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.
Jatin choudhary said:
9 years ago
f = R/2*3.14*L.
(1)
Laurent p .Eranga said:
10 years ago
First find fr = 1/6.28 times square root of LC.
You will got 71.2 hz.
Then find xl = 6.28 times fl you will got 2.23 k if xl = xc.
Then q = xl/r you will got 2.23 k/1 k = 2.235.
If the formula for bw = fr/q.
fr = 71.2 and q = 2.235 so bw = fr/q will be equals to 31.8 hz.
You will got 71.2 hz.
Then find xl = 6.28 times fl you will got 2.23 k if xl = xc.
Then q = xl/r you will got 2.23 k/1 k = 2.235.
If the formula for bw = fr/q.
fr = 71.2 and q = 2.235 so bw = fr/q will be equals to 31.8 hz.
Gaurav Kumar Yadav said:
1 decade ago
As Bandwidth = Resonant frequency/Quality Factor.
And Resonant Frequency = 1/(2*pi*Sqareroot LC).
And Quality factor = 1/R*squareroot(L/C).
Now BW = 1/ (2*pi*Suareroot LC)/1/R*squareroot (L/C) = R/(2*pi*L).
BW = 1000/(2*3.14*5) = 31.8 Hz.
And Resonant Frequency = 1/(2*pi*Sqareroot LC).
And Quality factor = 1/R*squareroot(L/C).
Now BW = 1/ (2*pi*Suareroot LC)/1/R*squareroot (L/C) = R/(2*pi*L).
BW = 1000/(2*3.14*5) = 31.8 Hz.
Krupal gowda jedrali said:
1 decade ago
= r/2.pai.L formula.
= 1000/2*pai*5 = 31.8105 hz.
= 1000/2*pai*5 = 31.8105 hz.
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.
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.
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
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.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers