RLC Circuit


by rissa_rue13
Tags: circuit
rissa_rue13
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#1
Aug24-08, 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
f0=resonant frequency
R = resistance
fv=frequency of varied voltage
Z = impedence
C = capacitance
L = inductance
XL = inductive reactance
XC = capacitive reactance
f0= 1/(2pi x sqrt (LC))
Z = sqrt(R2+(XL-XC)
XL = (2pi)fvL
XC = 1/ (2pi x fvC)

3. The attempt at a solution
I don't even know where to start!
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LowlyPion
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#2
Aug24-08, 01:52 PM
HW Helper
P: 5,346
Quote Quote by rissa_rue13 View Post
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
f0=resonant frequency
R = resistance
fv=frequency of varied voltage
Z = impedence
C = capacitance
L = inductance
XL = inductive reactance
XC = capacitive reactance
f0= 1/(2pi x sqrt (LC))
Z = sqrt(R2+(XL-XC)
XL = (2pi)fvL
XC = 1/ (2pi x fvC)

3. The attempt at a solution
I don't even know where to start!
You can start by taking stock of what you know.

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?
rissa_rue13
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#3
Aug24-08, 02:11 PM
P: 18
Since Z=R, then XL=XC
Using an equation I found in the book, this would mean that:
2pif0L=1/(2pif0C)
This gets me back to the equation:
f0=1/(2pi x sqrt(LC))

GTrax
GTrax is offline
#4
Aug24-08, 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..
rissa_rue13
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#5
Aug24-08, 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.
LowlyPion
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#6
Aug24-08, 03:02 PM
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P: 5,346
Quote Quote by rissa_rue13 View Post
Since Z=R, then XL=XC
Using an equation I found in the book, this would mean that:
2pif0L=1/(2pif0C)
This gets me back to the equation:
f0=1/(2pi x sqrt(LC))
Correct. And that allows you to calculate the relationship between L and C does it not - when you plug in the values?

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.)
LowlyPion
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#7
Aug24-08, 03:05 PM
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P: 5,346
Quote Quote by rissa_rue13 View Post
I'm not trying to find impedence
Correct. Because you are given that it is 1000 ohms at the 8khz frequency. (The 575 is the actual resistance in the RLC if you recall because that is the impedence at resonance.)
LowlyPion
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#8
Aug24-08, 03:10 PM
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Quote Quote by rissa_rue13 View Post
I don't understand how the inductive reactance= -capacitive reactance reveals the value of a resistor.
If you look at the phasor diagram:
http://hyperphysics.phy-astr.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.
rissa_rue13
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#9
Aug24-08, 03:21 PM
P: 18
So what I've done, and I have no idea if this is correct:

f0=1/(2pi x sqrt(LC))
C = sqrt(1/f02piL

Z = sqrt(R2 + (XL-XC)2)

simplifies to:
sqrt(Z2 + R2) x C = L

substituting C:
[sqrt (Z2 + R2) / f0 x 2piL] = L

I ended up getting 0.147 H = L
Is that heading in the right direction?
LowlyPion
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#10
Aug24-08, 03:27 PM
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Quote Quote by rissa_rue13 View Post
So what I've done, and I have no idea if this is correct:

f0=1/(2pi x sqrt(LC))
C = sqrt(1/f02piL
Careful. It looks like your math isn't quite right.

C will be equal to 1/(L*w^2) where w is 2pi*f
rissa_rue13
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#11
Aug24-08, 03:39 PM
P: 18
Okay, so making that adjustment, it becomes:
(sqrt Z2+R2) (1/L(2pi*f0)2) = L

To factor out:
sqrt Z2+R2 = (2pi*f0)2L2

Then:
(sqrt Z2+R2)/(2pi*f0)2 = L2

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 F0 or Fv for this equation?
LowlyPion
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#12
Aug24-08, 03:53 PM
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Quote Quote by rissa_rue13 View Post
Am I supposed to be using F0 or Fv for this equation?
Fo = 6000hz relates L to C
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).
rissa_rue13
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#13
Aug24-08, 04:10 PM
P: 18
Z2=R2+(wL-(1/wC))2
Z2-R2=(wL-1/wC)2
sqrt(Z2-R2)=wL-1/wC
sqrt(Z2-R2) + 1/wC = wL
[sqrt(Z2-R2) + 1/wC] /w = L

substituting in C:
[sqrt(Z2-R2) + 1/w(1/w02L)]/w = L

I think this would simplify to:
[sqrt(Z2-R2 + w02L/w]/w = L
LowlyPion
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#14
Aug24-08, 04:20 PM
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Quote Quote by rissa_rue13 View Post
Z2=R2+(wL-(1/wC))2
Z2-R2=(wL-1/wC)2
sqrt(Z2-R2)=wL-1/wC
sqrt(Z2-R2) + 1/wC = wL
[sqrt(Z2-R2) + 1/wC] /w = L

substituting in C:
[sqrt(Z2-R2) + 1/w(1/w02L)]/w = L

I think this would simplify to:
[sqrt(Z2-R2 + w02L/w]/w = L
Looks more like it. Plug in the values and solve.

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.
rissa_rue13
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#15
Aug24-08, 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.
LowlyPion
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#16
Aug24-08, 05:13 PM
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P: 5,346
Quote Quote by rissa_rue13 View Post
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.
How do you see it canceling out?

sqrt(Z2-R2)=wL-1/wC

When you replace C with 1/wo2*L

that gives:
sqrt(Z2-R2) = wL-wo2*L/w = wL*(1 -wo2/w2 )


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