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posted by  DrewF on 7/20/2008 11:38:22 PM  |  status: Closed  

Electric Circuits Question

Course Textbook Chapter Problem
Analog Circuits Fundamentals of Electric Circuits (3rd) by Alexander, Sadiku 14 30P
Question Details:
A circuit consisting of a coil with inductance 10mH and resistance 20ohms is connected in series with a capacitor and a generator with an rms voltage of 120V.  Find:
 
a.)the value of the capacitance that will cause the circuit to be in resonance at 15kHz.
b.)the current through the coil at resonance.
c.)the Q of the circuit.
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posted by Shree_vani on 7/21/2008 12:15:25 AM  |  status: Live
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DrewF's comment:
"Informative! Thanks!"
Response Details:
Given
L = 10mH, R = 20Ω, fr = 15kHz, V=120V
The resonant frequecny is given by
fr = 1/ 2π√LC
squring and rearranging
C = 1/ 2π2fr2L
C= 1/ (2)(3.14)2 (15000)2 (0.01)
C = 2.25*10-8 F
or
C = 22.5nF
 
b) At resonance the impedance of the circuit is
Z =R
the current is given by
I = V/R = 120 / 20 = 6A
 
c) The Q factor is given by
Q = [1/R] [√L/C]
Q = [1/20] [√( 0.01/ 22.5n)] = 33
 
 
 
 
posted by prince(EE-SME) on 7/21/2008 12:17:27 AM  |  status: Live
Asker's Rating: Lifesaver   
DrewF's comment:
"Very Helpful! Thanks!"
Response Details:
 
at resonance
Where f is resonant frequency
Given that f=15KHz

 
at resonance
Current through the network is
 
Q factor=
 
 
I hope this helps u.......................
 
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