Time Notes Evaluate Course Contents Set 7 (due Thurs 3/16 at 11:59PM) AC Power G

Question

Time Notes Evaluate Course Contents Set 7 (due Thurs 3/16 at 11:59PM) AC Power Generator An AC generator supplies an rms voltage of 115 V at 50.0 Hz. It is connected in series with a 0.250 H inductor, a 3.40 HF capacitor and a 306 2 resistor. What is the impedance of the circuit? 9.11x102 ohm You are correct Previous Tries What is the rms current through the resistor? 1.26 x 10 i A You are correct. Previous Tries What is the average power dissipated in the circuit? 14.5 W is dissipated in a resistor. Thus, you have to multiply the rms current times the rms voltage for the resistor Power only Submit Answer Incorrect. Tries 2/20 Previous Tries. What is the peak current through the resistor? 1.79X 10 i A You are correct Previous Tries What is the peak voltage across the inductor? 0.178 A Submit Answer Incompatible units No conversion found between "A" and the required units Tries o/20 Previous Tries What is the peak voltage across the capacitor? Submit Answer Tries o/20 The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency? Submit Answer Tries 0/20

Explanation / Answer

Inductance=0.25 H

Capacitance=3.4 F

Resistance=306

gen Voltage=115 V

XL=2**50*0.2578.53

XC=1/(2**50*3.4 E-6)936.2

a) (3062+(XL-XC)2)911 =Z

b) 115/9110.126 A=Itotal=IR

c) Pavg=I2*R38.556 W in steady state only

d) peak current is I*20.178 A

e) Ipeak*XL=13.97 V

f) Ipeak*XC=166.64 V

g) fo=1/(2**(L*C))172.62 Hz

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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