Originally Posted by **KingLee** Would please show how you work it all out, with some details about everything you use, please, i have been sitting with this question for a week now, i read the book over 10 times, it just wont enter my head |

It's basically just linear algebra. I'll write all the equations out for you that's a little easier to look at:

$\displaystyle R_1 I_1 - R_2 I + R_2 1_1 = -R_2 I + (R_1 + R_2 )I_1= 0$

$\displaystyle R_3 I_2 - R_4 I + R_4 I_2 = -R_4 I + (R_3 +R_4) I_2 = 0 $

$\displaystyle R_2 (I-I_1) + R_4 (I - I_2) + V_{battery} = (R_2 + R_4) I - R_2 I_1 - R_4 I_2 = -V_{battery}$

In terms of linear algebra and matrices this is just:

$\displaystyle \begin{bmatrix}

-R_2 & (R_1+R_2) & 0 \\

-R_4 & 0 & (R_3 + R_4) \\

(R_2 + R_4) & -R_2 & -R_4

\end{bmatrix}

\begin{bmatrix}

I \\

I_1 \\

I_2

\end{bmatrix} =

\begin{bmatrix}

0 \\

0 \\

-V_{battery}

\end{bmatrix}

$

Once you solve this matrix equation you will have I, I1 and I2 in terms of R1, R2, R3 and the battery voltage. Then current through AB is just I1 - I2 (or I2 - I1 depending on sign).

Someone should probably check my algebra in case I messed up signs somewhere and you should probably get used to doing that yourself anyway.

In this case it's probably just as easy to solve via substitution without all the matrix hullabaloo but you get the idea.