Electronics - RLC Circuits and Resonance - Discussion

26. 

What is the bandwidth in the given circuit. If the winding resistance of the coil is 12 omega.gif?

[A]. 6.4 kHz
[B]. 47.2 kHz
[C]. 53.7 kHz
[D]. 60.2 kHz

Answer: Option A

Explanation:

No answer description available for this question.

Deepti said: (May 18, 2011)  
R/2*3.141*L

Kiran said: (Nov 2, 2011)  
fr=1/2*pi*sqrt(LC),
q=1/r*sqrt(L/C),
r=150+12=162;
B.W= fr/q=6.4khz

Monalisha said: (Dec 21, 2013)  
We know, q=wL/R OR 2*pi*f*L/R = f/B.W.

f = 1/2*pi*sqrt(L*C) = 53650HZ = 53.65KHZ.

q = 2*Pi*f*L/R = 8.32.

B.W = f/q = 6.4KHZ.

Emil Lagman said: (Mar 5, 2015)  
R = 150+12 = 162 ohms.

BW = R/2*pi*L.

BW = 162/2*3.1416*4 mH = 6.44 kHz.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.