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.
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.
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.
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.
Jatin choudhary said:
9 years ago
f = R/2*3.14*L.
(1)
Ashish said:
8 years ago
Here, r/2 * π * L = 31.8.
(1)
Vijay said:
8 years ago
Q=(omega*L)/R.
BW=f/Q,
=> BW=(f*R)/(w*L)
=> BW=R/(2*pi*L)
=> Bw=31.8.
BW=f/Q,
=> BW=(f*R)/(w*L)
=> BW=R/(2*pi*L)
=> Bw=31.8.
Mody shaban said:
8 years ago
HOW?
Ayush Purwar said:
8 years ago
Bandwidth= R/(2*L)...Hz.
= 1000/(2*3.14*5).
= 31.84 Hz.
= 1000/(2*3.14*5).
= 31.84 Hz.
SUMA said:
7 years ago
Here, R/(2*π*L) = 31.8.
PRITAM said:
5 years ago
Q=XL/R.
BW=fr/Q.
fr=1/2*pi*√LC=71.17.
XL=WL=2235.87,
Q=XL/R=2.23.
B=71.17/2.23=31.83.
BW=fr/Q.
fr=1/2*pi*√LC=71.17.
XL=WL=2235.87,
Q=XL/R=2.23.
B=71.17/2.23=31.83.
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