ohm.

Total resistance in terms of vector = 50+70.96j

j is imaginary unit  number

Magnitude of this resistance = √ 50² + 70.96² = 86.80 ohm

current in resistance (rms) ( I ) = 99/86.80 = 1.14 A.

Power dissipated in resistor = I² R = 1.14 X 1.14 X 50 = 65 W( approx)

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