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

Explanation

Let `I_1,I_2` represent branch currents .
Let V be the common potential , so that ,
`I_1=(E_1-V)/r_1 , I_2=(E_2-V)/r_2`
hence, `I=(E_1-V)/r_1 + (E_2-V)/r_2`
i.e. `I=(E_1/r_1 + E_2/r_2) -V (1/r_1 +1/r_2)`
Comparing this with the terminal potential difference,
`V=E_(eq)-Ir_(eq)`
i.e., `V=E_(eq)-I/((I/r_(eq)))`
For two cells in parallel combination,
`(E_(eq))_p=(E_1/r_1 + E_2/r_2)/(1/r_1 + 1/r_2)=(E_1r_2 +E_2r_1)/(r_1+r_2)`
`1/(r_(eq))_p=1/r_1 +1/r_2` , and main current = `I=E_(eq)/(R+r_(eq))`
Note : (i) For n cells in parallel
`V=[(sum_(i=1)^n E_i/r_i)/(sum_(i=1)^n 1/r_1)]-[I/(sum_(i=1)^n 1/r_i)]`
`(E_(eq))_"parallel"=[(sum_(i=1)^n E_i/r_i)/(sum_(i=1)^n 1/r_i)]`
`(1/r_(eq))_p = sum_(i-1)^n (1/r_i)`
(ii) For .n. number of identical cells
E-emf of each cell
r-internal resistance of each cell.
`(E_(eq))_p=(n(E/r))/(n(1/r))`
i.e., `(E_(eq))_p=E`
and `(1/r_(eq))_p=n(1/r)`
i.e. `(r_(eq))_p=r/n`
Main current , `I=(nE)/(R+n/r)` and terminal potential difference across cells V=IR.