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 0.350 H inductor, a 3.10 ?F capacitor and a 256 ? resistor.
What is the impedance of the circuit?

What is the rms current through the resistor?

What is the average power dissipated in the circuit?

What is the peak current through the resistor?

What is the peak voltage across the inductor?

What is the peak voltage across the capacitor?

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

Explanation / Answer

1. Impedance
Z = sqrt(R^2 + (wL -1/wC)^2 ) = sqrt(256^2 + (2*pi*60*0.35 - 1/(2*pi*60*3.1*10^-6))^2) = 767.67 ohm

2. rms current.

Phase between current and voltage, say theta

tan(theta) = (wL-1/wC)/R = (2*pi*60*0.35 - 1/(2*pi*60*3.1*10^-6))/256 = -2.82705018

theta = arctan(-2.82705018) = -1.23080636 rad

Thus rms voltage across resitor = Vrms cos(theta) = 115 * cos(-1.23080636) = 38.3499278 V

Thus rms current (same across all elements) = 38.3499278/256 = 0.149804405 amp

3. Average power.

Power is dissipiated only in the resistor. Thus power dissipiated

Pavg = Irms^2*R = 0.149804405^2 *256 = 5.7449881 W

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

Ipeak = Irms * sqrt(2) = 0.149804405 * sqrt(2) = 0.211855421amp.

5. Peak voltage across inductor

Vpeak = Ipeak * X = 0.211855421 * 2*pi*60*0.35 = 27.9536642 V

6. Peak voltage across cap.

Vpeak = Ipeak*X = 0.211855421 /(2*pi*60*3.1*10^-6) = 181.278696 V

7. Resonant frequency

f = 1/(2*pi*sqrt(LC)) = 1/(2*pi*sqrt(0.35*3.1*10^-6)) = 152.793635 Hz

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.