Tetrahedral Current

The other two vertices are in same voltage.So the net resistance is (R/2+R/2)×R/2R=R/2 Where R=2ohm So that the current is 2amp

Using the diagram of the question, let the $2V$ battery be connected to the bottom two vertices $A$ and $B$ . The five $2\Omega$ resistors on top form two triangles sharing an edge.

Convert one of the triangle or delta resistor circuit to wye or star circuit.

Each of the wye-resistor will have a resistance of $\frac{2\times 2}{2+2+2}=\frac{2}{3}\Omega$ .

Now the equivalent resistance between $A$ and $B$ is:

$R_{AB}=2//\left(\frac{2}{3}+(\frac{2}{3}+2)//(\frac{2}{3}+2)\right)$

$\quad = 2//\left(\frac{2}{3}+(\frac{1}{2})(\frac{8}{3})\right)$

$\quad = 2//(\frac{2}{3}+\frac{4}{3})$

$\quad = 2//2=1 \space \Omega$

Therefore current through $A$ and $B$ $=\frac{2V}{1\Omega}=2\space A$

The problem here is to find equivalent resistance.there are 6 line and 4 point. we can choose our cell to be in between any two point so there remains 5 lines and 5 resistance. The two adjacent (I mean lines which have different origin but same conclusion) line can be thought as 2 resistance in series . In this picture we can see there are only two of this kind of combination. that makes up 4 lines . the middle line connects the two combination. AND THE MIDDLE LINE WILL NOT CONTRIBUTE TO NET RESISTANCE BECAUSE THE VOLTAGE DIFFERENCE BETWEEN ITS 2 END IS EQUAL (AS i*r=constant). The rest you can calculate by yourself.

(A picture was badly needed. And also my English is bad... anyways tnx :) )