Question

Two cells of emfs E₁ and E₂ and internal resistances r₁ and r₂ respectively are connected in parallel as shown in the figure. Deduce the expression for the

1. equivalent emf of the combination
2. equivalent internal resistance of the combination
3. potential difference between the points A and B.

Answer 1

Here, I = I₁ + I₂   ...(i)

Let V = Potential difference between A and B.

For cell ε₁

Then, V = ε₁ – I₁r₁

⇒ I₁ = `(ε_1 - V)/r_1`

Similarly, for cell ε₂ I₂ = `(ε_2 - V)/r_2`

Putting these values in equation (i)

I = `(ε_1 - V)/r_1 + (ε_2 - V)/r_2`

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

or V = `((ε_1r_2 + ε_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 = ε_eq – Ir_eq  ...(iii)

Then

1. ε_eq = `(ε_1r_2 + ε_2r_1)/(r_1 + r_2)`
2. r_eq = `(r_1r_2)/(r_1 + r_2)`
3. The potential difference between A and B
   V = ε_eq – Ir_eq