The circuit in Fig. shows two cells connected in opposition to each other. Cell `E_(1)` is of emf `6V` and internal resistance `2Omega,` the cell `E_(2)` is of emf `4 V` sand internal resistance `8omega`. Find the potential difference between the points `A` and `B`.

Explanation

Applying Ohm's law
Effective resistance `=2Omega+8Omega=10Omega` and effective emf of two cells `=6-4=2V`, so the electric current is given by
`I=(6-4)/(2+8)=0.2A`
Along anti-clockwise direction, since `E_(1) gt E_(2)`.
The direction of flow of current is always from high potential to low potential. Therefore
`V_(B) gt V_(A)`.
`impliesV_(B)-4V-(0.2)xx8=V_(A)`
Therefore, `V_(B)-V_(A)=3.6V`