Impedance and Admittance

  • #1
I have somehow worked myself into a mental loop that I need a push to break out of.

Consider an inductor in series with a resistor. In sinusoidal steady state, the combination has an impedance Z = R + jωL. The admittance is given by (1/Z) = (R-jωL)/(R2+(ωL)2), and if R is zero, it is simply -j/(ωL) as expected. But the admittance can be broken into conductance and suseptance, so the calculated admittance Y = G+jB = 0 - j/(ωL).
But I was expecting G = infinity!

Clearly I am confusing exactly how and when these quantities are defined. Any help would be appreciated.

-John
 

Answers and Replies

  • #2
5,571
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That's a good one. But think of if R=0, you only have an inductor

[tex]Z_L=jωL\;\Rightarrow Y_L=\frac 1 {jωL}=-\frac j {ωL}[/tex]

There should be no G. You don't even have R in the equation!!!
 
  • #3
So is a G only defined if you have an R, or is an R only defined when you have a finite G?

I guess I would like a clear and unambiguous mathematical definition of how and when R,X,G, and B are defined.

(I have read statements before such as one cannot always define an impedance and admittance matrix simultaneously, for example, a short circuit has no well defined admittance matrix, so I believe my question is a simpler case of this)
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
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Consider an inductor in series with a resistor. In sinusoidal steady state, the combination has an impedance Z = R + jωL. The admittance is given by (1/Z) = (R-jωL)/(R2+(ωL)2), and if R is zero, it is simply -j/(ωL) as expected. But the admittance can be broken into conductance and suseptance, so the calculated admittance Y = G+jB = 0 - j/(ωL).
But I was expecting G = infinity!
Let's see whether you can get me just as confused.

Admittances in parallel simply add. So G+jB is a resistor in parallel with a reactance.
If the resistor is of infinite Ohms (i.e., a perfect insulator) then G=0.
If G=infinity then you would be talking about a short circuit across the inductor.
 

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