Resistive network with two voltage sources

Question

Update: (I updated the description/Question for clarity)

I am currently stuck trying to solve a seemingly simple resistive network. The voltages U1 and U2, aswell as Rr are known. The unknown components are the resistances R1, R2, R3.

The dotted resistances "Rr" can be added or removed as needed. The current through them is known, but the current through the unknown resistances is of course unknown.

These restrictions let me form three equations, which I would not call linear dependant. Here is the step by step approach:

1)

$$Ip_{1} = \frac{U_{1}}{Rr \left(\frac{1}{Rr} + \frac{1}{R_{1}}\right) \left(\frac{R_{1} Rr}{R_{1} + Rr} + \frac{1}{\frac{1}{R_{3}} + \frac{1}{R_{2}}}\right)} - \frac{U_{2}}{Rr \left(R_{2} + \frac{1}{\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{1}}}\right) \left(\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{1}}\right)}$$

2)

$$Ip_{2} = - \frac{U_{1}}{Rr \left(R_{1} + \frac{1}{\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{2}}}\right) \left(\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{2}}\right)} + \frac{U_{2}}{Rr \left(\frac{1}{Rr} + \frac{1}{R_{2}}\right) \left(\frac{R_{2} Rr}{R_{2} + Rr} + \frac{1}{\frac{1}{R_{3}} + \frac{1}{R_{1}}}\right)}$$

3)

$$Im = \frac{U_{1}}{Rr \left(R_{1} + \frac{1}{\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{2}}}\right) \left(\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{2}}\right)} + \frac{U_{2}}{Rr \left(R_{2} + \frac{1}{\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{1}}}\right) \left(\frac{1}{Rr} + \frac{1}{R_{3}} + \frac{1}{R_{1}}\right)}$$

So I solve the equation system and get the following: Solving above equations for R3 gives: $$R3 = \left[ \frac{- Ip_{1} R_{1} R_{2} Rr + R_{1} R_{2} U_{1}}{Ip_{1} R_{1} R_{2} + Ip_{1} R_{1} Rr + Ip_{1} R_{2} Rr - R_{1} U_{1} + R_{1} U_{2}}\right]$$ $$R3 = \left[ \frac{- Ip_{2} R_{1} R_{2} Rr + R_{1} R_{2} U_{2}}{Ip_{2} R_{1} R_{2} + Ip_{2} R_{1} Rr + Ip_{2} R_{2} Rr + R_{2} U_{1} - R_{2} U_{2}}\right]$$ $$R3 = \left[ - \frac{Im R_{1} R_{2} Rr}{Im R_{1} R_{2} + Im R_{1} Rr + Im R_{2} Rr - R_{1} U_{2} - R_{2} U_{1}}\right]$$

Using these equations to solve for R2: $$R2 = \left[ \ \frac{R_{1} U_{2} \left(- Ip_{1} Rr + Ip_{2} Rr + U_{1} - U_{2}\right)}{Ip_{1} R_{1} U_{2} + Ip_{1} Rr U_{1} - Ip_{2} R_{1} U_{1} - Ip_{2} Rr U_{1} - U_{1}^{2} + U_{1} U_{2}}\right]$$

$$R2 = \left[\ \frac{R_{1} U_{2} \left(- Im Rr - Ip_{1} Rr + U_{1}\right)}{U_{1} \left(Im R_{1} + Im Rr + Ip_{1} Rr - U_{1}\right)}\right]$$

Using these equations to solve for R1: $$R1 = \left[ 0\right]$$

The result I get reduces to 0. There are more conditional solutions to this system, that I see, however. they do not produce reasonable results. Here are example values for testing purposes, generated in spice simulations.

