An AC generator supplies an rms voltage of 115 V at 60 Hz. It is connceted in se

Question

An AC generator supplies an rms voltage of 115 V at 60 Hz. It is connceted in series with a .300 H inductor, a 4.60 uF cpacitor and a 231 ohm resistor.

a) Impedance = 518 ohm

b) rms current = 0.222 A

c) Average Power dissapated in circuit = 11.4 W

d) Peak current through resistor =.314 A

e) What is the peak voltage across the inductor?

Hint: Use an equation similar to Ohm's law that includes the inductive reactance.

f) What is the peak voltage across the capacitor?

Hint: Use an equation similar to Ohm's law that includes the inductive reactance.

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

Please show work and explain solution.

Explanation / Answer

E. peak voltage across inductor

Ipeak = 0.314 Amp.

Vpeak = ipeak*XL

XL = wL = 2*pi*f*L

Vpeak = 0.314*2*3.14*60*0.3

Vpeak = 35.49 V

F.

Vpeak = ipeak*XC

XC = 1/wC = 1/(2*pi*f*C)

Vpeak = 0.314/(2*3.14*60*4.6*10^-6)

Vpeak = 181.16 V

G. for circuit in resonance

XC = XL

wL = 1/wC

w^2 = 1/LC

w = 2*pi*f

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

f = 1/[2*3.14*sqrt(0.3*4.6*10^-6)]

f = 135.55 Hz

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