# Finding current using Kirchhoff's circuit rules

1. Oct 29, 2012

### slaufer

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

2. Relevant equations

Kirchhoff's loop rule and (maybe) Kirchhoff's current law

3. The attempt at a solution

So I attempted to apply Kirchhoff's loop rule to the circuit, and ended up with a system of linear equations:

24 - 6I$_{1}$ - 3I$_{3}$ = 0 $\rightarrow$ 2I$_{1}$ - I$_{3}$ = 8
36 - 6I$_{1}$ - 6I$_{2}$ = 0 $\rightarrow$ I$_{1}$ + I$_{2}$ = 6
12 - 3I$_{3}$ - 6I$_{2}$ = 0 $\rightarrow$ 2I$_{2}$ + I$_{3}$ = 4

but every time I try to solve it, I end up with everything dropping out. I'm not sure what I'm missing here, but I've been banging my head against this one for about 4 hours now, so any help would be appreciated!

Last edited: Oct 29, 2012
2. Oct 29, 2012

### ehild

Check the signs.
You also have a relation between the currents. (Nodal Law) Use that and two loop equations.

ehild

Last edited: Oct 29, 2012
3. Oct 29, 2012

### slaufer

Thanks, that makes more sense now.

I might have eventually realized that my system was cancelling out because it was using mutually derived equations, but I probably never would have realized that I was treating I$_{3}$ as if it were flowing in two directions at once.

Using two of the equations from the original system in the nodal rule equation I$_{1}$ = I$_{2}$+ I$_{3}$ made it all fall together though. I'm still not sure how I'd figure out what direction I$_{3}$ was flowing in if it weren't given in the problem, but at least I'm FINALLY done with this problem.

tl;dr you are the best, thanks

4. Oct 29, 2012

### ehild

You need not find out the direction of the currents in advance. If you have not chosen the proper direction you would get negative value for the current.

ehild