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 2 of 4.
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 .
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
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
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
Manjunath said:
1 decade ago
Bandwidth = R/2*pi*L . BW= 1/2*3.14*5 = 31.8 Hz.
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.
Shah said:
1 decade ago
B' = R/L in radians.
B' = 1000/5.
B' = 200.
B = B'/2*pi.
B = 200/2*pi.
B = 31.83.
B' = 1000/5.
B' = 200.
B = B'/2*pi.
B = 200/2*pi.
B = 31.83.
Daks said:
1 decade ago
Easy way.
BW = Fr/Q
BW = 1/2*pi*L/R
BW = R/2*pi*L
BW = 31.83
BW = Fr/Q
BW = 1/2*pi*L/R
BW = R/2*pi*L
BW = 31.83
Ruben said:
1 decade ago
So, when you decrease the resistance in the circuit Q will increase and bandwidth and will decrease, but phase angle theta will increase. The circuit efficiency will decrease number.
Dhana said:
1 decade ago
XL=wL.
XL=(2*pi*f)(5H).
XL=2*22/7*5.
XL=31.42 app.
XL=(2*pi*f)(5H).
XL=2*22/7*5.
XL=31.42 app.
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