How to calculate the current across two points ? PLEASE HELP How to calculate the current across two points in this : https://lounge.kth.se/RequestHandler...17508f458bd73b So what i did is i used Ohms law to find the total current across the whole circuit which i computed to be 3 A and i am given the EMF of the battery and asked to neglect the internal resistance. So i am stuck at calculating the current across point AB. Please help 
Next step is to calculate the currents through R1 and R2. You know that the voltage drop across these two resistors is the same (do you see what that is?), so you can solve for I1 and I2 using: $\displaystyle R_1 I_1 = R_2 I_2$ and: $\displaystyle I_1 + I_2 = I_T$ Can you take it from here? 
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I am not after \I_t\ what i am after is \I_{ab}\ here is my work: since \R_1\ is parallel to \R_2\ and \R_3\\R_4\, hence: \R_t_1\=\frac{\frac{1}{R_1}\cdot \frac{1}{R_2} }{\frac{1}{R_1}+ \frac{1}{R_2} }\ = 2\Omega[math] Applying the same for R_3 and R_4 we get that the resultant R is 2 as well hence: I=\frac{V}{R}= \\frac{12}{2+2)} which gives that I_T =3 A but what i am after is the current floating between AB 
1 Attachment(s) I wouldn't try to solve this using resistor series / parallel formulas. That's making it difficult. Basically what ChipB was saying except he was probably drawing the current arrows a little differently (see my diagram). In the case shown where I redrew your diagram to make it a little clearer. $\displaystyle R_1 I_1 = R_2 (I1_1)$ $\displaystyle R_3 I_2 = R_4 (I  I_2) $ $\displaystyle R_2 (II_1) + R_4 (I  I_2) + E_{battery} = 0$ (Don't think you need the 3rd equation because that loop doesn't go through AB but included for completeness) Basically its just relying on the voltages added up around a circle should always equal zero. This gives you a set of linear equations which you can solve for I1 and I2 in terms of I. 
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See: https://en.wikipedia.org/wiki/Kirchh..._law_.28KVL.29 
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$\displaystyle R_1 I_1  R_2 I + R_2 1_1 = R_2 I + (R_1 + R_2 )I_1= 0$ $\displaystyle R_3 I_2  R_4 I + R_4 I_2 = R_4 I + (R_3 +R_4) I_2 = 0 $ $\displaystyle R_2 (II_1) + R_4 (I  I_2) + V_{battery} = (R_2 + R_4) I  R_2 I_1  R_4 I_2 = V_{battery}$ 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. 
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