Electrical Engineering - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 7)
A 6.8 k resistor, a 7 mH coil, and a 0.02 F capacitor are in parallel across a 17 kHz ac source. The coil's internal resistance, RW, is 30 . The equivalent parallel resistance, Rp(eq), is
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34 comments Page 1 of 4.

DARSH said:   8 months ago
Here, we didn't find the equivalent resister of parallel circuit. Without resistor place in series with capacitor.

Mahe said:   12 months ago
I'm not getting it. Anyone, explain me the right answer.

Anomi said:   4 years ago
Please explain the solution.

Swathi said:   4 years ago
Can anyone explain me in detail please?

Muhammad said:   4 years ago
By using this equation1/Z=(1/R)2+(1/XL2-1/XC2)2 we get the answer.

Arunkumar said:   5 years ago
What is X value in G and J value in Y?

Please explain clearly.

KONDAVENI SANTHOSH said:   5 years ago
It is for internal resistance present in the inductor and capacitor.

Genrally, we solve this problem in tank circuit model

After convert all parallel networks into conductance,suceptance,admitance values
Conductance(G)=(R)/(R^2 + X^2); R=30
In this scenario internal restance present in L&C.
so we are taking suceptance(B)=(X)/(R^2 + X^2); R=30.
Y=y1 + y2 + y3 Here, Y= admitance.

Y=G1+G2 + G3 + J(BL -Bc); G1 value take restor R1=6800 OHM; G2&G3 values take R=30.

After calculation,
Y=G + J(B),
finally R=1/G,
R=18,780 OHMS.

Rasel said:   5 years ago
Here, rp= (sq(Rw)+sq(Xl))/RW= 18665.

Revathi said:   5 years ago
I am not understanding please explain it.

Puja said:   6 years ago

Can you tell me what is Q? And by using your formula I get an answer is 18660.192 but I have a confusion this answer is right or wrong.

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