# 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,

*R*, is 60 . The circuit impedance is_{W}Discussion:

10 comments Page 1 of 1.
DARSH said:
8 months 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.

Kiko said:
2 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).

Tom said:
6 years ago

Z=Rw(Q^2+1).

Q=XL/Rw.

Q=XL/Rw.

Sid said:
6 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.

Pruthvi said:
6 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.

Stuti kushwaha said:
7 years ago

I got different answer.

Kuldeep kumar said:
8 years 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.

Parneet kaur said:
8 years ago

My answer is 431.78. Can anybody explain the answer in detail?

Vishnu said:
8 years 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.

Rafi said:
8 years ago

Please any one give the solution.

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