A series R–L–C circuit of R = 165 , L = 0.885 H and C = 1.80 F carries an rms cu

Question

A series R–L–C circuit of R = 165 , L = 0.885 H and C = 1.80 F carries an rms current of 0.285 A with a frequency of 100 Hz . A) What is the average rate at which electrical energy is converted to heat in the resistor? Express your answer in watts to three significant figures. B) What average power is delivered by the source? Express your answer in watts to three significant figures. C) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor and in the inductor? Enter your answers, separated by a comma, to the nearest integer.

Explanation / Answer

A)

Pavg,R = I^2*R = 0.285^2*165 = 13.4 W

B)

Pavg,S = Irms*Vrms

Now, Vrms = Irms*Z

Z = sqrt(165^2 + (2*pi*100*0.885 - 1/(2*pi*100*1.8*10^-6))^2)

= 367.3 ohm

So, Pavg,S = Irms^2*Z = 0.285^2*367.3 = 29.8 W <-----answer

c)

Pavg,CI = sqrt(29.8^2 - 13.4^2) = 26.6 W

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