
As shown in figure two cells are connected in parallel with the load resistance of R=10Ω. Let current i1,i2 drawn from both cells as shown, from K.C.L. i=i1+i2...(1)
Apply K.V.L. for loop ACDF,
E1−iR−i1r1=0⇒12−10i−i1=0...(2)
From equations (1) and (2)
12−11i1−10i2=0...(3)
Now, for loop BCDE,
Apply K.V.L.
E2−10i−i2r2=0⇒13−10i−2i2=0⇒(∵i=i1+i2)∴13−10i1−12i2=0...(4)
From equation (3) and (4),
i1=167A,i2=3223A
So, net current in external load R=10Ω is
I=i1+i2=167+3223=3237A
Then potential difference for a given load,
V=IR=3237×32=11⋅56V
So, the potential difference lies between, 11⋅5V and 11⋅6V.