The circuit in Fig. shows two cells connected in opposition to each other. Cell `epsi_1`, is of emf 6 V and internal resistance 2`Omega` and the cell `epsi_2` is of emf 4 V and internal resistance 8 `Omega`. Find the potential difference between the points A and B.

Explanation

Since two cells are connected in opposition, hence net emf `epsi = epsi_1 - epsi_2 = 6-4 = 2 V `and total resistance `r = r_1 + r_2 = 2 + 8 = 10 Omega`
` therefore ` Circuit current `I = epsi/r = 2/10 = 0.2 A`
` therefore ` Potential difference between the points A and `B = |V_A-V_B | = epsi_2 + r_2I = 4 + 8 xx 0.2 = 5.6 V`
Of course point B is at higher potential than the point A.