# Electrical Engineering - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 7)
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
1,878 18,780 18,750 626 Explanation:
No answer description is available. Let's discuss.
Discussion:
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

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

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?

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
XL=2*3.14*f*L.
Xc=1/(2*3.14*f*C).

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
@Mrakovic.

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.