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

Question

A series RLC circuit driven by a source with an amplitude of 120.0 and a frequency o 50.0 Hz has an inductance of 76?? a resistance of 234 n and a capacitance of a (a) What are the maximum current and the phase angle between the current and the source emf in this circuit? max (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

part A:

angular frequency W = 2pif = 2*3.14* 50 = 314 rad/s

Iamx = V/Z

Z is impedence = Z^2 = R^2 + (XL-Xc)^2

XL = wL = 314 * 0.767 = 240.838 ohms

XC = 1/wC = 1/(314 *47.1*10^-6)

Xc = 67.6 ohms

Z^2 = (234^2 + (240.84 - 67.6)^2

Z = 291.14 ohms

so Imax = V/Z = 120/291.14 = 0.412 Amps

phase angle at this current is tan phi = (XL-Xc)/R  

tan phi = (240.84 - 67.6)/234  

phase angle = 36.51 deg

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B)

MAx PD across Inductor = VL = Imax * XL

VL max = 0.412 * 314 * 0.767

VLmax = 99.22 Volts

phase angle = 90 deg

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part C :

MAx PD across Resistor = VL = Imax * R

VR max = 0.412 * 234  

VLmax = 96.41 Volts

phase angle = 0 deg

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part D :

max PD across Capacitor X

Vc = Imax* Xc

Vc = 0.412 * 1/wC

Vc = 0.412/(314* 47.1*10^-6)

vc = 27.85 Volts

phase angle = -90 deg

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