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 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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.