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?
138 MHz
10,866 Hz
1,907 Hz
138 kHz
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
16 comments Page 1 of 2.

MADHUKUMAR S said:   2 years ago
BW = R/L for a series RLC circuit.

HAris said:   3 years ago
Bandwidth= Resonant Frequency/q factor.

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.

SHWETHA said:   8 years ago
BW = R/2*3.142*L HZ.

Arun said:   8 years ago
Pico farad. Pico = 10^-12.

Surya said:   8 years ago
You mentioned 60 of capacitor. Pf means what?

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)

Rani S said:   8 years ago
BW = R/(2*Pi*L).
i.e, BW = 10851.47 = 10852 Hz.

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.

Rajendar said:   1 decade ago
BW=f/Q
=R/(2*pi*L) (since Q=(2*pi*L)f/R)


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