Find the `emf` and the internal resistance of a source which is equivalent to two batteries connected in parallel whose `emf's` are equal to `E_(1)` and `E_(2)` and internal resistances to `R_(1)` and `R_(2)`

Explanation

Let us connect a load of resistance `R` between the points `A` and `B` (fig)
From the loop rule, `Delta varphi = 0`, we obtains
`i R = xi_(1) - i_(1) R_(1)`....(1)
and `iR = xi_(2) - (i - i_(1)) R_(2)`
or `i (R + R_(2)) = xi + i_(1) R_(2)` ....(2)
Solving Eqs. (1) and (2), we get
`i = (xi_(1) R_(1) + xi_(2) R_(2))/(R_(1) + R_(2))//R + (R_(1) R_(2))/(R_(1) + R_(2)) = (xi_(o))/(R + R_(0))` ....(3)
where `xi_(0) = (xi_(1) R_(1) + xi_(2) R_(2))/(R_(1) + R_(2))` and `R_(0) = (R_(1) R_(2))/(R_(1) + R_(2))`
Thus one can replace the given arrangement of the cells by a single cell having the emf `xi_(0)` and internal resistance `R_(0)`.