
#1
Aug2408, 01:14 PM

P: 18

1. The problem statement, all variables and given/known data
A series RLC circuit has a resonant frequency of 6.00 kHz and a resistance of 575 ohms. When connected to an AC power supply that varies the voltage at 8.00 kHz, its impedence is 1.00 kOhms. What are the capacitance and inductance of the circuit? 2. Relevant equations f_{0}=resonant frequency R = resistance f_{v}=frequency of varied voltage Z = impedence C = capacitance L = inductance X_{L} = inductive reactance X_{C} = capacitive reactance f_{0}= 1/(2pi x sqrt (LC)) Z = sqrt(R^{2}+(X_{L}X_{C}) X_{L} = (2pi)f_{v}L X_{C} = 1/ (2pi x f_{v}C) 3. The attempt at a solution I don't even know where to start! 



#2
Aug2408, 01:52 PM

HW Helper
P: 5,346

At resonance what do you know about the relationship between L and C? Is there a relationship that you can derive from knowing that the impedence of an RLC at resonance is equal to R? 



#3
Aug2408, 02:11 PM

P: 18

Since Z=R, then X_{L}=X_{C}
Using an equation I found in the book, this would mean that: 2pif_{0}L=1/(2pif_{0}C) This gets me back to the equation: f_{0}=1/(2pi x sqrt(LC)) 



#4
Aug2408, 02:51 PM

P: 127

RLC Circuit
You have several equations there, and two given situations for you to put in some known values.
The thing to know about resonance is that the relation between L and C given by f0= 1/(2pi x sqrt (LC)) becomes true at that one frequency where all the reactive bits cancel out, leaving us only with a resistive part. The inductive reactance becomes equal in value to the capacitive reactance, but of opposite sign. The sign part is not shown in your capacitive reactance formula. Anyways, it reveals the value of the resistor. So now move on to the new condition at 8kHz. You now have enough information to figure the new impedance. They are all in series. 2 equations with 2 unknowns. Have another try.. 



#5
Aug2408, 02:58 PM

P: 18

I'm sorry, but what you just wrote makes absolutely no sense to me. I'm not trying to find impedence and I don't understand how the inductive reactance= capacitive reactance reveals the value of a resistor.




#6
Aug2408, 03:02 PM

HW Helper
P: 5,346

Now going back to your equation for impedence (Z) you know the value of R, from the resonance condition and you know the value of L in terms of C, or C in terms of L your choice, which now allows you to calculate what you are asked. (Note this equation depends on the new frequency in calculating the L and C reactances.) 



#7
Aug2408, 03:05 PM

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P: 5,346





#8
Aug2408, 03:10 PM

HW Helper
P: 5,346

http://hyperphysics.phyastr.gsu.edu/hbase/hframe.html What he is trying to explain to you is that the Inductive Reactance AT RESONANCE is equal in magnitude and opposite in sign to the Capacitive Reactance. Hence the impedence is equal to the Resistance and just the Resistance. 



#9
Aug2408, 03:21 PM

P: 18

So what I've done, and I have no idea if this is correct:
f_{0}=1/(2pi x sqrt(LC)) C = sqrt(1/f_{0}2piL Z = sqrt(R^{2} + (X_{L}X_{C})^{2}) simplifies to: sqrt(Z^{2} + R^{2}) x C = L substituting C: [sqrt (Z^{2} + R^{2}) / f_{0} x 2piL] = L I ended up getting 0.147 H = L Is that heading in the right direction? 



#10
Aug2408, 03:27 PM

HW Helper
P: 5,346

C will be equal to 1/(L*w^2) where w is 2pi*f 



#11
Aug2408, 03:39 PM

P: 18

Okay, so making that adjustment, it becomes:
(sqrt Z^{2}+R^{2}) (1/L(2pi*f_{0})^{2}) = L To factor out: sqrt Z^{2}+R^{2} = (2pi*f_{0})^{2}L^{2} Then: (sqrt Z^{2}+R^{2})/(2pi*f_{0})^{2} = L^{2} And then take the square root of the left side to get: 7.58 x 10^{4} H = L Am I supposed to be using F_{0} or F_{v} for this equation? 



#12
Aug2408, 03:53 PM

HW Helper
P: 5,346

And the Reactances XL and XC depend on the f1 = 8000 hz. Your equation should be: Z^2 = R^2 + (XL  XC)^2 Where XL = w1L and XC = 1/w1C Then you can use C = 1/(wo^2L) to come up with an expression for XC in terms of L. Then you can solve for L. Then C = 1/(wo^2L). 



#13
Aug2408, 04:10 PM

P: 18

Z^{2}=R^{2}+(wL(1/wC))^{2}
Z^{2}R^{2}=(wL1/wC)^{2} sqrt(Z^{2}R^{2})=wL1/wC sqrt(Z^{2}R^{2}) + 1/wC = wL [sqrt(Z^{2}R^{2}) + 1/wC] /w = L substituting in C: [sqrt(Z^{2}R^{2}) + 1/w(1/w_{0}^{2}L)]/w = L I think this would simplify to: [sqrt(Z^{2}R^{2} + w_{0}^{2}L/w]/w = L 



#14
Aug2408, 04:20 PM

HW Helper
P: 5,346

I haven't calculated it ( and won't). So I will leave you to the calculation of it. I would note you've come a long way from not having any idea what to do. Good Luck. 



#15
Aug2408, 04:53 PM

P: 18

Just one more quick thing, I'm not sure how I would factor L out of the left hand side without cancelling out the L on the right hand side.




#16
Aug2408, 05:13 PM

HW Helper
P: 5,346

sqrt(Z^{2}R^{2})=wL1/wC When you replace C with 1/w_{o}^{2}*L that gives: sqrt(Z^{2}R^{2}) = wLw_{o}^{2}*L/w = wL*(1 w_{o}^{2}/w^{2} ) 


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