R1=1e3
R2=3e3
R3=2e3
Ip1 = 12.068965e-3
Ip2 = 2.5862069e-3
Im = 25.86207e-3
U1 = 400
U2 = 300
Rr = 10e3

here is a python snipped to follow along:

import sympy as sym
U1,U2,R1,R2,R3,Rr,Ip1,Ip2,Im = sym.symbols('U1,U2,R1,R2,R3,Rr,Ip1,Ip2,Im')

a= -Ip1  +  ((U1/(((Rr*R1)/(Rr + R1))+(1/(1/R2 + 1/R3)))))*((1/Rr)/(1/Rr + 1/R1))-((U2/(R2+(1/(1/R1 + 1/R3 + 1/Rr))))*((1/Rr)*(1/(1/R1 + 1/R3 + 1/Rr))))    
b= -Ip2 + ((U2/(((Rr*R2)/(Rr + R2))+(1/(1/R1 + 1/R3))))*((1/Rr)/(1/Rr + 1/R2)))- ((U1/(R1+(1/(1/R2 + 1/R3 + 1/Rr))))*((1/Rr)*(1/(1/R2 + 1/R3 + 1/Rr))))    
c= -Im + ((U1/(R1+(1/(1/R3 + 1/R2 + 1/Rr))))*((1/Rr)/(1/Rr + 1/R2 + 1/R3))+ (U2/(R2+(1/(1/R1 + 1/R3 + 1/Rr))))*((1/Rr)/(1/Rr + 1/R3 + 1/R1)))

q01 = sym.solve(a,R3)
q02 = sym.solve(b,R3)
q03 = sym.solve(c,R3)
q11 = sym.solve(q01[0]-q02[0],R2)
q12 = sym.solve(q01[0]-q03[0],R2)
q21 = sym.solve(q11[1]-q12[1],R1)

What could I be missing?

Regarding the answer of @colintd

This are the innitial equations: $$\frac{R_{1} \left(- I_{1} Rr + U_{2}\right) + R_{2} \left(- I_{1} Rr + U_{1}\right)}{R_{1} R_{2}} = \frac{I_{1} R_{3} Rr^{2}}{R_{3} + Rr}$$

$$\frac{- R_{1} R_{2} Rr \left(I_{2} Rr - U_{1}\right) + \left(R_{1} + Rr\right) \left(- I_{2} Rr + U_{2}\right)}{R_{2} \left(R_{1} + Rr\right)} = \frac{I_{2} R_{3} Rr^{2}}{R_{3} + Rr}$$

$$\frac{- R_{1} R_{2} Rr \left(I_{3} Rr - U_{2}\right) + \left(R_{2} + Rr\right) \left(- I_{3} Rr + U_{1}\right)}{R_{1} \left(R_{2} + Rr\right)} = \frac{I_{3} R_{3} Rr^{2}}{R_{3} + Rr} $$

Here are the initial equations solved for R3: $$R3 = \left[ - \frac{I_{1} R_{1} R_{2} Rr}{I_{1} R_{1} R_{2} + I_{1} R_{1} Rr + I_{1} R_{2} Rr - R_{1} U_{2} - R_{2} U_{1}}\right]$$

$$R3 = \left[ - \frac{I_{2} R_{1} R_{2} Rr^{2}}{2 I_{2} R_{1} R_{2} Rr + I_{2} R_{1} Rr^{2} + I_{2} R_{2} Rr^{2} - R_{1} R_{2} U_{1} - R_{1} Rr U_{2} - R_{2} Rr U_{1}}\right]$$

$$R3 = \left[ - \frac{I_{3} R_{1} R_{2} Rr^{2}}{2 I_{3} R_{1} R_{2} Rr + I_{3} R_{1} Rr^{2} + I_{3} R_{2} Rr^{2} - R_{1} R_{2} U_{2} - R_{1} Rr U_{2} - R_{2} Rr U_{1}}\right]$$

here are the two simplified equations with R3 eliminated: $$R2 = \left[ 0, \ \frac{R_{1} Rr U_{2} \left(I_{1} - I_{2}\right)}{I_{1} I_{2} R_{1} Rr - I_{1} R_{1} U_{1} - I_{1} Rr U_{1} + I_{2} Rr U_{1}}\right]$$

$$R2 = \left[ 0, \ \frac{R_{1} Rr U_{2} \left(I_{1} - I_{3}\right)}{I_{1} I_{3} R_{1} Rr - I_{1} R_{1} U_{2} - I_{1} Rr U_{1} + I_{3} Rr U_{1}}\right]$$

The end result is however the same, as these two equations end up being linearly dependent. This can be seen, when test values are used and the the functions are plotted.

