# Electrical Engineering - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 4)
4.
A resistor of 3 k, a 0.05 F capacitor, and a 120 mH coil are in series across a 5 kHz, 20 V ac source. What is the impedance, expressed in polar form?
636
3,769
433
4,337
Explanation:
No answer description is available. Let's discuss.
Discussion:
4 comments Page 1 of 1.

Girmay Abadi said:   9 years ago
Given parameters; L = 120 mH, R = 3*10^3 ohm,

c = 0.05 uF and F = 50 kHz.

XL = 2*3.14*5*10^4*1.2*10^-1 = 3728 ohm.

Xc = 1/2 (3.14*5*10^4*5*10^-8) = 636.94 ohm.

Magnitude |Z| = (R^2+(XL-XC)^2)^1/2 = 4337.3 ohm.

Tan (Q) = (XL-XC)/R = (3728-636.94)/3000 = 3091.06/3000 = 1.03035333.

Q = tan^-1 (1.03035333) = 45.85 degree.

Therefore the impedance as polar form is Z=|Z|at angle of Q = 4337.3 ohms at angle of 45.85 degree.

(1)

Bashishth singh said:   1 decade ago
Give that,

L = 120mH = 1.2*10^-1H and R = 3*10^3 ohm.
c = 0.05uF = 5*10^-8F.
F = 50kHz = 5*10^4Hz.

XL = 2*3.14*5*10^4*1.2*10^-1 = 3728 ohm.
Xc = 1/2(3.14*5*10^4*5*10^-8) = 636.94 ohm.

Z = (R^2+(XL-XC)^2)^1/2.
= 4337.3 ohm.
(1)

Find XL=2*3.14*(5*10^3)*(120*10^-3)
=3728 ohm.

Xc=1/2*3.14*(5*10^3)*(0.05*10^-6)
=636.94 ohm.

Z=sqrt[R^2+(XL-Xc)^2]
=4,337.3 ohm.

Aleksandr said:   9 years ago
Also must be angle 46.
(1)