Solving Kirchhoff's Problem: Finding Current in a Directed Graph

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Discussion Overview

The discussion revolves around solving a problem related to Kirchhoff's laws in a directed graph, specifically finding the current in the edge ac. Participants explore the application of Kirchhoff's First and Second Laws, share their attempts at solving the equations, and seek clarification on their assumptions regarding current direction and the implications of negative values in their results.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines their assumptions about current direction, stating that currents connected to the sink flow into it and those connected to the source flow away from it.
  • The same participant expresses uncertainty about the validity of their assumptions and the inability to solve the equations simultaneously.
  • Another participant reports successfully solving the equations using a matrix solver, providing specific current values, but questions the correctness of their method and the meaning of negative values.
  • A third participant reassures that incorrect assumptions about current direction will result in negative values but will not affect grading, emphasizing that the graph must be directed.
  • There is a suggestion that with six unknowns and six equations, a systematic approach should yield correct results.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the assumptions about current direction and the implications of negative current values. There is no consensus on the correctness of the initial assumptions or the method used to solve the equations.

Contextual Notes

The discussion includes limitations related to the assumptions made about current flow direction and the potential for negative current values indicating a need for reevaluation of those assumptions. The mathematical steps taken to arrive at the reported current values are not fully resolved.

Dench

Homework Statement


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.

Homework 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.

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 kirchhoff 1 i have:
Icb - 10 - Ibs = 0
Itc - Ica - Icb = 0
Ita + Ica + 10 - Ias = 0

And from kirchhoff 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!
 
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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.
 
So are my assumptions about the direction of current correct?
and why are these values all negative?
I mean I'm not even sure I am using the correct method! :S
 
@ tiny-tim : OK, thanks a lot! I'll work on that!
 
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Welcome to PF!

Hi Dench! Welcome to PF! :smile:

{have an omega: Ω :wink:)
Dench said:
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.

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! :wink:

(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.)
So from kirchhoff 1 i have:
Icb - 10 - Ibs = 0
Itc - Ica - Icb = 0
Ita + Ica + 10 - Ias = 0

And from kirchhoff 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

You have 6 unknowns and 6 equations, so if you just slug your way through, everything should come out fine. :smile:

(The Electrician, please don't give out full answers!)
 

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