An AC generator supplies an rms voltage of 120 V at 60.0 Hz. It is connected in

Question

An AC generator supplies an rms voltage of 120 V at 60.0 Hz. It is connected in series with a 0.400 H inductor, a 5.30 ?F capacitor and a 251 ohm 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?

D.) What is the peak current through the resistor?

Explanation / Answer

XL=2*pi*60*0.4=150.72 ohms
XC=1/(2*pi*60*5.30E-6F)=500.7 ohms
R = 251 ohms

Z=sqrt(R^2+(XL-XC)^2)= 430.7 ohms

IR=120 V/430.7 ohms=0.278 amps rms

avg power=V*I*R/Z=120 V*120 V/430.7 ohms*386/430.7= 33 .433 W

IR peak=0.278*sqrt(2)=0.393 amp peak

VL peak=0.393 amps*150.72 ohms=59.25 volts

VC peak=0.393*500.7 ohms=196.77 volts

f = 1/(2*pi*sqrt(L*C))=1/(2*pi*sqrt(0.4*5.30 E-6F))=109.36 Hz

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