Solve this:
10V = I2*R2 + I4*R4
10V = I1*R1 + I5*R5
10V = I1*R1 + I3*R3 + I4*R4
Edit:
You need set 2 resistors to solve the problem
You have not fully characterized the solution. You should say more than "set 2 resistors to solve the problem". The reader might think that all that is needed is to set
any two resistors.
You began to derive the rest of the full characterization in reply #11, but you didn't carry it through to a conclusion.
The problem cannot be solved by setting the pair R1/R5 or the pair R2/R4 to arbitrary values, but any other pairs will do.
It's possible to find 5 equations for the 5 variables. There are 4 paths from the top node to the bottom node, and 1 path around the whole thing. The first 4 paths yield equations set equal to 10 volts, and the 5th path yields an equation set to zero volts. See this image; the paths are numbered 1 through 5:
Here are the 5 equations:
These equations can be put in matrix form:
The reader is wondering at this point how we can possibly have 5 equations for 5 unknowns when earlier it was pointed that we need to set two resistors to arbitrary values, then we can solve for the other 3 resistors. The choice of values for the initial two resistors is not completely arbitrary--some choices can lead to negative values for the remaining 3.
Had we followed this procedure from the beginning, the first thing to do is to determine if enough of our 5 equations are independent to give a unique solution to the problem. We can use a linear algebra concept for this. We simply compute the rank of the 5x5 coefficient matrix, which would be 5 if we had 5 independent equations, but this matrix only has a rank of 3. This means there are an infinite number of solutions, and two of the variables are free variables and can be chosen arbitrarily (with the aforementioned limitation).
Ordinarily, this would mean that we could choose any pair of resistor values arbitrarily and then our equations would give a solution for the rest. But this problem is peculiar in that the pairs R1 R5 and R2 R4 cannot be chosen arbitrarily because when you choose one of the pair the other one is immediately determined. The values of R1 R5 and R2 R4 are not independent as Picuino mentioned.
The linear algebra procedure that can show this is to set a pair of variables in the matrix to a constant and see if the rank of the matrix increases to 5. This happens for all pairs of variables except the R1 R5 pair and the R2 R4 pair.
I used a math program to find all the solutions to the system and here's the result. The two resistors on the left of each line are the two arbitrarily chosen, and the rest of the line shows the calculation for the other resistors.
Notice that all pairs are present except the R1 R5 and R2 R4 pairs:
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