# Find effective resistance using kirchoff's laws

1. Jan 20, 2013

### brushman

1. The problem statement, all variables and given/known data
Find the effective resistance between the circled nodes in the following network (use a computer to solve the system of equations you get). Note that bump in the crossing wires meaning they're not connected.

2. Relevant equations
Sum of currents at a node is 0. Sum of voltages around a closed loop is 0.

3. The attempt at a solution

I don't know how to tell if currents are the same by symmetry, so as far as I can tell there's 5 different currents I1, I2, I3, I4, and I5. But I only see 2 nodes to create current equations with, leaving me with 3 more unknowns. I would guess that the current through the R1 resistors are the equal, and the current through the R2 resistors are equal, but I don't know how to tell the direction the current would be flowing.

Secondly, I only see 3 loops, one of them being a figure 8. So basically I can make 5 equations, but I have 7 unknowns.

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2. Jan 20, 2013

### Staff: Mentor

There are really only three independent currents, the other currents can be made up of combinations of those three thanks to KCL at the junctions. When currents split off at a junction, try to create as few new currents as you can. For example, suppose a current I1 flows into a junction and two paths leave the junction. Label one departing current as I2, and the other as I1 - I2. Then only one new current is created at the split. When currents meet at a junction, sum already known incoming currents to determine the outgoing current whenever possible.

Three currents, three loops, no problem.

if you're looking to write node equations, then if you put a current source driving the input and ground the lower input terminal (reference node), I can see three nodes (including the top input terminal).

3. Jan 20, 2013

### ehild

To find the effective resistance between A and B, imagine a battery with emf E connected between these points and find the current I1 flowing through this battery. You can also redraw the circuit so as easier to follow the six currents and three loops. There are enough equations!

ehild

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4. Jan 20, 2013

### brushman

Thanks. So to finish the problem I need to set up my equations and solve for I1, the current going through the imaginary voltage source. Then with the simplified circuit consisting of my Emf, Ref, and I1, I can solve for Ref in terms of I1 and Emf (where I1 is in terms of R1, R2, and R3). Is that correct? So my answer depends on what voltage source you attach? That doesn't seem right.

edit: nevermind, it looks like you can solve for Emf in terms of Rs and Is

5. Jan 21, 2013

### ehild

But I1 will be proportional to E, so E cancels.

ehild