Electrical Engineering - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 19)
19.
A 9 mH coil is in parallel with a 0.015 
F capacitor across an 18 kHz ac source. The coil's internal resistance, RW, is 60 
. The circuit impedance is
F capacitor across an 18 kHz ac source. The coil's internal resistance, RW, is 60 . The circuit impedance is17,340
1,734
290
1,018
Discussion:
11 comments Page 1 of 2.
Kuldeep kumar said:
1 decade ago
Internal resistance of coil is in series with inductance and these are parallel with capacitance.
Z = (R+jwL)||(1/jwC). But the correct option is not given.
Z = (R+jwL)||(1/jwC). But the correct option is not given.
(1)
DARSH said:
3 years ago
This question is solved by the admittance method and the process is lengthy and time-consuming.
In this question the capacitor is alone so the impedance of the second branch is not determined, so we can not find the impedance of the given question also voltage is not given, so the given question is the wrong statement.
If you can solve you assume the resistor value which will place in series with the capacitor than and then you can solve the problem otherwise it is not possible. Only inductor branch impedance is calculated.
In this question the capacitor is alone so the impedance of the second branch is not determined, so we can not find the impedance of the given question also voltage is not given, so the given question is the wrong statement.
If you can solve you assume the resistor value which will place in series with the capacitor than and then you can solve the problem otherwise it is not possible. Only inductor branch impedance is calculated.
(1)
Rafi said:
1 decade ago
Please any one give the solution.
Vishnu said:
1 decade ago
xl = 2 pi f l.
xc = 1/2 pi f c.
z = sqrt(rw2+(xl-xc)2).
Since internal impedance appears in series which converts ideal source into practical source
ideal v+series resistance = Practical v.
xc = 1/2 pi f c.
z = sqrt(rw2+(xl-xc)2).
Since internal impedance appears in series which converts ideal source into practical source
ideal v+series resistance = Practical v.
Parneet kaur said:
1 decade ago
My answer is 431.78. Can anybody explain the answer in detail?
Stuti kushwaha said:
10 years ago
I got different answer.
Pruthvi said:
9 years ago
It is very easy. Here r is internal resistance.
X = XL + r; & XC is given both are in parallel.
1/z = √(1/X - 1/XC)^2 = √(1/X - 1/XC) find z and it is 17340 ohm.
Then do the calculation you will get the answer.
X = XL + r; & XC is given both are in parallel.
1/z = √(1/X - 1/XC)^2 = √(1/X - 1/XC) find z and it is 17340 ohm.
Then do the calculation you will get the answer.
Sid said:
8 years ago
Can anyone explain this question? I am unable to find the answer.
My answer comes out to be around 250.
My answer comes out to be around 250.
Tom said:
8 years ago
Z=Rw(Q^2+1).
Q=XL/Rw.
Q=XL/Rw.
Kiko said:
5 years ago
Internal resistance can be transformed into its equivalent theoretical ckt:
Rp = ((Rw^2+XL^2)/Rw) = 17,327Ω.
Since Z=Rp=Rw(1+Q^2).
Rp = ((Rw^2+XL^2)/Rw) = 17,327Ω.
Since Z=Rp=Rw(1+Q^2).
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