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

EMF of battery is equal to potential difference across the terminals, when no current is drawn from battery (for extrenal circuit) [Here all the elements in the circuit are is series]
Current in internal circuit `=i`
`thereforei=("Netemf")/("total resistance") or i=(V_(1)+V_(2))/(r_(1)+r_(2))`
`thereforeV_(1)-V_(B)=V_(1)-ir_(1)` [`therefore` cell is discharging]
or `V_(A)-V_(B)=V_(1)-((V_(1)+V_(2))/(r_(1)+r_(2)))r_(1)`
or `V_(A)-V_(B)=(V_(1)r_(2)-V_(2)r_(1))/(r_(1)+r_(2))`
`therefore` equivalent emf of the battery `=V`
`V=(V_(1)r_(2)-V_(2)r_(1))/(r_(1)+r_(2))`
(ii) internal resistance of equivalent battery `r_(1)` and `r_(2)` and in parallel
`(1)/(r)=(1)/(r_(1))+(1)/(r_(2)) or r=(r_(1)r_(2))/(r_(1)+r_(2))`