# Urgent help with R-L-C series resonant circuit!

#### Tristanorr

Joined Jan 27, 2011
2
Sorry I posted this in the wrong part earlier today, was in a rush at uni.

After proforming a test at college, I am left to write a report about it. I am now stuck and need help explaining when when you double R-L-C in turn in the circuit you have the same, larger or smaller bandwidth.

Here are my results, i cant show you the circuit as I dont have multisim8 on this PC.

RESULTS Resonant frequency Hz Amplitude at resonance
dB Bandwidth Hz
f2 - f1
R = 80 ohms
L = 40 mH
C = 400 nF 1.254K -38.191 321
Resistance doubled
R = 160 ohms
L = 40 mH
C = 400 nF 1.254K -44.142 645.761
Inductance doubled
R = 80 ohms
L= 80mH
C = 400nF 887.692 -38.212 161.742
Capacitance doubled
R = 80 ohms
L = 40 mH
C = 800nF 887.692 -38.18 324.985

So as you can see, I can tell where the bandwidth is the same or increased / decreased, but I don't understand why if you double the inducatance you get a lower Bandwidth, and if you double the capaistance you get the same... I believe with the resistance doubled you have a much larger bandwidth as you have much less gain (if you plot gain vs. frequncy).

Any help tonight would be very helpful.

#### edgetrigger

Joined Dec 19, 2010
133
The ratio of center frequency(resonant frequency) to its bandwidth is called as q factor

Q = fc /(f2-f1) or f2-f1 = fc/Q --- eq1

Q is also = √L/ (R√C) ----- eq2

with increase in L, Q increases square root times L as per eq2 and as per eq1 bandwidth will decrease.

If C is increased Q will reduce quare root times C as per eq2 so bandwidth will increase.