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 1 of 4.
Sachin said:
3 years ago
W2-W1 = R/L.
BandWidth = F2 - F1 = R/(2 * π * L).
BandWidth = F2 - F1 = R/(2 * π * L).
(1)
Jyothi said:
5 years ago
BW = R/(2*π*L),
BW = (1*10^3)/(2*π*5),
=31.83Hz.
BW = (1*10^3)/(2*π*5),
=31.83Hz.
(1)
Sap said:
1 decade ago
Hussain you are right.
(1)
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)
Mody shaban said:
8 years ago
HOW?
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.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers