Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 2)
2.
What is the bandwidth of the circuit?
Discussion:
34 comments Page 3 of 4.
Govind Singh Chouhan said:
1 decade ago
Data Given:
R = 1k.
C = 1uF.
L = 5H.
So, resonance freq. is = 1/2*pi*sqrt(LC).
(fo is a resonance freq.)
fo = 1/2*3.14*sqrt(5*1*10^-6)).
fo = 1/2*3.14*sqrt(0.000005).
fo = 71.42Hz.
we know about the Quality factor "Q",
Q = WoL/R.
Q = 2*pi*fo*5/1k.
Q = 2.24.
The relationship between Q,fo and BW is:
Q = fo/BW.
So BW = fo/Q.
BW = 71.42/2.24.
BW = 31.88Hz.
R = 1k.
C = 1uF.
L = 5H.
So, resonance freq. is = 1/2*pi*sqrt(LC).
(fo is a resonance freq.)
fo = 1/2*3.14*sqrt(5*1*10^-6)).
fo = 1/2*3.14*sqrt(0.000005).
fo = 71.42Hz.
we know about the Quality factor "Q",
Q = WoL/R.
Q = 2*pi*fo*5/1k.
Q = 2.24.
The relationship between Q,fo and BW is:
Q = fo/BW.
So BW = fo/Q.
BW = 71.42/2.24.
BW = 31.88Hz.
Manjunath said:
1 decade ago
Bandwidth = R/2*pi*L . BW= 1/2*3.14*5 = 31.8 Hz.
Alka said:
1 decade ago
BW=fr/Q
Q=(2*pi*fr*L)/R
Bw=R/(2*pi*L)
=31.83
Q=(2*pi*fr*L)/R
Bw=R/(2*pi*L)
=31.83
Sreeyush Sudhakaran said:
1 decade ago
B.W=fr/Q
Q=(2*Pi*fr*L)/R
B.W = fr/((2*Pi*fr*L)/R) = R/2*Pi*L = 1000/2*3.14*5 = 31.8Hz
Q=(2*Pi*fr*L)/R
B.W = fr/((2*Pi*fr*L)/R) = R/2*Pi*L = 1000/2*3.14*5 = 31.8Hz
FADI said:
1 decade ago
B.W=FR/Q
Q=(1/R)SQRT(L/C)
FR=1/2*PI*SQRT(LC)
B.W=[1/2*PI*SQRT(LC)]/(1/R)SQRT(L/C)
B.W=R*SQRT C / 2*PI*L*SQRT C
B.W=R/2*PI*L
Q=(1/R)SQRT(L/C)
FR=1/2*PI*SQRT(LC)
B.W=[1/2*PI*SQRT(LC)]/(1/R)SQRT(L/C)
B.W=R*SQRT C / 2*PI*L*SQRT C
B.W=R/2*PI*L
Amrita said:
1 decade ago
BANDWIDTH= Resonant freq/ Quality Factor
Resonant Frequency= 1/(2*pi*sqrt of(L*C))
Quality Factor= 1/R*(sqrt(L/C))
Using above fomula we can determine, Bandwidth=31.83 .
Resonant Frequency= 1/(2*pi*sqrt of(L*C))
Quality Factor= 1/R*(sqrt(L/C))
Using above fomula we can determine, Bandwidth=31.83 .
M.V.KRISHNA said:
1 decade ago
BW=fr/Q;
fr=1/(2*pi*underroot(L*C));
Q=(1/R)*underroot(L/C);
BW=R/(2*pi*L);
=>1Kohm/(2*pi*5H);
=32.3Hz
fr=1/(2*pi*underroot(L*C));
Q=(1/R)*underroot(L/C);
BW=R/(2*pi*L);
=>1Kohm/(2*pi*5H);
=32.3Hz
Sandeep said:
1 decade ago
Q=Fr/BW;
BW=Fr/Q;
Fr=1/2*PI*ROOT(L*C);
Q=(1/R)(ROOT(L/C));
BW=Fr/Q;
Fr=1/2*PI*ROOT(L*C);
Q=(1/R)(ROOT(L/C));
Seenivasagan said:
1 decade ago
Bandwidth(BW)=fr/Q
Q=(2*pi*fr*L)/R
fr=1/(2*pi*root LC)
by applying this ,
we can get fr=71.21
Q=2.236
BW=fr/Q
=71.21/2.236
=31.84
approximately 31.8
Q=(2*pi*fr*L)/R
fr=1/(2*pi*root LC)
by applying this ,
we can get fr=71.21
Q=2.236
BW=fr/Q
=71.21/2.236
=31.84
approximately 31.8
Oneidensis said:
1 decade ago
My answer is 200... iIve been solving this so many times.. here's the formulas I used,
f=1/(square roof of (LC))
then,
Q= (1/fRC) ; Q= (fL/R)
then,
B=f/Q... B=200
f=1/(square roof of (LC))
then,
Q= (1/fRC) ; Q= (fL/R)
then,
B=f/Q... B=200
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