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Old 03-15-2004, 08:05 PM
jermermery jermermery is offline
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I have a problem.... if you have a circuit with an inductor and resistor in series and a capacitor across them in paralell, with an AC source, is there a condition where the total impedance is a pure real number? I have got as far as getting the total impedance algebraically, but can't see where to go from here. I know that resonance in a series LCR circuit can cause this case to occur (when angular frequency=1/root LC), but i don't know if this works for paralell. Any help would be appreciated....
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Old 03-16-2004, 05:35 AM
Battousai Battousai is offline
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Yes there definitley is. If I'm not mistaken any weird combination of LCR will result in at least once resonant frequency where the inductor and capacitor cancel eachother out, for your particular case, the overall impedance is:

Zcap//(Zinductor+R) = (1/sC)//(R+sL) = (R+sL)/[1+sRC+(s^2)LC]

Ok I seem to have hit a mental block... But there is a resonant frequency and it's most likely w=1/(LC)^0.5
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Old 03-19-2004, 01:44 AM
Optikon Optikon is offline
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Quote:
Originally posted by Battousai@Mar 16 2004, 01:35 AM
Yes there definitley is. If I'm not mistaken any weird combination of LCR will result in at least once resonant frequency where the inductor and capacitor cancel eachother out, for your particular case, the overall impedance is:

Zcap//(Zinductor+R) = (1/sC)//(R+sL) = (R+sL)/[1+sRC+(s^2)LC]

Ok I seem to have hit a mental block... But there is a resonant frequency and it's most likely w=1/(LC)^0.5
Here you are:

Take for example a series network of R, L & C

The impedance looking into this network is Zin = R + sL + 1/sC (They are in series)

Now, at the resonant frequency, the impedance of this network looks purely resistive right? Well in mathematical terms this means that Im{Zin(s)} = 0
The imaginary part of the impedance equals 0 and only the Real part remains (resistive)

So let s --> jw (for sinusoids)

Zin(jw) = R + jwL + 1/(jwC)

recall, 1/jwC = -j/wC

=> Zin(jw) = R + j(wL - 1/wC)

Now Set Im{Zin(jw)} = 0 and the Imaginary part is wL - 1/wC

So we solve wL - 1/wC = 0
=> wL = 1/wC
=> w^2 = 1/LC
=> w = Sqrt(1/LC)

And that's it.

When this network experiences a sinusoidal input with frequency w = sqrt(1/LC), The voltage and current will be in phase and the magnitudes predicted as if only the R were present.


NOW, following along with this same method, consider a network of an L in series with R, both in parallel with a C.

The Zin looking into this network is Zin(s) = [(sL+R)(1/sC)] / [(1/sC) + sL + R]

After some simplifying and s --> jw,

Zin(jw) = (jwL + R) / [ 1 - w^2LC + jwRC]

After much algebra, I finally get an expression for Im{Zin(jw)} and I set that equal to zero. Since the denominator is real and positive for all real values of w,R,L & C, I only need to solve for what makes the numerator zero in this expression. When I do that, I get:

w = Sqrt[ (L - R^2C) / L^2C ]

I checked the answer with a few dummy values and it checked out ok.
BUT an interesting observation a real valued w only exists for this resonant frequency if L > R^2*C which would imply that I can choose L, R and C in such a way as to make a resonant frequency nonexistent...
which to me says there is something wrong because, I think any second order system has a "natural" frequency. So I prolly goofed somewhere in all the algebra.

but anyhow, in general that is how one solves resonant frequency problems algebraically.

Now that I think about it, EVERY second order system has a natural frequency. Because second order systems can always be written as a polynomial of second degree in the denominator (2 poles) and the fundamental theorem of algebra states that every polynomial of degree n (n>1) has n zeros in the complex plane.
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