Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 26)
26.
What is the bandwidth in the given circuit. If the winding resistance of the coil is 12 
?
Discussion:
4 comments Page 1 of 1.
Emil Lagman said:
1 decade ago
R = 150+12 = 162 ohms.
BW = R/2*pi*L.
BW = 162/2*3.1416*4 mH = 6.44 kHz.
BW = R/2*pi*L.
BW = 162/2*3.1416*4 mH = 6.44 kHz.
Monalisha said:
1 decade ago
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.
f = 1/2*pi*sqrt(L*C) = 53650HZ = 53.65KHZ.
q = 2*Pi*f*L/R = 8.32.
B.W = f/q = 6.4KHZ.
Kiran said:
1 decade ago
fr=1/2*pi*sqrt(LC),
q=1/r*sqrt(L/C),
r=150+12=162;
B.W= fr/q=6.4khz
q=1/r*sqrt(L/C),
r=150+12=162;
B.W= fr/q=6.4khz
DEEPTI said:
1 decade ago
R/2*3.141*L
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