How Do Kirchhoff's Laws Determine Currents in Electrical Circuits?

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Kirchhoff's laws are applied to determine the currents I_C, I_B, and I_L in an electrical circuit. The first law indicates that I_C minus I_B minus I_L equals zero, while the second law provides equations relating the currents to voltage drops across resistors. A discrepancy arises when comparing solutions, as the correct equations yield different values for the currents. The direction of current I_B affects the potential drop across the resistor, which must be accounted for in the equations. Ultimately, using the correct sign conventions leads to consistent results in solving for the unknown currents.
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Homework Statement



http://img181.imageshack.us/img181/9554/cktpz6.jpg


Find the values of I_C,I_B,I_L

Homework Equations


\sum V=\sum IR
\sum I=0


The Attempt at a Solution



By Kirchoff's first law:
I_C-I_B-I_L=0

by Kirchoff's 2nd law:
0.1I_C+1.2I_L=14

0.011I_B+1.2I_L=12

When I used these to solve for the unknowns I got
I_C=21.25A \ \ I_B=11.35A \ \ I_L=9.9A

Now the solution says to use the first two equations and this one:
0.1I_C +0.011I_B=2

and that gives a different answer to what I had calculated. Did I take the incorrect loops?
 
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Your third equation should be -0.011*ib+1.2*il=12. Do you know why?
 
The only way I see that would work is if I reversed the direction of the current I_B which I think should not really matter because when solved if the direction is wrong the answer will be negative.
 
rock.freak667 said:
The only way I see that would work is if I reversed the direction of the current I_B which I think should not really matter because when solved if the direction is wrong the answer will be negative.

What (dau)^4 is saying is that the current I_B is passing through the 0.011 ohm resistor in the same direction that you are tracing currents in that loop. So the change in potential across that resistor will be a "drop" of 0.011·I_B in that equation. (The way you have potential changes traced in your other loop is consistent.)

You'll then find that

0.1I_C+1.2I_L=14
and
-0.011I_B+1.2I_L=12

leads to

0.1I_C +0.011I_B=2
 
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