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How is G = R / R^2 + X^2 derived?

  1. May 2, 2012 #1
    If you don't already know, this is an equivalent to conductance. What I can't figure out is why reactance (X) would have any place in this equation? Remember, we're talking about conductance here, not susceptance. In an equation for susceptance it would make sense... being that it's comprised of both resistive and reactive parts... but in an equation for conductance (which is also equal to 1 / R remember) I just don't understand how it could be derived. What am I missing here?

    Again the equation is G = R / R^2 + X^2

    Here's a link to Wikipedia's page on Admittance. This equation is near the top under the header "Conversion from impedance to admittance".

    Thanks in advance for anyone who shares their thoughts on this.
  2. jcsd
  3. May 3, 2012 #2
    Hi impaJah,

    This is a consequence of how complex numbers work. The derivation is mixed in with some narration to help.

    If we start with the impedance: [itex]Z = R + jX[/itex], we know that the impedance is a sum of its real resistance, R, and its imaginary resistance, X, called reactance. By only considering the real part, we simply ignore the jX term.

    And we say that the admittance is the inverse of the impedance:
    [itex]Y = Z^{-1} = \frac{1}{R + jX}[/itex]

    Now, when you look at this equation, can you separate the real part from the imaginary part so simply? Nope, because the imaginary part is in the denominator, so its is "mixed" with the real part.

    You can multiply the numerator and the denominator both by the complex conjugate, to get an equivalent expression that brings the imaginary part into the numerator only, which is what we need to do to find just the real part:

    [itex] Y = \left(\frac{1}{R + jX}\right)\left(\frac{R-jX}{R-jX}\right) = \frac{R-jX}{R^{2}+X^{2}} = \frac{R}{R^{2}+X^{2}} + j\left(\frac{-X}{R^{2}+X^{2}}\right)[/itex]

    Now we have an equivalent term for the inverse of impedance, but now the imaginary component is entirely in the numerator which allows us to subtract it out to find just the real part, or conductance:

    [itex]G = \Re e\left(\frac{R}{R^{2}+X^{2}} + \frac{-jX}{R^{2}+X^{2}}\right) = \frac{R}{R^{2}+X^{2}}[/itex]
    Last edited: May 3, 2012
  4. May 3, 2012 #3
    First of all, thank you so much for taking the time to break this down for me DragonPetter!!!

    Okay forgive me if this next question is kind of dumb. In the last equation, what is that fancy R you decided to multiply the whole equation by?
  5. May 3, 2012 #4
    It isn't a multiplication, its more of an operation or function, although my math vocabulary is pretty bad, so it might have a more appropriate name. I probably should have written it out as [itex]\Re e[/itex]

    It basically means, take the expression, X, in my parenthesis [itex]\Re e\left(X\right)[/itex], and give me only the real part of X.

    So, for any complex number [itex]a + jb[/itex], you can use the function to get just the real part of it, [itex]\Re e\left(a +jb\right) = a[/itex]

    Here is wikipedia's information on it:
  6. May 3, 2012 #5
    So when we take the imaginary j parts from a reactance, what are we left with? Whatever real resistance it has? This would mean that every reactance contains at least some real resistance, which would make sense because every coil or capacitor must have at least some resistance. Is this correct?
  7. May 3, 2012 #6
    I think you might be slightly confused with this question, or at least your question has confused me.

    An impedance, Z, is a complex number made of a real part that we call resistance, and the imaginary part that we call reactance.

    So when you ask "when we take the imaginary parts from a reactance", you seem to be implying that reactance has a real part also. Reactance is ONLY the imaginary part of the impedance.
  8. May 3, 2012 #7
    When you showed how to separate out the j (or imaginary parts) from the equation by multiplying by its complex conjugate I'm now left wondering what X^2 is representing here? It used to represent imaginary reactance but since there's no imaginary to it anymore, is it only whatever real resistance is left? (because capacitors and coils do always have at least some real resistance)

    Does this explain the question better?
  9. May 3, 2012 #8
    The reactance squared term in the denominator is a result of the pure imaginary reactance term from the impedance. I know you're thinking that this represents the equivalent series resistance in reactive components, but that is not the case. Any practical model including ESR would add the ESR term to the real component of the impedance, not the imaginary term. Consider the ESR as another resistor in series with the reactive component, rather than some property that is built in to the reactance value.
  10. May 3, 2012 #9
    I think I just realized what the X^2 is doing in this equation. When reactance is present in a circuit there is either a current or voltage lag. Because of this there will be a discrepancy between the voltages and currents across each of the components and the total voltage and current of the circuit (at least if you use ohm's law). I've read that vectorially the voltages and currents match but I've been unable to see how this works myself... if you know of any resources that explain this I'm all ears.

    So maybe the X^2 has something to do with this? What do you think?
  11. May 4, 2012 #10
    Yes, I think that when you take the inverse of impedance, some phase contribution from the reactance component will be present in the real part. That is going out on a limb for me to say though, because I have never explicitly worked that out mathematically.
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