In the circuit in figure `E_1=3V, E_2=2V, E_3=1V` and `R=r_1=r_2=r_3=1Omega`

Find the potential differece between the points `A` and `B` and the currents through each branch.

Explanation

(i) Equivalent emf of three batteries would be
`E_(eq)=(sum (E//r))/(E(l//r))=((3//1+2//1+1//1))/((1//1+1//1+1//1)=2V`
Further `r_1,r_2 and r_3` each are `1Omega` of Therefore, internal resistance of the equivalent battery will be `1/3 Omega` as all three are in parallel.
The equivalent circuit is therefore shown in the given figure. Since, no current is taken from the battery. `V_(AB)=2V` (From V=E-ir)
Further `V_(AB)=V_A-V_B=E_1-i_1E_1, therefore i_1=(V_B-V_A+E_1)/r_1=(-2+3)/1=1A`
Similarly `i_2=(V_B-V_A+E_2)/r_2=(-2+2)/1=0 and i_3=(V_B-V_A+E_3)/r_3=(-2+1)/1=-1A`

(ii) `r_2` is short circuited means resistance of this branch becomes zero. Making a closed circuit with a battery and resistance R. Applying Kirchhoff’s second law in three loops so formed.
`3-i_1-(i_1+j_2+j_3)=0....(i) `
`2-(j_1+j_2+j_3)=0`......(ii)
`1-j_3-(j_1+j_2+j_3)=2A`..........(iii)
From Eq. (ii) `j_1+j_2+j_3=2A`
`therefore Substituting in Eq (i) we get `j_1=1A` , Substituting in Eq (iii) we get `j_3=-1A`
`therefore i_2=2A`