Question

A voltage source, Vs = 180sin760t is in series with an inductor L = 435 mH, a capacitor C = 2 μF, and a resistor R = 240 Ω.

c) What is the phase angle of this circuit? Does the current lead the total voltage or lag behind it?

d) What is the average power delivered by the battery and dissipated by each component? e) What is the instantaneous current in the circuit, I, Vs, VR, VL, VC at t = 3 seconds?

e) What is the instantaneous current in the circuit, I, Vs, VR, VL, VC at t = 3 seconds?

h) Same questions but at the resonance frequency.

Answers according to my professor should be:

c) φ = -.938 rads, The current leads the source voltage by .938 rads or the source voltage lags behind the current by .938 rads.

d) PbatteryAvg = PresistorAvg = 23.60 watts, PinductorAvg = PcapacitorAvg = 0 watts

e) I(3) = +.0626 amps, Vsource = -128.65 volts, VR = +15.04 volts, VC = -288.85 volts, VL = +145.15, Vtot = -128.65 volts [Note: Vsource = Vtot = VL + VC + VR] For instantaneous voltages, they simply add up like a regular addition at any time t!

h) φ = 0 rads, The current is in phase with the source voltage.

PbatteryAvg = PresistorAvg = 67.42 watts,

PinductorAvg = PcapacitorAvg = 0 watts

I(3) = -.478 amps, Vsource = -114.74 volts, VR = -114.74 volts, VC = -269.51 volts, VL = +269.51 volts, Vtot = -114.74 volts

Answer 1

Given, Vs = 180sin760t ; L = 435 mH, C = 2 μF, R = 240 Ω.

The inductive reactance is . XL = 0.435 x 760 = 330.6 Ω.

The capacitive reactance is .  XC = 106 / 2x760 = 657.89 Ω

The phase angle is .=> =>

The circuit is capacitive as XC is greater than XL .
Then phase angle is - 0.938 rad . If XC is greater than XL then current leads the source voltage.

d) The impedance in the circuit is . Z = 405.86 Ω

Vrms = V0 / = 127.28 V

The average power delivered by the battery is P = irms x Vrms cos = Vrms2 / Z cos

P = (127.282 / 405.86 ) x (0.59) = 23.55 W

The power will be dissipated only through resistor and there is no power dissipation through inductor and capacitance.

Power dissipated through resistor is 23.56W. Power dissipated through inductor is zero. Power dissipated through capacitor is zero.

e) Instantaneous current is I = 180sin(760t +) / 405.86

At t=3s, I3 = 180sin(760x3 + 0.938) / 405.86 = 0.06265 A

Instantaneous voltage of source Vs = 180sin(760t) then at t = 3s, V = -128.655 V

Instantaneous voltage through resistor VR = I3 x R = 0.06265 x 240 = 15.036 V

Instantaneous voltage through capacitor VC = XC x 180sin(760x3 - 0.938​​​​​​​) / Z = 657.89 x (-177.6) / 405.86

VC = -288.86 V

Instantaneous voltage through inductor VL = - XL x 180sin(760x3 - 0.938​​​​​​​) / Z = - 330.6 x (-177.6) / 405.86

VL = 145.15 V

h) At resonance frequency Z = R and phase angle is zero. = 0 and ;

Power dellivered by battery P = Vrms2 / R = 127.282 / 240 = 67.5 W = Power dissipated through resistor.

Power dissipation through inductor and capacitor is zero.

Instantaneous current is I3 = 180sin(1072.1125 t +) / 240 = 180 sin (1072.1125 x 3) / 240 = - 0.478 A

The instantaneous voltage of the source is V = 180sin(1072.1125 t) = 180 sin (1072.1125 x 3) = - 114.74 V

The instantaneous voltage across resistor is VR = I3 x R = - 0.478 x 240 = - 114.74 V

The instantaneous voltage across capacitor is VC = I3 x XC = - 0.478 / (2x10-6x1072.1125) = - 269.51 V

The instantaneous voltage across inductor is VL = - I3 x XL = - 0.478 x 435 x 10-3 x 1072.1125 = 269.51 V