A Challenging Question about Resistor's equivalent

studiot

Joined Nov 9, 2007
4,998
I have five black boxes that have unspecified impedances in them.
As a matter of interest you cannot be suggesting that anything of the form (a+jb) where both a and b are non zero real numbers can be greater or less than another similar thing eg (c+jd)?

It is a pretty fundamental piece of complex analysis that the complex numbers have no ordering relation.
 

WBahn

Joined Mar 31, 2012
33,198
Perhaps this will help (or perhaps not).

Wouldn't it be nice to have relationships that are uncomplicated and are simple analogies to the case for DC circuits?

Wouldn't it be nice to be able to use a relationship that says:

V = I*Z

and it doesn't matter whether Z is an inductor, capacitor, resistor, or some combination of all them?

Wouldn't it be nice to say that

Z = R + jX

and be done and not have to say

Z = "R + jX if the reactance is inductive and R - jX if the reactance is capacitive"

and then substitute what is in quotes into every place that we need Z?

Wouldn't it be nice to say that, in the case of zero reactance, that

V=I*R

and in the case of zero resistance that

V=I*X

and not

V = I*X if the reactance is inductive and -I*X is the reactance is capacitive ?

We use transforms to make our lives simpler. Why not use a transform that makes our lives simpler.


Finally, the very notion of impedance and reactance come from the Laplace transform of the differential equations describing the relationship between the voltage and the current in a circuit or circuit component. The minus sign on the capacitive reactance comes not from any arbitrary definition, but directly out of the Laplace transform, though it only appears explicitly when we translate from the more general Laplace domain to the special case of the Fourier domain by setting s=jω.

\(
i_c(t) = C \frac{dv_c(t)}{dt}
I_c(s) = sC V_c(s)
Z_c(s) = \frac{1}{sC}
Z_c(j \omega ) = \frac{1}{j \omega C}
Z_c(j \omega ) = j \( \frac{-1}{ \omega C} \)
\)

If you want to insist that capacitive reactance is positive in the Fourier domain (i.e., when s=jω) and that we have to track a separate operator depending on the flavor of the reactance, how are you going to reconcile that with the fact that capacitance and inductance are both simply impedances in the Laplace domain and, for a capacitance, which is just 1/(sC)?
 

WBahn

Joined Mar 31, 2012
33,198
I have five black boxes that have unspecified impedances in them.
As a matter of interest you cannot be suggesting that anything of the form (a+jb) where both a and b are non zero real numbers can be greater or less than another similar thing eg (c+jd)?

It is a pretty fundamental piece of complex analysis that the complex numbers have no ordering relation.
So, what...? Are you saying that I can't have five black boxes that have unspecified impedances in them? Or that somehow having five black boxes that have unspecified impedances in them somehow implies an ordering of the complex numbers?

Where am I suggesting any ordering? I've already stated that any ordering is based on the magnitude and NOT the signs.
 
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studiot

Joined Nov 9, 2007
4,998
makes our lives simpler.
Of course it's simpler, of course it's convenient.
I already said that.

Just as writing 1/0 = ∞ is convenient.

But the whole V = IZ is a simplification of what actually happens in AC circuits.
If you assemble the full differential (or integral) equations this is the particular, not the general solution.

We normally ignore the gneral solution but waiting for "steady state" conditions, ie to allow the transient part of the solution to die away.

But I'm sure you know this.

I just try to avoid saying anything in a simplification that is actually incompatible with more complete theory.

I have no issue with Z = R +j(XL-XC), though you may have noticed it's not a formula I use very often. There are also other methods.
 

WBahn

Joined Mar 31, 2012
33,198
Of course it's simpler, of course it's convenient.
I already said that.

Just as writing 1/0 = ∞ is convenient.

But the whole V = IZ is a simplification of what actually happens in AC circuits.
What happens in AC circuits is the simplication. Remember, V=IZ comes from the Laplace transform while what happens in AC circuits is only the special case of sinusoidal steady state.

I just try to avoid saying anything in a simplification that is actually incompatible with more complete theory.
But having XL and XC both be positive is incompatible with the more complete theory (namely the use of Laplace transforms).

I have no issue with Z = R +j(XL-XC), though you may have noticed it's not a formula I use very often. There are also other methods.
So what is the reactance of Z?
 
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