A series RLC circuit driven by a source with an amplitude of 120.0 V and a frequ

Question

A series RLC circuit driven by a source with an amplitude of 120.0 V and a frequency of 50.0 Hz has an inductance of 767 mH, a resistance of 212 , and a capacitance of 48.5 µF.

(a) What are the maximum current and the phase angle between the current and the source emf in this circuit?

(b) What are the maximum potential difference across the inductor and the phase angle between this potential difference and the current in the circuit?

VL, max

(c) What are the maximum potential difference across the resistor and the phase angle between this potential difference and the current in this circuit?

VR, max

(d) What are the maximum potential difference across the capacitor and the phase angle between this potential difference and the current in this circuit?

VC, max

Explanation / Answer

A) Inductive reactanceis XL = 2*pi*f*L = 2*3.142*50*0.767 = 241 ohm

Capacitive inductance is XC = 1/(2*pi*f*C) = 1/(2*3.142*50*48.5*10^-6) = 65.62 ohm

R = 212 ohm

then Impedence is Z = sqrt((XL-XC)^2+R^2) = sqrt((241-65.62)^2+212^2) = 275.14 ohm

maximum current is Imax = Vmax/Z = 120/275.12 = 0.436 A

phase angle is phi = tan^(-1)((XL-XC)/R) = tan^(-1)((241-65.62)/212) = 39.6 degrees

b) VL,max= Imax*XL = 0.436*241 = 105.076 V

current I leads Voltage V by a phase angle 90 degrees

C) VR,max = Imax*R = 0.436*212 = 92.432 V

Current I and Voltage V are in same phase

D) VC,max = Imax*XC = 0.436*65.62 = 28.6 V

Current I lags behind the Volatge V by a phase difference of 90 degrees

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