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.

Question

Bryan Hunt

Physics

4 November, 14:45

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.

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Answers (1)

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Tanner Noble · Answer 1

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)