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## §EXPAND: Geophysical interpretation

This article does not mention the use of admittance in Geophysics. Verisimilus T 10:15, 16 May 2007 (UTC)

Done, and tag removed. But I think that this should have an article of its own, along with mechanical admittance, rather than being tacked on to this article. Although the mechanical and electrical concepts are clearly strongly analogous, they are still not the same thing. SpinningSpark 22:58, 29 March 2008 (UTC)

## §context needed

<tag removed>

69.140.152.55 (talk) 05:34, 29 February 2008 (UTC)
Done, and tag removed. I have firstly, moved all the equation stuff (and added one that was missing) to a sub-heading. The equations are really about how to convert to and from impedance and do not really add anything to the understanding of the concept. The sub-heading will now alert the casual reader that there is no need to go into it if not desired. Secondly, I have added a paragraph trying expand on what the quantity represents. However, IMHO the first line says it all, it is the inverse of Z. That's all there is to it really. SpinningSpark 23:05, 29 March 2008 (UTC)

## §Question to "Admittance in mechanics": relation of position to force?

Quote from section "Admittance in mechanics":

"... would have inputs of force and would have outputs such as position or velocity ..."

Is position really correct? In the article to impedance only velocity is mentioned. 160.85.104.90 (talk) 15:59, 14 April 2008 (UTC)

Maybe mechanical science does not define these terms so strictly as electrical science. The impedance article states that mobility is also the inverse of impedance. Unfortunately, there is no mechanical mobility article and I don't know their definition, but in electronics, mobility of a charge is defined as velocity/electric field (electric field is force per charge). I am only guessing, but possibly mobility is the preferred term where velocity is the output and admittance where distance is the output. We need some good references here really. SpinningSpark 20:23, 14 April 2008 (UTC)

## §Is the math wrong?

How can you have a quantity that is the inverse of another quantity, and both quantities break up so that their real and imaginary components are also inverses? That makes no sense to me.

Using the math on the page to demonstrate my complaint, the page says that:

G = R / (R^2 + X^2)

But it also says that G is the conductance and R is resistance, and these are inverses, so we have:

1 / R = R / (R^2 + X^2)

cross-multiplying, we get:

R^2 = (R^2 + X^2)

therefore,

X^2 = 0

and so

X = 0.

Could someone please fix this? 72.177.12.71 (talk) 06:51, 12 December 2009 (UTC)

The maths is not wrong, but you have posted in the wrong place - new posts go to the bottom on Wikipedia. The error you are making is that the "R"s you are equating are not the same R. It is not correct that $\scriptstyle \Re(Z)=1/\Re(Y)$. Perhaps the easiest way of explaining why not is to view it from a circuit connection point of view. For instance, the impedance Z=4+j2 can be realised as a resistor of resistance = 4 in series with an inductor of reactance = 2. Converted to an admittance this is Y=0.2-j0.1. If you take the inverses of the components you get a resistor of value 5 and an inductor of value 10 but the impedance is not Z=5+j10. Why? because Y=G+jB does not represent G and B connected in series, but G and B connected in parallel (or at least that is the most straightforward way of realising it as a circuit). The Z using these values must also be treated as a parallel circuit and the correct expression is Z=5//j10. evaluating that using the usual "product over sum" formula for parallel networks, $\scriptstyle (5 \times j10)/(5+j10)$ returns the original values of 4+j2 showing that the parallel realisation is an equivalent circuit. Hope that's clear. SpinningSpark 11:21, 12 December 2009 (UTC)