A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance $1<mspace\; width="1em"></mspace>\Omega$ (as shown in the figure). The total current in the network is:

$1.<mspace\; width="1em"></mspace>6<mspace\; width="1em"></mspace>A$
$2.<mspace\; width="1em"></mspace>12<mspace\; width="1em"></mspace>A$
$3.<mspace\; width="1em"></mspace>4<mspace\; width="1em"></mspace>A$
$4.<mspace\; width="1em"></mspace>2<mspace\; width="1em"></mspace>A$

The current in the branch AB in the given network is:

 
1. 5/8 A

2. 5/2 A

3. 15/8 A

4. 3/2 A