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?
Discussion:
4 comments Page 1 of 1.
Jay said:
1 decade ago
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.
=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:
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.
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)
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.
So as me the answer is not given here; no answer!
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!
(1)
Aleksandr said:
9 years ago
Also must be angle 46.
(1)
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