# Electrical Circuits - Linear Algebra

1. Jan 22, 2014

### tg22542

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

Here is a picture of the diagram.

http://gyazo.com/f1b7051fda5b9e1d3a185c53abde1211

I must use the Gauss Jordan elimination method and solve for X1, X2 and X3

I am having problems setting up my equations

2. Relevant equations

3. The attempt at a solution

The way my professor showed an example of a question similar to this was by setting up 2 equations for each junction, and 2 for each loop.

My attempt:

∑ Flows Entering = 0

Junction A (Top of the line inbetween loops) : X1 - X2 - X3 =0 ??
Junction B (Bottom of line inbetween loops) : X1 + X2 - X3 =0 ??

Loop1 (Left side) : 10X1 - 5X2 + X3 = 0
Loop2 (Right Side): X1 + 5X2 - X3 = -52

Can someone correct my equations?

2. Jan 22, 2014

### E7.5

You have 3 unknowns, therefore need a minimum of 3 different equations. However, I think there is a deeper, underlying issue with you misunderstanding Kirchoff's rules, or circuit theory in general based on how you are setting up your equations. For example, your "Loop1" equation implies that there are currents, but they are flowing without a power source (the 52V).

3. Jan 22, 2014

### Curious3141

This part is based on Kirchoff's current law. The current flowing into a junction must be equal to the current flowing out of a junction. Which currents are flowing in and which out?

This is based on Kirchoff's voltage law. But the equations don't seem to correspond to the circuit in the image. Loop 1 (as indicated) spans the cell, and the 1Ω and 5Ω resistances. Loop 1 doesn't include the 10Ω resistance. But your first equation has a coefficient of 10, and I don't know where that's coming from. Same issue with the second equation. The 52V (cell voltage) term also seems out of place.

I think you need to very carefully spell out how you're taking your loops before we can help you check your equations. For example, "loop 1 starts with the negative terminal of the cell, goes across the cell to the positive terminal, across the 1Ω resistor in the direction of current x1, down the 5Ω resistor in the direction of current x2 and ends back at the negative terminal".