Find the inductance and capacitance of an RLC/LC circuit


by Parad0x88
Tags: capacitance, circuit, inductance, rlc or lc
Parad0x88
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#1
Nov13-12, 08:45 AM
P: 74
1. The problem statement, all variables and given/known data
The energy of the RLC circuit decreases by 1% during each oscillation when R=2 Ohms. If this resistance is removed, the resulting LC circuit oscillates at a frequency of 1 kHz. Find the inductance and the capacitance. Hint: be aware of the difference between regular and angular frequencies.


2. Relevant equations
R = 2Ω
LC frequency, ω = 1000 Hz
Q0 = I can assume Q0 to be 100 by he way the problem is stated
ω = 1 / √(LC)

3. The attempt at a solution
I'm really confused by this problem. I'm given R, I can find Q0 (which I did above) and I have the angular frequency of the LC circuit, but I feel like I am missing some information.
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gneill
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#2
Nov13-12, 09:23 AM
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Is it a series or parallel RLC circuit?

The Q of the circuit is given by: ##Q = 2\pi \frac{Stored\;Energy}{Dissipated\;Energy}##

There is a relationship between the Q of a circuit and the damping factor ##\zeta## (or the attenuation factor, ##\alpha##) that may help you. The attenuation factor is also related to the inductance L and the resistance R. Pay attention to the given hint.
Parad0x88
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#3
Nov13-12, 09:30 AM
P: 74
Quote Quote by gneill View Post
Is it a series or parallel RLC circuit?

The Q of the circuit is given by: ##Q = 2\pi \frac{Stored\;Energy}{Dissipated\;Energy}##

There is a relationship between the Q of a circuit and the damping factor ##\zeta## (or the attenuation factor, ##\alpha##) that may help you. The attenuation factor is also related to the inductance L and the resistance R. Pay attention to the given hint.
I copy/pasted the whole problem, so it's not mentioned if it's series or parallel

I could find Q with the formula you gave me, if the system loses 1%, then it would be:

Q = 2pi X 100/1 = 2pi X 100

Attenuation factor, that's not even something we covered in class... Do you simply mean the resistance?

gneill
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#4
Nov13-12, 09:43 AM
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Find the inductance and capacitance of an RLC/LC circuit


Quote Quote by Parad0x88 View Post
I copy/pasted the whole problem, so it's not mentioned if it's series or parallel
Then you can probably assume it's a series RLC configuration, as it's the most common one used in problem sets
I could find Q with the formula you gave me, if the system loses 1%, then it would be:

Q = 2pi X 100/1 = 2pi X 100

Attenuation factor, that's not even something we covered in class... Do you simply mean the resistance?
Attenuation factor and damping factor are names of convenience assigned to parameters of the differential equation describing the circuit. Take a look at the wikipedia article "RLC Circuit", and in particular the section "Series RLC Circuit". Then look at the article on "Q Factor" and the section "Physical Interpretation of Q" for some handy relationships.
Parad0x88
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#5
Nov13-12, 09:55 AM
P: 74
Quote Quote by gneill View Post
Then you can probably assume it's a series RLC configuration, as it's the most common one used in problem sets


Attenuation factor and damping factor are names of convenience assigned to parameters of the differential equation describing the circuit. Take a look at the wikipedia article "RLC Circuit", and in particular the section "Series RLC Circuit". Then look at the article on "Q Factor" and the section "Physical Interpretation of Q" for some handy relationships.
I don't think I have the proper tools to answer that question just yet, all of this is Chinese to me... The teacher didn't really cover it, he just skimmed over, guess I'll have to read the book. Even the concepts you mentioned, or what's in wikipedia, have not been covered in class (and actually don't seem to be covered in the book, but that's what wikipedia and youtube are for!)
WallyP
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#6
Nov13-12, 07:14 PM
P: 5
After calculating the quality factor using the above listed equation for Q, wouldn't it be possible to calculate the inductance by solving for L in the equation ## Q= {\frac{w_oL}{R}} ## and then using the value of inductance to find the capacitance using the equation ## w_o = {\frac{1}{\sqrt{LC}}} ## ? This way just seems too easy.
gneill
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#7
Nov13-12, 07:21 PM
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Quote Quote by WallyP View Post
After calculating the quality factor using the above listed equation for Q, wouldn't it be possible to calculate the inductance by solving for L in the equation ## Q= {\frac{w_oL}{R}} ## and then using the value of inductance to find the capacitance using the equation ## w_o = {\frac{1}{\sqrt{LC}}} ## ? This way just seems too easy.
What's wrong with easy?
WallyP
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#8
Nov13-12, 07:28 PM
P: 5
Quote Quote by gneill View Post
What's wrong with easy?
That is true!
Parad0x88
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#9
Nov13-12, 07:30 PM
P: 74
Quote Quote by WallyP View Post
After calculating the quality factor using the above listed equation for Q, wouldn't it be possible to calculate the inductance by solving for L in the equation ## Q= {\frac{w_oL}{R}} ## and then using the value of inductance to find the capacitance using the equation ## w_o = {\frac{1}{\sqrt{LC}}} ## ? This way just seems too easy.
That's an interesting formula you got there! I didn't see that in my book o_O
WallyP
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#10
Nov13-12, 07:45 PM
P: 5
It definitely is an interesting one.
Parad0x88
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#11
Nov13-12, 11:18 PM
P: 74
Quote Quote by WallyP View Post
It definitely is an interesting one.
Ok well if I do that, this is what I get (I left out the units on here because it gets confusing and I don't know how to do it like you guys:

