I'm designing a system that uses a grid of resistive sensors, and I'm having trouble figuring out the solution to a set of equations that is the output of this system. I haven't done serious maths for a long time, so go easy on me!
The equations are:
$$\frac{\frac{1}{\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}{R1+\frac{1}{\frac{1}{R2}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}=Y_1$$
$$\frac{\frac{1}{\frac{1}{R1}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}{R2+\frac{1}{\frac{1}{R1}+\frac{1}{R3}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}=Y_2$$
$$\frac{\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}{R3+\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R4}...\frac{1}{R_x}+\frac{1}{A}}}=Y_3$$
$$...$$
$$\frac{\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}...\frac{1}{R_{x-1}}+\frac{1}{A}}}{R_x+\frac{1}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}...\frac{1}{R_{x-1}}+\frac{1}{A}}}=Y_x$$
With the number of equations equal to the number of RR terms (so for $R1−R4$, there would actually be $4$ equations). $A$ is a constant, and $Y_1,Y_2,Y_3$ etc are all known.
I have no idea really how to begin solving this. I need a general solution because the values will be constantly changing over time. Does anyone have any pointers?
edit: put back A
edit: here is a diagram of the circuit that these equations represent.
$Y1$ is the output voltage with the top switch connected to +V and the others connected to 0V. $Y2$ is the output voltage with the second switch connected to +V and the others connected to 0V. $Y3$ is the output voltage with the third switch connected to +V and the others connected to 0V, etc.
Actually, more exactly, Y is the output voltage divided by +v, to give a ratio. The circuit can be simulated using this link.
Let $\;\dfrac{1}{R} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\ldots+ \dfrac{1}{R_x}+\dfrac{1}{A}\;$ then the equations can be written as:
$$\require{cancel} \dfrac{\;\;\dfrac{1}{\dfrac{1}{R}-\dfrac{1}{R_k}}\;\;}{\;\;R_k+\dfrac{1}{\dfrac{1}{R}-\dfrac{1}{R_k}}\;\;} = Y_k \;\;\iff\;\; R_k\left(\dfrac{1}{R}-\cancel{\dfrac{1}{R_k}}\right)+\cancel{1}=\frac{1}{Y_k} \;\;\iff \dfrac{1}{R_k} = \dfrac{Y_k}{R} \tag{1} $$
Summing up $\,(1)\,$ for $\,k=1,2,\ldots x\,$ gives $\,\dfrac{1}{R}-\dfrac{1}{A}= \dfrac{Y}{R}\,$ where $\displaystyle\,Y = \sum_{k=1}^x Y_k\,$, so $\,R = (1-Y)A\,$ then it follows from $\,(1)\,$ that $\,R_k=\dfrac{(1-Y)A}{Y_k}\,$.