Obtain an expression for the equivalent emf and internal resistance of two cells connected in parallel.

Explanation

Consider two cells cnnected in parallel between two point A and C as shown in figure.

Let `E_(1), E_(2)` be the emfs of the two cells and `r_(1), r_(2)` be their internal resistance respectively. Since the cells are connected in parallel, the terminal potential difference V across each is same. Let `I_(1), I_(2)` be the currents from first and the second cell respectively.
For the first cell,
`V = E_(1) - I_(1)r_(1)` or `I_(1) = (E_(1)- V)/(r_(1)) = ((E_(1))/(r_(1)) - (V)/(r_(1)))`
For the second cell,
`V = E_(2) - I_(2) r_(2)` or `I_(2) = (E_(2)-V)/(r_(2)) = (E_(2))/(r_(2)) - (V)/(r_(2))`
Then the total current I is `I = I_(1) + I_(2)` ......(1)
Substituting `I_(1)` & `I_(2)` in eqn (1)
`I = (E_(1))/(r_(1)) -(V)/(r_(1)) + (E_(2))/(r_(2)) - (V)/(r_(2))`
`I = ((E_(1))/(r_(1)) + (E_(2))/(r_(2))) - V((1)/(r_(1)) + (1)/(r_(2))) rArr I = ((E_(1)r_(2) + E_(2) + r_(1))/(r_(1)r_(2))) - V((r_(1) + r_(2))/(r_(1)r_(2)))`
`V = ((r_(1) + r_(2))/(r_(1)r_(2))) = ((E_(1)r_(2) + E_(2)r_(1))/(r_(1)r_(2))) - I`
Multiplying throughout by `((r_(1)r_(2))/(r_(1) + r_(2)))`
`V((r_(1) + r_(2))/(r_(1)r_(2)))((r_(1)r_(2))/(r_(1) + r_(2))) = ((E_(1)r_(2) + E_(2)r_(1))/(r_(1)r_(2)))((r_(1)r_(2))/(r_(1) + r_(2))) -1 ((r_(1)r_(2))/(r_(1) + r_(2)))`
`V = ((E_(1)r_(2) + E_(2)r_(1))/(r_(1) + r_(2)))-I((r_(1)r_(2))/(r_(1) + r_(2)))` .....(2)
If `E_(eq)` is the equivalent emf and `r_(eq)` is the equivalent internal resistance of two cells connected in parallel, then
`V = E_(eq) - I r_(eq)` ...(3)
On comparing (2) and (3)
`E_(eq) = ((E_(1)r_(2) + E_(2)r_(2))/(r_(1) + r_(2)))`
`r_(eq) = ((r_(1)r_(2))/(r_(1) + r_(2)))`