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 potentiai, 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/((1/(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_(1)r_(2)+E_(2)r_(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_(i)))]-[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)))]` and `(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))`