Electrical Engineering - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 3)
3.
A 15 
resistor, a 220 
H coil, and a 60 pF capacitor are in series across an ac source. What is the bandwidth of the circuit?
resistor, a 220 H coil, and a 60 pF capacitor are in series across an ac source. What is the bandwidth of the circuit?Discussion:
16 comments Page 1 of 2.
MADHUKUMAR S said:
4 years ago
BW = R/L for a series RLC circuit.
HAris said:
5 years ago
Bandwidth= Resonant Frequency/q factor.
Abhi s g said:
6 years ago
XL=2 * π * f * l.
XC=1/(2*π*f*c),
when XL=XC.
f=1/2* π *(LC)^1/2,
And;
Q=1/R * (L/C)^1/2.
BW=f/Q.
XC=1/(2*π*f*c),
when XL=XC.
f=1/2* π *(LC)^1/2,
And;
Q=1/R * (L/C)^1/2.
BW=f/Q.
(1)
SHWETHA said:
9 years ago
BW = R/2*3.142*L HZ.
(1)
Arun said:
9 years ago
Pico farad. Pico = 10^-12.
(1)
Surya said:
9 years ago
You mentioned 60 of capacitor. Pf means what?
Rammohan rao said:
10 years ago
If it would be the question regarding series RLC resonance circuit then the bandwidth of the circuit should be R/L. Here the question is not clear.
(2)
Rani S said:
10 years ago
BW = R/(2*Pi*L).
i.e, BW = 10851.47 = 10852 Hz.
i.e, BW = 10851.47 = 10852 Hz.
Rimjhim said:
1 decade ago
BW = f/Q.
Q = (1/R)(L/C)^1/2.
= 127.65.
f = 1/2*3.14*(L/C)^1/2.
= 1385265.971
f/Q = 10852 = Answer.
Q = (1/R)(L/C)^1/2.
= 127.65.
f = 1/2*3.14*(L/C)^1/2.
= 1385265.971
f/Q = 10852 = Answer.
(4)
Rajendar said:
1 decade ago
BW=f/Q
=R/(2*pi*L) (since Q=(2*pi*L)f/R)
=R/(2*pi*L) (since Q=(2*pi*L)f/R)
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