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Translation first!
The figure shows an electric circuit.
a) Use Kirchhoffs on the part-circuit abed and show that $I_3$ = 0,50A.
b) Explain that $I_2 + I_5 = 2,0A$. Use Kirchhoffs on the part-curcuit befc to find another equation with $I_2$ and $I_5$ as unknows. Solve the set of equations.
c) Find $I_1$ and $I_4$.
d) Find the resitance of R.

a) Assume clockwise rotation of circuit. That makes 8v positive and 5,5v negative. 8-5,5=2,5v. This would be the V over the resitance at $I_3$. $\frac{2,5v}{5 ohm} = 0,5A$ I think this is correct.

b) My problem is that I can't seem to find the equation properly. If I assume a clockwise current, that makes V at $I_3$ negative. And V at $I_2$ positive and V at $I_5$ negative.
$-2,5 = 5I_2 - 2,5I_5$
Solution in the book is $-2.5 = - 5I_2 + 2,5I_5 $ with $I_2 + I_5 =2$ It gives $I_2=1$ and $I_5 = 1$
My logic is wrong, but wrong how?

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Fixed it myself. The book defined the in a stupid way. Wiki helped me out. – Algific May 5 '12 at 12:04
Hi Algific, and welcome to Physics Stack Exchange! It would be great if you write an answer describing what you were doing wrong and the solution you found, and perhaps even describing what's wrong with the book's description of the 2nd law. We encourage self-answers here. (You may have to wait a few hours before doing so because of technical measures imposed by the system for new users.) – David Z May 5 '12 at 15:26

The book defined the as follows.
The sum of all emf around a closed circuit is equal to the voltage over all the resistors.

I just think it fails to press that if you do just sum up all the voltages it should indeed be 0. Which is a nice thing if you don't really have an emf in a part-circuit you are analyzing. And it should be mentioned that an emf value is positive if you pass trough it from negative to positive. And negative if you go the other way around.

It also helps writing the part equations in the right order so you don't plot the wrong variables into the calculator. Something I didn't do for starts. Bad me.

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