Q = 2pi * 100/1 = 628.319
and R = 2Ω

We have Q = (ω0L) / R
628.319 = (1000 X L) / 2
1256.638 = 1000L
L = 1.2566

And then

ω0 = 1 / √(LC)
1000 = 1 / √(1.2566C)
1,000,000 = 1 / 1.2566C
1,256,638 = 1 / C
C = 1 / 1,256,638
C = 7.96 X 10-7

Would that be it?
gneill
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#12
Nov13-12, 11:31 PM
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Pay attention to the hint! :

"Hint: be aware of the difference between regular and angular frequencies."
Parad0x88
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#13
Nov14-12, 12:38 AM
P: 74
Quote Quote by gneill View Post
Pay attention to the hint! :

"Hint: be aware of the difference between regular and angular frequencies."
Isn't that what I did? The formula for frequency is the one for the LC circuit, and the information for the frequency is given for the LC circuit.

Edit: Oh wait, is it on the first formula that lies the problem, since I'm finding L in an RLC circuit with the frequency for an LC circuit?

If that is the case, I have to find a way to convert the frequency from LC data to RLC data, or the Q data from RLC to LC data. Is that it?
aralbrec
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#14
Nov14-12, 01:34 AM
P: 296
Quote Quote by Parad0x88 View Post
That's an interesting formula you got there! I didn't see that in my book o_O
These come from the standard form for the response to a second order differential equation.

If you find the voltage across the capacitor in a series RLC circuit you will find the transfer function looks like:

##\frac{\frac{1}{LC}}{s^2+s\frac{R}{L}+\frac{1}{LC}}##

Here you can see if R=0, the response has poles on the imaginary axis so there will be oscillation of frequency ##\frac{1}{\sqrt LC}##


The standard normalized (gain=1 at dc) second order response is:

##\frac{wn^2}{s^2+s(\frac{wn}{Q})+wn^2}##

Comparing the factor multiplying s, you can find Q in terms of R, L and wn.


You have an equation for Q in terms of energy but if you did not know that you could look at this the long way. The natural response in the time domain (impulse response, inverse laplace) is a damped sinusoid meaning e-atsin(wt) where a and w are functions of Q and wn (look it up if needed).

Since this response is the voltage across the capacitor in a series circuit, or so I have assumed, the current through the circuit is the current through the capacitor i=Cdv/dt. When the voltage waveform is at a peak, i=0. This means the total energy stored in the circuit at these peak times is just 0.5CV2 since 0.5Li2=0. So energy at some time t0 will be 0.5CV2 and then one period later, 10% less. The e-at term will tell you how much the voltage on the capacitor has decreased one period later so that you will have another equation for L,C. This is where that energy equation for Q comes from.
gneill
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#15
Nov14-12, 06:36 AM
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Quote Quote by Parad0x88 View Post
Isn't that what I did? The formula for frequency is the one for the LC circuit, and the information for the frequency is given for the LC circuit.

Edit: Oh wait, is it on the first formula that lies the problem, since I'm finding L in an RLC circuit with the frequency for an LC circuit?

If that is the case, I have to find a way to convert the frequency from LC data to RLC data, or the Q data from RLC to LC data. Is that it?
Nope. That's not it. What the hint is trying to tell you is that f ≠ ω . f expresses cycles per second, while ω expresses radians per second (angular frequency).
Parad0x88
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#16
Nov14-12, 07:02 AM
P: 74
Quote Quote by gneill View Post
Nope. That's not it. What the hint is trying to tell you is that f ≠ ω . f expresses cycles per second, while ω expresses radians per second (angular frequency).
Oh! I see! Well in that case, let's see... There's 2 X pi radians in a cycle, so basically ω = 6283.19... Would that be what I have to use in those forumlas?

Q = 2pi * 100/1 = 628.319
and R = 2Ω

We have Q = (ω0L) / R
628.319 = (6283.19 X L) / 2
solve for L like I previously did

And then

ω0 = 1 / √(LC)
6283.19 = 1 / √(1.2566C)
solve for C like I previously did
gneill
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#17
Nov14-12, 07:11 AM
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Quote Quote by Parad0x88 View Post
Oh! I see! Well in that case, let's see... There's 2 X pi radians in a cycle, so basically ω = 6283.19... Would that be what I have to use in those forumlas?
Yup, that's what the hint was all about; Often we're given frequency f when ω is more convenient for the maths of the problem.
Q = 2pi * 100/1 = 628.319
and R = 2Ω

We have Q = (ω0L) / R
628.319 = (6283.19 X L) / 2
solve for L like I previously did

And then

ω0 = 1 / √(LC)
6283.19 = 1 / √(1.2566C)
solve for C like I previously did
The method looks good.
Parad0x88
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#18
Nov14-12, 07:25 AM
P: 74
Quote Quote by gneill View Post
Yup, that's what the hint was all about; Often we're given frequency f when ω is more convenient for the maths of the problem.


The method looks good.
Sa-weet, thanks dude!


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