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

a. Find the potential differece between the points `A` and `B` and the currents through each branch.
b. If `r_2` is short circuited and the point `A` is connected to point `B`, find the currents through `E_1,E_2, E_3` and the resistor `R`

Explanation

a. Equivalent emf of the battries would be
`E_(eq)=(Sigma(E/r))/(Sigma(1/r))=((3/1+2/1+1/1))/((21/1+1/1+1/1))=2V`
Further `r_1, r_2` and `r_3` each are of `1Omega`. Therefore, internal resistance of the equivalent battery will be `1/3 Omega` as alll three are in parallel.

The equivalent circuit is therefore shown in the figure.
Since no current is taken from the battery.
`V_(AB)=2V` (From `V=E-ir)`
Further `V_A-V_B=E_1-i_1r_1`
`:. i_1=(V_B-V_A+E_1)/r_1=(-2+3)/=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`
b. `r_2` is short circuirted means resistance of this branch becomes zero. Making closed circuit with a battery and resistance `R`. Applying Kirchhoff's second law in three loops so formed.
`3-i_1-(i_1+i_2+i_30=0` ..............i
`2-(i_1+i_2+i_3)=0`.............ii
`1-i_3-(i_1+i_2+i_3)=0`.................iii
From equ. ii `i_1+i_2+i_3=2A`
`:.` Substitutig in equ. i we get `i_1=1A`
Substitution in eq. iii we get `i_3=-1A`
`:. i_2=2A`