# RLC Circuits question

1. Jan 22, 2012

### charger9198

A series RLC circuit is connected to a 5 V supply, the frequency of the supply is adjusted to give a maximum current of 11.9 mA at 2.5 kHz. The Q factor is 70. Determine the component values of the circuit.

R= 5/.0119=420.2 Ohm
Q = (1/R)*(sqrt L/C)
70 = (1/420.2)*(sqrt L/C)
70/(1/420.2)= 29414
L/C = 865183396
L = 865183396*C
Resonate frequency
=1/[(2pi)*(sqrt L*C)]
2500=1/[(6.28)*(sqrt ((865183396*C)*C)]
2500=.159/(29414*C)
29414*C=.159/2500
29414*C=.0000636
C=.002uF
L=865183396*C
L=1.73 H

Is this an ok way to calculate this? Does it appear correct?

Last edited: Jan 22, 2012
2. Jan 22, 2012

### rude man

Your values for L and C are close but off just a bit: I got L = 1.87 Hy and C = 2.17nF

I think you went about it the hard way: start with R as you did; realize that f = 2500 Hz so ω = 2pi*2500 rad/s; then use relation between f, L, R and Q to get L, then use relationship between L, C and ω to get C.

3. Jan 23, 2012

### charger9198

Thanks, i agree i did think that. do you mind me asking what equation you transposed for L? All I can come up with is that involving capacitance

4. Jan 23, 2012

### rude man

Q = ωL/R so L = QR/ω.

ω = 2πf and f = 2.5e3 Hz.

Then LC = 1/ω2 to get C.

5. Jan 24, 2012

### charger9198

Thanks rude man, I now can confirm I get the same as you
L=1.872 Hy
C=2.165 nF

6. Jan 24, 2012

### rude man

Yer' welcome!

7. Sep 17, 2013

### Luk-e-2012

Hi people. Did you get this question correct?

8. Sep 17, 2013

### rude man

Yes. Look at the last two posts.

9. Nov 14, 2013

### saucysaunders

How did you find the equation for C

Hi rude man.

I've been looking at this question for a while now and am yet to see how you found the equation LC =1/ω2

I assume you found it from 2πfL/R = 1/R x (√L/c)
But I cannot confirm this is the case.
I tried to find the capacitance using L/QR2 but I get a different answer.

Can you help me? The answer is there I know but I want to understand.

10. Nov 14, 2013

### rude man

A series R-L-C circuit has impedance √[R^2 + (wL - wC)^2] which is mimimized when wL = 1/wC. Minimum impedance occurs at the resonant frequency.

11. Nov 20, 2013

### saucysaunders

Another way to complete

Thanks! I understand where you're coming from now.

I also managed to calculate the capacitor value using the resonance formula.

Resonance -fo/r = 1 / 2π√LC

∴ C = (1/-fo/r)2 / ((2π)2 * L)

C = (1/2500)2 / ((2)2 * 1.87)

C = 2.167ηF

12. Nov 20, 2013

### rude man

Without actually checking your numbers you seem to be on-course.