An AC generator supplies an rms voltage of 115 V at 60.0 Hz. It is connected in

Question

An AC generator supplies an rms voltage of 115 V at 60.0 Hz. It is connected in series with a 0.500 H inductor, a 4.30 ?F capacitor and a 401 ? resistor. What is the impedance of the circuit? 5.87x10 ohm You are correct. PreviousTries What is the rms current through the resistor? Submit Answer Tries 0/20 What is the average power dissipated in the circuit? Submit Answer Tries 0/20 What is the peak current through the resistor? Submit Answer Tries 0/20 What is the peak voltage across the inductor? Submit Answer Tries 0/20 What is the peak voltage across the capacitor? Submit Answer Tries 0/20 The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?

Explanation / Answer

Problem 1. Solved

2. Phase between current and voltage be theta

tan(theta) = (wL-1/wC)/R = (2*pi*60*0.50 - 1/(2*pi*60*4.3*10^-6))/ 401= -1.068

theta = arctan(-1.068) = -0.818 rad

rms voltage across resistor = Vrms cos(theta) = 115 * cos(-0.818) = 78.62 V

Thus rms current (same across all elements) = 78.62/401 = 0.196 A

3. Power is dissipiated only in the resistor. Thus power dissipiated

Pavg = Irms^2*R = 0.196^2 *401 = 15.42 W

4. Ipeak = Irms * sqrt(2) = 0.196 * sqrt(2) = 0.277 A

5. Vpeak = Ipeak * X = 0.277 * 2*pi*60*0.50 = 52.25 V

6. Vpeak = Ipeak*X = 0.277 /(2*pi*60*4.3*10^-6) = 170.88 V

7. f = 1/(2*pi*sqrt(LC)) = 1/(2*pi*sqrt(0.5*4.3*10^-6)) = 108.54 Hz

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