An R L C circuit consists of a 200- resistor, a300- F capacitor, and a 3.00-H in

Question

An RLC circuit consists of a 200- resistor, a300-F capacitor, and a 3.00-H inductor. The circuit is connected to a power outlet that has a peak emf of 270 V and oscillates at 400 Hz.

Part A

What is the average rate at which energy is dissipated in this circuit?

Express your answer with the appropriate units.

4.83W

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I tried using this method

Xc = 1/2pifC = 1 / (2pi x 400 x 300 x 10^-6)

Xc = 1.33 ohm

XL = 2pifL = 2 x pi x 400 x 3 = 7539.82 ohm

R = 200 ohm

Impedanced of circuit Z = sqrt(R^2 + (XL - Xc)^2)

Z = 7541.14 ohm

Pav = Vrms^2 / Z   = (270/sqrt(2))^2 / 7541.14   = 4.83W but is was incorrect

4.83W

Explanation / Answer

the power dissipated in the RLC circuit is equal to the power dissipated by the resistor,since the voltage across a resistor and the current through it are in phase

then power dissipation is P = irms^2*R

Xc = 1/2pifC = 1 / (2pi x 400 x 300 x 10^-6)

Xc = 1.33 ohm

XL = 2pifL = 2 x pi x 400 x 3 = 7540.8 ohm

R = 200 ohm

Impedanced of circuit Z = sqrt(R^2 + (XL - Xc)^2)

Z = 7542.12 ohm

irms = Vrms/Z = (270/1.414)/7542.12 = 0.0253 A

then power dissipated is P= irms^2*R = 0.0253*0.0253*200 0.128 W

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