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 812 mH, a resistance of 212 ohm, and a capacitance of 44.3 mu F. (a) What are the maximum current and the phase angle between the current and the source emf in this circuit? I_max = A phi = degree (b) What are the maximum potential difference across the inductor and the phase angle between this potential difference and the current in the circuit? V_L, max = V phi = degree (c) What are the maximum potential difference across the resistor and the phase angle between this potential difference and the current in this circuit? V_R, max = V phi = degree (d) What are the maximum potential difference across the capacitor and the phase angle between this potential difference and the current in this circuit? V_C, max = V

Explanation / Answer

given

Vmax = 120 V
f = 50 Hz
L = 812 mH = 0.812 H
R = 212 ohms
C = 44.3*10^-6 F

a) capative reactance, XC = 1/(2*pi*f*C) = 1/(2*pi*50*44.3*10^-6) = 71.8 ohms

inductive reactnce, XL = 2*pi*f*L = 2*pi*50*0.812 = 255 ohms

impedance of the ckt, z = sqrt(R^2 + (XL - XC)^2)

= sqrt(212^2 + (255 - 71.8)^2)

= 280 ohms

Imax = Vmax/z

= 120/280

= 0.428 A

phase angle, phi = tan^-1( (XL - XC)/R )

= tan^-1( (255 - 71.8)/212)

= 40.8 degrees

b) VL_max = XL*Imax

= 255*0.428

= 109 V

phi = 90 degrees

c) VR_max = R*Imax

= 212*0.428

= 90.7 V

phi = 0 degrees

d) VC_max = Xc*Imax

= 71.8*0.428

= 30.7 V

phi = 90 degrees

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