A resistor with R = 320 and an inductor are connected in series across an ac sou

Question

A resistor with R = 320 and an inductor are connected in series across an ac source that has voltage amplitude 500 V . The rate at which electrical energy is dissipated in the resistor is 316 W .

Part A

What is the impedance Z of the circuit?

Express your answer to three significant figures and include the appropriate units.

Part B

What is the amplitude of the voltage across the inductor?

Express your answer to three significant figures and include the appropriate units.

Part C

What is the power factor?

Express your answer using three significant figures.

Explanation / Answer

power P = RI^2 and from here you have that

I = sqrt(P/R) = 0.99

A which is effective value of current.

effective value of voltage =500*sqrt(2)/2 = 353.56 V

impedance = effective value of voltage /effective value of current

Z = V/I = 357.13 .
Now that you have impedance and

Z^2 = R^2 + X^2

X = 158.56 so effective value of voltage across inductor is U1 = X*I = 156.97 V
and power factor is cos = R/Z = 0.896

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