In the given circuit in Fig. all batteries have emf 10V and internal negligble. All resitors are in ohm Calcualte the current in the rightmost `2Omega` resistor.

Explanation

The simplified circuit is shown in fig. We have to find I.

Let the potehntial of point P be zero. Potential at other points is
shown in fig. Applying Kirchhoff's current law at `X-X,`
we get
`(x -10)/(4) +(x - 10)/(2) +(x-20)/(4) +((x-10))/(2) = 0 or x (35)/(3)V`
`I=(20-(35)/(3))/(4)=(25)/(12)A`