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.
Rammohan rao said:
8 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.
(1)
Amrit kumar mohanta said:
1 decade ago
BW = R/(2*Pi*L)
Shivani said:
1 decade ago
Whats this Pi in last formula?
AJEET KUMAR SHUKLA said:
1 decade ago
Will there be no effect of capacitor on the bandwidth?
Sauvik biswas said:
1 decade ago
Q=1/R*(L/C)^1/2
f=1/(2*pi*(LC)^1/2)
BW=f/Q
f=1/(2*pi*(LC)^1/2)
BW=f/Q
PARTHU said:
1 decade ago
BW= 1/[Q*(2*3.14*(LC)^0.5]
NETHRA said:
1 decade ago
FOR SERIES CIRCUIT Q FACTOR=1/R*(C/L)^1/2
F=1/92*PI*(LC)^1/2
BW=F/Q
F=1/92*PI*(LC)^1/2
BW=F/Q
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)
Rimjhim said:
9 years 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.
Rani S said:
8 years ago
BW = R/(2*Pi*L).
i.e, BW = 10851.47 = 10852 Hz.
i.e, BW = 10851.47 = 10852 Hz.
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