An ac generator supplies an rms (not peak) voltage of 180 V at 60.0 Hz. The gene

Question

An ac generator supplies an rms (not peak) voltage of 180 V at 60.0 Hz. The generator is connected in series with a 0.50 H inductor, a 6.0 x 10^-6 F capacitor and a 300 ohms resistor.

a. What is the peal (not rms) current through the resistor?

b. What is the phase angle for this circuit?

c. What is the average power supplied to the circuit?

d. What is the peak (not rms) voltage across the inductor?

e. What is the peak (not rms) voltage across the resistor?

f. What is the peak (not rms) voltage across the capacitor?

Explanation / Answer

Assume the generator outputs a sinusoidal waveform.
The peak voltage of the generator, the capacitive reactance, the inductive reactance
Vp = 180V x 2^(1/2) = 254.6V
XC = -1 / (2 x pi x f x C) = -2pi60x6.0uF = -442.1ohm
XL = 2 x pi x f x L = 2pi60x0.5H = 188.5ohm

The impedance of the circuit

Z = 300 + j(-442.1+188.5) = 300-j253.6 ohm

a)What is the peal (not rms) current through the resistor

|Z|=(300^2+253.6^2)^(1/2)=392.8ohm

Irms = E/R = 180V/392.8ohm = 0.4582A
Ip = 0.4582A x 2^(1/2) = 0.6480A

b)What is the phase angle for this circuit?

angle = arctan(-253.6/300)=-40.21deg

c) What is the average power supplied to the circuit?

Pave=Irms^2 x R = 0.4582A^2 x 300ohm=62.98W

VpL = 0.6480A x 188.5ohm = 122.1V
VpR = 0.6480A x 300ohm = 194.4V
VpC = 0.6480A x -442.1ohm = -286.5V
where the minus sign means the voltage has polarity opposite to the voltages of the inductor and resistor.

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