# Equivalent resistance, using node voltage analysis

**1. The problem statement, all variables and given/known data**

(I added in the R values for reference)

**2. Relevant equations**

**GV=I**

KCL

**3. The attempt at a solution**

I am not concerned with the MATLAB portion of the problem, only the setup.

I think what the problem wants me to do is:

1) connect a 1A test current from the ground source to v_{1} (like shown in the picture), find v_{1} using node voltage analysis, then find the equivalent resistance by dividing the test voltage (v_{1}) by the test current, which is 1A.

2) connect a 1A test current from the ground source to v_{2}. Repeat steps, where now v_{2} is the test voltage, to get equivalent resistance.

3) same as above, except with v_{4}.

When I set up my system of equations I get too many unknowns.

For example, trying KCL on case 1 (where 1A is connected to v_{1}), I get this system of equations:

v_{1}/R_{1} + (v_{1}-v_{2})/R_{2} + (v_{1}-v_{3})/R_{6} = i_{test}

(v_{2}-v_{1})/R_{2} + (v_{2}-v_{4})/R_{7} + (v_{2}-v_{7})/R_{3} = 0

(v_{4}-v_{2})/R_{7} + (v_{4}-v_{3})/R_{11} + (v_{4}-v_{6})/R_{12} = 0

Plugging in 1 for i and for all R values still gives me 6 different voltages to solve for, and only 3 equations.

Using the same method for case 2 and 3 (where v_{test} is v_{2} and v_{4} respectively), I get nearly identical systems of equations, the only difference being which equation is set equal to i instead of 0.

http://ift.tt/1gbcY2D

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