A series RLC circuit driven by a source with an amplitude of 120.0 V and a frequ
ID: 1452064 • Letter: A
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 782 mH, a resistance of 300 Ohm, and a capacitance of 42.2 mu F. What are the maximum current and the phase angle between the current and the source emf in this circuit? What are the maximum potential difference across the inductor and the phase angle between this potential difference and the current in the circuit? What are the maximum potential difference across the resistor and the phase angle between this potential difference and the current in this circuit? What are the maximum potential difference across the capacitor and the phase angle between this potential difference and the current in this circuit?Explanation / Answer
a)
R = 300 ohm
Vo = 120 V
f = 50 Hz
XL = 2*pi*f*L
=2*pi*50*(782*10^-3)
= 245.7 ohm
XC = 1/(2*pi*f*C)
= 1/(2*pi*50*42.2*10^-6)
=75.4 ohm
Z = sqrt (R^2+ (XC-XL)^2)
=sqrt (300^2 + (75.4 - 245.7)^2)
= 345 ohm
Io = Vo/Z
= 120/345
=0.348 A
angle = atan {(XC-XL)/R}
=atan {(75.4-245.7)/300}
= - atan {(245.7-75.4)/300}
= -29.6 degree
Answer:
0.348 A
-29.6 degree
b)
VLmax = Io*XL
= 0.348*245.7
=85.5 V
angle = -90 degree
c)
VRmax = Io*R
= 0.348*300
= 104.4 V
angle = 0 degree
d)
VLmax = Io*Xc
= 0.348*75.4
=26.2 V
angle = 90 degree