Question

Two cells of emf E₁ and E₂ and internal resistances r₁ and r₂ are connected in parallel, with their terminals of the same polarity connected together. Obtain an expression for the equivalent emf of the combination.

Answer 1

Two cells of emf ε₁ and ε₂ and internal resistance r₁ and r₂ connected in parallel,

Total current will be I = I₁ + I₂ ..............(i)

Let V = Potential difference between A and B. ε₁ 

For cell ε₁,

Then `V = ε_1 - I_1r_1`

`I_1 = ((ε_1 - V))/r_1`

Similarly, for cell ε₂, I₂ = `((ε_2 - V))/r_1`

Putting these values in equation (i),

I = `((ε_1 - V))/r_1 + ((ε_2 - V))/r_2`

I = `(ε_1/r_1 + ε_2/r_2) - V(1/r_1 + 1/r_2)`

or V = `((epsilon_1r_2 + epsilon_2r_1))/((r_1 + r_2)) - I((r_1r_2)/(r_1 + r_2))` ...........(ii)

Comparing the above equation with the equivalent circuit of emf 'ε_eq' and internal resistance 'r_eq' then,

V = `epsilon_{eq} - Ir_{eq}` ............(iii)

Then

so, the equivalent emf will be `epsilon_{eq} = ((epsilon_1r_2 + epsilon_2r_1))/((r_1 + r_2))`.