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Finding current using Kirchhoff's circuit rules

  • Thread starter slaufer
  • Start date
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1. Homework Statement

Nt0Cm.png



2. Homework 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[itex]_{1}[/itex] - 3I[itex]_{3}[/itex] = 0 [itex]\rightarrow[/itex] 2I[itex]_{1}[/itex] - I[itex]_{3}[/itex] = 8
36 - 6I[itex]_{1}[/itex] - 6I[itex]_{2}[/itex] = 0 [itex]\rightarrow[/itex] I[itex]_{1}[/itex] + I[itex]_{2}[/itex] = 6
12 - 3I[itex]_{3}[/itex] - 6I[itex]_{2}[/itex] = 0 [itex]\rightarrow[/itex] 2I[itex]_{2}[/itex] + I[itex]_{3}[/itex] = 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:

ehild

Homework Helper
15,363
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1. Homework Statement

Nt0Cm.png



2. Homework 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[itex]_{1}[/itex] - 3I[itex]_{3}[/itex] = 0 [itex]\rightarrow[/itex] 2I[itex]_{1}[/itex] - I[itex]_{3}[/itex] = 8 wrong
36 - 6I[itex]_{1}[/itex] - 6I[itex]_{2}[/itex] = 0 [itex]\rightarrow[/itex] I[itex]_{1}[/itex] + I[itex]_{2}[/itex] = 6
12 - 3I[itex]_{3}[/itex] - 6I[itex]_{2}[/itex] = 0 wrong[itex]\rightarrow[/itex] 2I[itex]_{2}[/itex] + I[itex]_{3}[/itex] = 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!
Check the signs.
You also have a relation between the currents. (Nodal Law) Use that and two loop equations.

ehild
 
Last edited:
2
0
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[itex]_{3}[/itex] 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[itex]_{1}[/itex] = I[itex]_{2}[/itex]+ I[itex]_{3}[/itex] made it all fall together though. I'm still not sure how I'd figure out what direction I[itex]_{3}[/itex] 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
 

ehild

Homework Helper
15,363
1,770
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
 

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