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
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), is1,878
18,780
18,750
626
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
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.
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
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.
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.
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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