Answer RLC Circuit Q Factor 70, 5V Supply at 2.5kHz

[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?

 

[rude man]

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 correct
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?

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.

 

[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

 

[rude man]

Q = ωL/R so L = QR/ω.

ω = 2πf and f = 2.5e3 Hz.

Then LC = 1/ω² to get C.

 

[charger9198]

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

 

[rude man]

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

Yer' welcome!

 

[Luk-e-2012]

Hi people. Did you get this question correct?

 

[rude man]

Hi people. Did you get this question correct?

Yes. Look at the last two posts.

 

[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/ω²I 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.

 

[rude man]

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.

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.

 

[saucysaunders]

Another way to complete

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.

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

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

Resonance -f_o/r = 1 / 2π√LC

∴ C = (1/-f_o/r)² / ((2π)² * L)

C = (1/2500)² / ((2)² * 1.87)

C = 2.167ηF

 

[rude man]

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