An AC generator supplies an rms voltage of 240 V at 50.0 Hz. It is connected in

Question

An AC generator supplies an rms voltage of 240 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 3.90 F capacitor and a 281 resistor.

a) What is the impedance of the circuit?

b) What is the rms current through the resistor?

c) What is the average power dissipated in the circuit?

d) What is the peak current through the resistor?

e) What is the peak voltage across the inductor?

f) What is the peak voltage across the capacitor?

g) The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency

Explanation / Answer

a) impedance of the circuit

We know that

Z = sqrt(R^2 + (wL -1/wC)^2 ) =sqrt(281^2+ (2*pi*50*0.65 - 1/(2*pi*50*3.9*10^-6))^2) = 673.41 ohm

b) rms current through the resistor

tan(theta) = (wL-1/wC)/R=(2*pi*50*0.65 - 1/(2*pi*50*3.9*10^-6))/281=-2.1778

theta = arctan(-2.1778) =-1.1405 rad

Thus rms voltage across resitor = Vrms cos(theta) =240*cos (-1.1405 )=100.1136 V

Thus rms current (same across all elements) = 100.1136 /281 =0.3563 amp

c) average power dissipated in the circuit

Pavg = Irms^2*R =0.3563^2*281=35.6729W

d) . Peak current through resistor (same across all elements by the way).

Ipeak = Irms * sqrt(2) =0.3563* sqrt(2) =0.5039 amp

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