Relative contribution of parallel resistors to final resistance If you have two resistors A and B in parallel, how could you represent their individual percent contribution to the equivalent resistance? I thought of this question while trying to map the parallel resistances A and B in a pie chart to prioritize which resistance to try to reduce in value, for the lowest possible equivalent resistance value. 
For example:


*

*A has value 50$\Omega$ while B has value of $\infty$; with an equivalent resistance of 50$\Omega$ I can see resistor A is responsible for 100% of the final value respectively while B has 0%. 

*A has value 50$\Omega$ while B has value of 50$\Omega$; with an equivalent resistance of 25$\Omega$, then A and B both have 50% contribution to the final value. 

*A has value $\infty$ while B has value of 50$\Omega$; with an equivalent resistance of 50$\Omega$ then A is 0% and B is 100%. 


I imagine this should be represented by some kind of function with two asymptotes. For a fixed value of B, Y is asymptotic to 100 when A is near 0 but also asymptotic to 0 when A is near infinity. The only functions I know that can satisfy this is a translation of tanh(x) or logistic/sigmoid function, with new bounds. However, I'm not able to successfully map the sigmoid nor a family related to to my bounds, while also satisfying the criteria that at A=B (=40), I reach a percentage of 50% in a nice closed form solution. 
The function I've come up with so far is:
$\dfrac{200e^{-x/40}}{e^{-x/40}+1}$
Here is my progress in sagemath.
 A: Percentages make sense when there is something additive.  In this situation, the conductance $1/R$ of a parallel resistor network is the sum of the conductances:
$$\frac{1}{R}=\frac{1}{R_A}+\frac{1}{R_B}.$$
This gives that the percentage contribution of resistor $A$ to the conductance is
$$\frac{\frac{1}{R_A}}{\frac{1}{R_A}+\frac{1}{R_B}} = \frac{R_B}{R_A+R_B}.$$
Without any good justification, we might reverse the percentages and say that the "percentage contribution to total resistance" by resistors $A$ and $B$ are then
$$P_A=\frac{R_A}{R_A+R_B}\text{ and }P_B=\frac{R_B}{R_A+R_B},$$
respectively.  Notice that as $R_B$ tends to $0$, $P_A$ tends to $100\%$, and as $R_B$ tends to $\infty$, $P_A$ tends to $0\%$.  Furthermore, when $R_A=R_B$, then $P_A=P_B=50\%$.

Another approach might be to calculate the percent marginal contribution to resistance, if what you are wanting is to know how much changing one of the resistors will affect the total resistance.
We can calculate the partial derivatives
\begin{align*}
\frac{\partial R}{\partial R_A} &= \frac{R_B^2}{(R_A+R_B)^2} \\
\frac{\partial R}{\partial R_B} &= \frac{R_A^2}{(R_A+R_B)^2}.
\end{align*}
and then (since these are all positive quantities) calculate a percentage like
$$Q_A = \frac{\frac{\partial R}{\partial R_A}}{\frac{\partial R}{\partial R_A} + \frac{\partial R}{\partial R_B}} = \frac{R_B^2}{R_A^2+R_B^2}.$$
This says that when $R_A$ is much larger than $R_B$, changing $R_A$ by an ohm doesn't change the total resistance as much as changing $R_B$ by an ohm would.  (This appears to be backwards from the function you might want, but it might be justified reversing the roles of $Q_A$ and $Q_B$ --- I'll leave that to you.)
