Find the inductance and capacitance of an RLC/LC circuitby Parad0x88 Tags: capacitance, circuit, inductance, rlc or lc 

#1
Nov1312, 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 Q_{0} = I can assume Q_{0} 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 Q_{0} (which I did above) and I have the angular frequency of the LC circuit, but I feel like I am missing some information. 



#2
Nov1312, 09:23 AM

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P: 11,446

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
Nov1312, 09:30 AM

P: 74

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
Nov1312, 09:43 AM

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P: 11,446

Find the inductance and capacitance of an RLC/LC circuit 



#5
Nov1312, 09:55 AM

P: 74





#6
Nov1312, 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.




#7
Nov1312, 07:21 PM

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P: 11,446





#8
Nov1312, 07:28 PM

P: 5





#9
Nov1312, 07:30 PM

P: 74





#10
Nov1312, 07:45 PM

P: 5

It definitely is an interesting one.




#11
Nov1312, 11:18 PM

P: 74

Q = 2pi * 100/1 = 628.319 and R = 2Ω We have Q = (ω_{0}L) / R 628.319 = (1000 X L) / 2 1256.638 = 1000L L = 1.2566 And then ω_{0} = 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
Nov1312, 11:31 PM

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P: 11,446

Pay attention to the hint! :
"Hint: be aware of the difference between regular and angular frequencies." 



#13
Nov1412, 12:38 AM

P: 74

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? 



#14
Nov1412, 01:34 AM

P: 296

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^{at}sin(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 0.5CV^{2} since 0.5Li^{2}=0. So energy at some time t0 will be 0.5CV^{2} 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
Nov1412, 06:36 AM

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#16
Nov1412, 07:02 AM

P: 74

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
Nov1412, 07:11 AM

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P: 11,446





#18
Nov1412, 07:25 AM

P: 74




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