Question

An AC generator supplies an rms voltage of 230 V at 60.0 Hz. It is connected in series with a 0.400 H inductor, a 3.40 μF capacitor and a 206 Ω resistor.

What is the impedance of the circuit? Tries 0/20

What is the rms current through the resistor? Tries 0/20

What is the average power dissipated in the circuit? Tries 0/20

What is the peak current through the resistor? Tries 0/20

What is the peak voltage across the inductor? Tries 0/20

What is the peak voltage across the capacitor? Tries 0/20

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

Answer 1

w = angular frequency = 2*pi*f

capacitive reactance Xc = 1/(wC) = 1/(2*pi*f*C) = 1/(2*pi*60*3.4*10^-6)=780 ohm

inductive reactance XL = wL = 2*pi*f*L = 2*pi*60*0.4= 151 ohm

impedance Z = sqrt(R^2 + (Xc - XL)^2 )

Z = sqrt(206^2 + (780-151)^2)

Z = 662 ohm   <<<--------------answer

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rms current Irms = Vrms/Z = 230/662 = 0.347 A

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average power = Irms^2 * R = 0.347^2*206=24.8 = 25 W

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peak current Ipeak = Irms*sqrt2 = 0.49 A

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peak voltage across inductor VL = Ipeak*XL = 0.49*151 = 74

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peak voltage across capacitor Vc = Ipeak*Xc = 0.49*780 = 382.2

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at resonance

Xc = XL

1/(wc) = wL

w^2 = 1/(LC)

(2*pi*f)^2 = 1/(LC)

frequency f = 1/(2*pi*sqrt(LC) )

f = 1/(2*pi*sqrt(0.4*3.4*10^-6)) = 136.5 Hz