# 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?

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)

Abhi s g said:
5 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.

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.

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

AJEET KUMAR SHUKLA said:
1 decade ago

Will there be no effect of capacitor on the bandwidth?

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.

Surya said:
8 years ago

You mentioned 60 of capacitor. Pf means what?

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)

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

HAris said:
3 years ago

Bandwidth= Resonant Frequency/q factor.

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