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

[rock.freak667]

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?

 

[daudaudaudau]

Your third equation should be -0.011*ib+1.2*il=12. Do you know why?

 

[rock.freak667]

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.

 

[dynamicsolo]

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

$0.1I_C +0.011I_B=2$