Question

Two cells of emf E₁ and E₂ and internal resistances r₁ and r₂ are connected in parallel. Derive the expression for the (i) emf and (ii) internal resistance of a single equivalent cell which can replace this combination.

Answer 1

where

E₁, E₂ = Emf of two cells

r₁, r₂ = Internal resistances of cell

I₁, I₂ = Current due to the two cells

Terminal potential difference across the first cell is

V = E₁ − I₁r₁

`=>I_1=(E_1-V)/r_1`

For the second cell, terminal potential difference will be equal to that across the first cell. So,

V = E₂ − I₂r₂

`⇒I_2=(E_2−V)/r_2`

Let E be effective emf and r is effective internal resistance. Let I be the current flowing through the cell.

I=I1+I2

`⇒I=(E1−V)/r_1+(E2−V)/r_2`

`⇒Ir_1r_2=E_1r_2+E_1r_1−(r_1+r_2)V`

`=>V=(E_1r_2+E_2r_1)/(r_1+r_2)−(Ir_1r_2)/(r_1+r_2)`

Comparing the equation with V=E−Ir, we get

`E=(E_1r_2+E_2r_1)/(r_1+r_2)`

`r=(r_1r_2)/(r_1+r_2)`