# Electrical Engineering - RLC Circuits and Resonance - Discussion

### Discussion :: 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?

 [A]. 636 [B]. 3,769 [C]. 433 [D]. 4,337 Explanation:

No answer description available for this question.

 Jay said: (Oct 17, 2012) 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.

 Bashishth Singh said: (Apr 14, 2014) 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.

 Girmay Abadi said: (Jul 22, 2015) 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. So as me the answer is not given here; no answer!

 Aleksandr said: (Feb 8, 2016) Also must be angle 46.