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I'm having trouble solving a system of 3 equations. The set of equations in question is shown below

$C_a=\frac{R_a}{\frac{R_a}{r_a}+\frac{R_b}{r_b}+\frac{R_c}{r_c}}, \quad C_b=\frac{R_b}{\frac{R_a}{r_a}+\frac{R_b}{r_b}+\frac{R_c}{r_c}}, \quad C_c=\frac{R_c}{\frac{R_a}{r_a}+\frac{R_b}{r_b}+\frac{R_c}{r_c}}$

where $C_a, C_b, C_c, r_a,r_b,$ and $r_c$ are the known quantities. I want to solve for $R_a,R_b,$ and $R_c$. Doing so analytically, I obtain for $R_a$

$C_a=\frac{R_a}{\frac{R_a}{r_a}+\frac{R_b}{r_b}+\frac{R_c}{r_c}}$

$R_a=C_a \Big(\frac{R_a}{r_a}+\frac{R_b}{r_b}+\frac{R_c}{r_c} \Big)$

$R_a-C_a \Big(\frac{R_a}{r_a}\Big)=C_a \Big(\frac{R_b}{r_b}+\frac{R_c}{r_c} \Big)$

$R_a\Big(1-\frac{C_a}{r_a}\Big)=C_a \Big(\frac{R_b}{r_b}+\frac{R_c}{r_c} \Big)$

$R_a=\frac{C_a \Big(\frac{R_b}{r_b}+\frac{R_c}{r_c} \Big)}{\Big(1-\frac{C_a}{r_a}\Big)}$

And similarly for $R_b$ and $R_c$

$R_b=\frac{C_b \Big(\frac{R_a}{r_a}+\frac{R_c}{r_c} \Big)}{\Big(1-\frac{C_b}{r_b}\Big)}$ $R_c=\frac{C_c \Big(\frac{R_a}{r_a}+\frac{R_b}{r_b} \Big)}{\Big(1-\frac{C_c}{r_c}\Big)}$

Now this all seems fine and dandy until I try and solve these equations with specific values. When I plug these equations into Matlab or Mathematica, I get the empty set as solutions (in fact, I get the empty set as solutions when I plug the original symbolic equations into Matlab or Mathematica and have them try and solve it). Did I do something wrong in solving the equations? Is there something inherent to the system that makes it possible to solve for the C's (with the R's known) but not the R's (with the C's known)? Any help is much appreciated!

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Let $$D=\frac{R_a}{r_a}+\frac{R_b}{r_b}+\frac{R_c}{r_c}\tag1$$ Then, we have $$C_a=\frac{R_a}{D},\quad C_b=\frac{R_b}{D},\quad C_c=\frac{R_c}{D}\tag 2$$ From $(2)$, we have $$D=\frac{R_a}{C_a}=\frac{R_b}{C_b}=\frac{R_c}{C_c}\tag3$$ So, from $(1)(3)$, we have

$$(D=)\frac{R_a}{C_a}=\frac{R_a}{r_a}+\frac{1}{r_b}\cdot\frac{C_bR_a}{C_a}+\frac{1}{r_c}\cdot\frac{C_cR_a}{C_a},$$ i.e. $$R_a\left(\frac{1}{C_a}-\frac{1}{r_a}-\frac{C_b}{r_bC_a}-\frac{C_c}{r_cC_a}\right)=0$$ Here suppose that $\frac{1}{C_a}-\frac{1}{r_a}-\frac{C_b}{r_bC_a}-\frac{C_c}{r_cC_a}\not=0$, then $R_a=0$. However, this leads that $D=0$, a contradiction. So, $$\frac{1}{C_a}-\frac{1}{r_a}-\frac{C_b}{r_bC_a}-\frac{C_c}{r_cC_a}=0$$ has to hold in order for $R_a,R_b,R_c$ to exist.

As a result, if $\frac{1}{C_a}-\frac{1}{r_a}-\frac{C_b}{r_bC_a}-\frac{C_c}{r_cC_a}\not=0$, then there exist no such $R_a,R_b,R_c$.

If $\frac{1}{C_a}-\frac{1}{r_a}-\frac{C_b}{r_bC_a}-\frac{C_c}{r_cC_a}=0,$ then $R_a$ can be any real number except $0$ and $R_b=\frac{C_b}{C_a}R_a,R_c=\frac{C_c}{C_a}R_a$.

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  • $\begingroup$ Thanks so much for your help! $\endgroup$ – PurdueCHE Jul 28 '15 at 18:57

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