Electrical Engineering - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 11)
11.
If the resistance in parallel with a parallel resonant circuit is reduced, the bandwidth
Discussion:
16 comments Page 2 of 2.
LAXMI said:
8 years ago
Q= R/(2pifL) when R decreases Q also decrease its value. And BW= fr/Q so decrease in the value of Q would increase the BW.
Gangadhar said:
7 years ago
Here;
Q = (XL/R)
BW = Fr/Q there fore
BW = (Fr*R) / XL
From the above equation, it is clear the BW is directly proportional to Resistor(R).
Q = (XL/R)
BW = Fr/Q there fore
BW = (Fr*R) / XL
From the above equation, it is clear the BW is directly proportional to Resistor(R).
Ashwini said:
6 years ago
BW=fr/Q.
Q=2πfrL/R.
BW=frR/2πfrL.
=R/2πL.
So Resistance is proportional to Bandwidth.
As Resistance decreases Bandwidth also decreases.
Q=2πfrL/R.
BW=frR/2πfrL.
=R/2πL.
So Resistance is proportional to Bandwidth.
As Resistance decreases Bandwidth also decreases.
HAris said:
6 years ago
Simply= BW=F/Q.
Q=1/R.√(L/C).
Put in BW,
BW=F/(1/R).
B.W= F*R.
SO, BW is directly proportional to Resistance.
Q=1/R.√(L/C).
Put in BW,
BW=F/(1/R).
B.W= F*R.
SO, BW is directly proportional to Resistance.
Gopal Gupta said:
5 years ago
@All.
For those who are saying Q=XL/R. I want to correct them. This Q is for series RLC ckt. Given question is about parallel. So here Q=XC*R.
For those who are saying Q=XL/R. I want to correct them. This Q is for series RLC ckt. Given question is about parallel. So here Q=XC*R.
Pavan said:
4 years ago
B.W = R/2 * π * L.
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