Here is a python code snipped for convenience:

import matplotlib.pyplot as plt
import numpy as np
import sympy as sym

U1,U2,R1,R2,R3,Rr,Ip1,Ip2,Im,Im2,Im3,Im4,I1,I2,I3 = sym.symbols('U1,U2,R1,R2,R3,Rr,Ip1,Ip2,Im,Im2,Im3,Im4,I1,I2,I3')

v1a = U1 - I1*Rr 
v2a = U2 - I1*Rr 
v3a = I1*Rr

v1b = U1 - I2*Rr 
v2b = U2 - I2*Rr 
v3b = I2*Rr

v1c = U1 - I3*Rr 
v2c = U2 - I3*Rr 
v3c = I3*Rr

t1 = sym.simplify(v1a/(R1) + v2a/(R2) - v3a*(1/R3 + 1/Rr))
t2 = sym.simplify(v1b*(1/R1 + 1/Rr) + v2b/(R2) - v3b*(1/R3 + 1/Rr))
t3 = sym.simplify(v1c/(R1) + v2c*(1/R2 + 1/Rr) - v3c*(1/R3 + 1/Rr))

q01 = sym.solve(t1,R3)
q02 = sym.solve(t2,R3)
q03 = sym.solve(t3,R3)

q11 = sym.solve(q01[0]-q02[0],R2)
q12 = sym.solve(q01[0]-q03[0],R2)
q21 = sym.solve(q11[1]-q12[1],R1)

Answer 1

I think having more than one of the Rr resistors in circuit allows solution of the problem using a relatively straight forward approach.

The outline approach is as follows:

1. Insert the R3 Rr resistor.
2. The current through this resistor will give you the voltage across both Rr and R3 (if measured current through Rr is I, then voltage V3 across Rr and R3 must be I*Rr).
3. Given knowledge of U1 and U2, loop analysis allows you to calculate the voltages across R1 & R2, as V1=U1-V3 and V2=U2-V3.
4. Measure the R3 Rr current for the 3 cases -"a) no R1 Rr and no R2 Rr", b) "R1 Rr but no R2 Rr", c) "no R1 Rr but R2 Rr"
5. As shown in 2. and 3. these measurements can be used to workout the voltages across R1, R2 & R3 in each of case a,b and c. Lets call these voltages V1a, V2a, V3a, V1b, V2b, V3b, V1c, V2c & V3c.
6. Given the voltages we can then calculate the currents through R1, R2, R3 and any Rr for each of case a,b and c. As the currents through the R3 branch is the sum of the currents through the R1 and R2 branches we can write 3 equations.

\$V1a/R1+V2a/R2=V3a(1/R3+1/Rr)\$

\$V1b(1/R1 +1/Rr)+V2b/R2=V3b(1/R3+1/Rr)\$

\$V1c/R1+V2c(1/R2+1/Rr)=V3c(1/R3+1/Rr)\$

The righthand side of the case a equation can be used to eliminte R3, from the case b and c equations leaving two truly independent equations containing just R1 and R2.

These two equations can then be solved to give R1 & R2. Once you have R1 and R2, you can then substitute in to solve for R3.

It is worth noting that if U1=U2, then R1 & R2 are effectively in parallel, and whilst you can still work out R1+R2, the R1 and R2 equations become degenerate, and you can't work out R1 and R2 individually.

The same voltage analysis works if you only have one Rr in each case (as in the OP's question). You use the measured current to work out V1, V2 or V3 (depending on which Rr is in place), which then allows you to work out the remaining voltages.

However the elimination of R3 is messier, and I am less certain the two remaining equations are actually independent. It may be you have to measure the total current in each case (proportional to the R3 Rr current), to get the solution to work.