Find the inductance and capacitance of an RLC/LC circuit

In summary: I don't really understand them):Ok well if I do that, this is what I get (I left out the units on here because it gets confusing and I... I don't really understand them):L = (100 * sqrt(LC)) / (2 * R)C = 100 * sqrt(LC)
  • #1
Parad0x88
74
0

Homework Statement


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.


Homework 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)

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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  • #2
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.
 
  • #3
gneill said:
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?
 
  • #4
Parad0x88 said:
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 :smile:
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.
 
  • #5
gneill said:
Then you can probably assume it's a series RLC configuration, as it's the most common one used in problem sets :smile:


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!)
 
  • #6
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.
 
  • #7
WallyP said:
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? :wink:
 
  • #8
gneill said:
What's wrong with easy? :wink:

That is true! :approve:
 
  • #9
WallyP said:
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
 
  • #10
It definitely is an interesting one.
 
  • #11
WallyP said:
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?
 
  • #12
Pay attention to the hint! :

"Hint: be aware of the difference between regular and angular frequencies."
 
  • #13
gneill said:
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?
 
Last edited:
  • #14
Parad0x88 said:
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.
 
  • #15
Parad0x88 said:
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).
 
  • #16
gneill said:
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
 
  • #17
Parad0x88 said:
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.
 
  • #18
gneill said:
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!
 

1. What is an RLC/LC circuit?

An RLC/LC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C). These three components are connected in either series or parallel, and their interactions create a resonant frequency.

2. Why is it important to find the inductance and capacitance of an RLC/LC circuit?

Knowing the inductance and capacitance of an RLC/LC circuit is crucial in understanding its behavior and performance. It helps in designing and optimizing the circuit for specific applications, such as in filtering or frequency selection.

3. How do you calculate the inductance and capacitance of an RLC/LC circuit?

The inductance and capacitance of an RLC/LC circuit can be calculated using the formulas L=XL/ω and C=1/ω^2C, where XL is the inductive reactance, ω is the angular frequency, and C is the capacitance. XL and ω can be determined using the values of the inductor and capacitor in the circuit, while C can be measured using a capacitance meter.

4. What factors affect the inductance and capacitance of an RLC/LC circuit?

The inductance of an RLC/LC circuit is affected by the number of turns in the inductor, the core material, and the physical dimensions of the inductor. On the other hand, the capacitance is affected by the distance between the plates of the capacitor, the type of dielectric material used, and the size of the plates.

5. Can the inductance and capacitance of an RLC/LC circuit be changed?

Yes, the inductance and capacitance of an RLC/LC circuit can be changed by altering the physical properties of the components. For example, the inductance can be increased by increasing the number of turns in the inductor or using a ferromagnetic core. The capacitance can be increased by using a larger distance between the plates or a material with a higher dielectric constant.

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