A series RLC circuit, driven with = 120 V at frequency f_d = 60 Hz, contains a r

Question

A series RLC circuit, driven with = 120 V at frequency f_d = 60 Hz, contains a resistance R = 180 ohm, an inductor with inductive reactance X_L = 74 ohm, and a capacitor with capacitive reactance X_C = 210 ohm. a) What is the impedance of this circuit? b) What are the power factor and phase constant (phi) of this circuit? c) Draw all four phasors (V_R, V_L, V_C, and). d) What is the average rate P_avg at which energy is dissipated in the resistance? c) What new capacitance is needed to maximize P_avg if the other parameters of the circuit arc not changed?

Explanation / Answer

a)We know that impedence is given by

Z = sqrt [(Xl - Xc)^2 + R^2]

Z = sqrt [(74 - 210)^2 + 180^2] = 225.6 Ohm

Hence, Z = 225.6 Ohm

b)Power factor is

P = R/Z = 180/225.6 = 0.798

phi = tan-1[(Xl - Xc)/R] = tan-1 (-136/180) = -37.1 deg

Hence, P = 0.798 and phi = -37.1 deg

c)I = V/Z = 120/225.6 = 0.532 A

Pav = V I cos(phi)

Pav = 120 x 0.532 x cos-37.1 = 50.92 W

Hence, Pav = 50.92 W

d)Cn = 1/2 pi f Xl

Cn = 1/2 x 3.14 x 60 x 74 = 35.86 x 10^-6 F

Hence, Cn = 35.86 microF.

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