# Mesh Circuit/Kirchhoff's Rule

1. Feb 16, 2013

### StarMan1234

1. The problem statement, all variables and given/known data
In the figure ε = 9.55 V, R1 = 2090 Ω, R2 = 3360 Ω, and R3 = 4440 Ω. What are the potential differences (in V) (a) VA - VB, (b) VB - VC, (c) VC - VD, and (d) VA - VC?

This figure is NOT the same as my problem, although the wires are the same. The resistors are in a different order. R2 is on the TOP LEFT and BOTTOM RIGHT, R3 is in the MIDDLE, and R1 is on the TOP RIGHT and BOTTOM LEFT. Hope this makes sense. R4 and R5 are only in the image I provided.

In the problem I define currents (I1 throught I5) as corresponding to the resistors in the diagram.

2. Relevant equations
Junction Rule
Kirchhoff's Rule

3. The attempt at a solution

I2=I3+I5
I4=I1+I3

Loop Equations:
E-I1R2-I4R1=0
E-I2R1-I3R3-I4R1=0
E-I1R2-I3R3-I5R2=0
E-I2R1-I5R2=0
Anyone familiar with this type of problem?

2. Feb 16, 2013

### Staff: Mentor

Wow, a problem that's a problem to decipher; Your description is a puzzle in itself.

From what I've been able to figure, you've got something like this?

It's not good practice to have multiple components with the same name (two R1's, two R2's), since it will make following the intent of equations written more difficult.

Regarding the equations that you wrote, note that once you've fixed a current's direction that the potential drop across resistors that it passes through is also fixed in polarity. When you pass though that component doing a "KVL walk" around a loop, be sure to take into account the correct sign for the resulting potential change. Specifically, you've passed through R3 in different directions for two of the equations that you've written, but ascribed the same polarity of potential drop in both cases.

Note also that you only need to select and write equations for just enough loops so that every component is included at least once. In the case of this circuit you can get away with a minimum of 3 loops. And finally, not every loop has to include the voltage source -- it would be fine to use the "triangular" loops too, if you wished.

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3. Feb 16, 2013

### StarMan1234

Your diagram is perfect.

Okay, I see what you mean.

My equations:
-I2R1-I3R3+I1R2=0 (upper triangle)
-I3R3-I4R1+I5R2=0 (lower triangle)
E-I1R2-I4R1=0 (simplest path that includes the battery)

This part is where I'm absolutely stuck. I have 5 unknown currents, and 3 equations. My job is to use the junction rules to simplify them. How do I use the following?

I2=I3+I5
I4=I1+I3

4. Feb 16, 2013

### Staff: Mentor

Temporarily add a current that flows through the battery, then you can do KCL at the top and bottom nodes...

5. Feb 16, 2013

### StarMan1234

I1+I2=I4+I5

Is this what you mean?

6. Feb 16, 2013

### Staff: Mentor

Yup. You should now have enough equations to find your unknowns.

7. Feb 16, 2013

### StarMan1234

Can you recommend any simple way to solve all the equation to find, say, I2?

I don't know where to begin with them. I could use a calculator with matrix functions to help, but is that the only way. It looks as if the equations are all interdependent.

8. Feb 16, 2013

### Staff: Mentor

The usual procedure is to eliminate variables by substitution. Isolate a variable in one equation (giving an expression for that variable in terms of the remaining variables), then substitute that expression for the variable into the remaining equations. The remaining equations then have one less unknown. Repeat until you are left with a single equation in one variable and solve for it. Then back-substitute to find the rest.

These equations are going to be a bit of a pain because of the number of variables involved; Lots of current variables were introduced and need to be whittled down by substitutions. So the solution by hand will be rather tedious. If the circuit analysis method had been a "true" mesh current method (presumably something you will cover later), then there would be only three equations in three unknowns.