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 3 of 4.
Stuti kushwaha said:
9 years ago
Answer is not getting please help me.
Likhitha said:
9 years ago
Anyone please tell me simplest and understandable answer.
Vani.c said:
9 years ago
Anyone please explain this problem with proper equation.
Bharath said:
9 years ago
Please show how to make the conversion without calculator.
Puja said:
9 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.
Revathi said:
8 years ago
I am not understanding please explain it.
(1)
Rasel said:
7 years ago
Here, rp= (sq(Rw)+sq(Xl))/RW= 18665.
(1)
KONDAVENI SANTHOSH said:
7 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.
Arunkumar said:
7 years ago
What is X value in G and J value in Y?
Please explain clearly.
Please explain clearly.
Muhammad said:
7 years ago
By using this equation1/Z=(1/R)2+(1/XL2-1/XC2)2 we get the answer.
(1)
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