# Homework Help: Kirchhoff Problem

1. Nov 13, 2009

### Dench

1. The problem statement, all variables and given/known data
I have been given the above information and asked to find the current in the edge ac. s is the sink, t is the source.

2. Relevant equations
Kirchoff's First Law, current in = current out.
Kirchoff's Second Law, the sum of resistance.current for all vertices in a cycle = 0.

3. The attempt at a solution
Firstly I have assumed that all currents connected to the sink go into the sink, and all connected to the source go away from the source. I have also assumed the current flows from c to b, and c to a. I have made these assumptions to make sure there cannot be an infinite cycle in the graph and I simply cannot see a way to solve this if the graph is not directed.

So from kirchoff 1 i have:
Icb - 10 - Ibs = 0
Itc - Ica - Icb = 0
Ita + Ica + 10 - Ias = 0

And from kirchoff 2:
cycle tca : 6Itc + Ica - 3Ita = 0
cycle cab : Ica - 20 - 3Icb = 0
cycle bas : 20 + Ias - 2Ibs = 0

However I cannot seem to solve these simultaneously to get any closer to my result,and I'm not sure my initial assumptions about the direction of flow of current are valid, any hints/tips would be appreciated!

Last edited by a moderator: Nov 14, 2009
2. Nov 14, 2009

### The Electrician

I was able to solve them simultaneously using the matrix solver on my HP50 calculator:

Ibs = -21.6667
Ias = -63.3333
Icb = -11.6667
Ica = -15.0000
Ita = -58.3333
Itc = -26.6667

The magnitude of the sink and source currents is 85.

3. Nov 14, 2009

### Dench

So are my assumptions about the direction of current correct?
and why are these values all negative?
I mean I'm not even sure im using the correct method! :S

4. Nov 14, 2009

### Dench

@ tiny-tim : OK, thanks a lot! I'll work on that!!

Last edited by a moderator: Nov 14, 2009
5. Nov 14, 2009

### tiny-tim

Welcome to PF!

Hi Dench! Welcome to PF!

{have an omega: Ω )
Yes, the graph does have to be directed, but there's no need to make the correct assumptions … if you guess wrong, the current will come out negative instead of positive … you won't lose any marks for it!

(and obviously, if you use I1 I2 I3 etc, you have to draw arrows, but it you use Icb etc, meaning from c to b, then you don't.)
You have 6 unknowns and 6 equations, so if you just slug your way through, everything should come out fine.