Find the emf (`V`) and internal resistance `(R)` of a single battery which is equivalent toa parallel combination of two batteries of emf `V_1` and `V_2` and internal resistances `r_1` and `r_2` respectively, with polrities as shown in figure

Explanation

(a) Equivalent emf (V) of the battery
PD across the terminals of the battery is equal to its emf when current drawn from the battery is zero. In the given circuit, , Current in the internal circuit,
`j=(Net emf)/(Total resistance)=(V_1+V_2)/(r_1+r_2)`
Therefore, potential difference between A and B would be
`V_A-V_B=V_1-ir_1`
`therefore V_A-V_B=V_1-((V_1+V_2)/(r_1+r_2)) r_1=(V1r_2-V_2r_1)/(r_1+r_2)=(20 times 3-5 times 6)/9=30/9=3.33`
So, the equivalent emf of the battery is 3.33 V, Note that if `V_1r_2=V_2r_1,V=0`
If `V_1r_2 gt V_2r_1,V_A-V_B=` Positive i.e., A side of the equivalent battery will become the positive terminal and vice-versa.

(b)Internal resistance (r) of the battery
`r_1 and r_2` are in parallel. Therefore, the internal resistance r will be given by
`1//r=1//r_1+1//r_2 or r=(r_1r_2)/(r_1+r_2)=(6 times 3)/(6+3)=2 Omega`