Question

Two cells of emf E₁, E₂ and internal resistance r₁ and r₂ respectively are connected in parallel as shown in the figure.

Deduce the expressions for

(1) the equivalent e.m.f of the combination

(2) the equivalent resistance of the combination, and

(3) the potential difference between the point A and B.

Answer 1

Let I₁ and I₂ be the current in two cells with emf E₁ and E₂ and internal resistance r₁ and r₂

So I = I₁ + I₂

Now let V be the potential difference between the points A and B. Since the first cell is connected between the points A and B.

V = potential difference across first cell

V = E₁ − I₁r₁

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

Now, the second cell is also connected between the points A and B. So,

`or I_2 =(E_2-V)/r_2`

Thus, substituting for I₁ and I₂

`I =(E_1-V)/r_1  + (E_2-V)/r_2`

`or, I (E_1 /r_1 + E_2 /r_2)-V(1 /r_1  +1 /r_2)`

`V =((E_1r_2 + E_2r_1)/(r_1+r_2)) -I((r_1r_2)/(r_1 +r_2))  ..... (1)`

If E is effective e.m.f and r, the effective internal resistance of the parallel combination of the two cells then V = E − Ir …..(2)

Comparing (1) and (2)

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

This is equivalent e.m.f of the combination

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

This is equivalent resistance of the combination.

(3) The potential difference between the point A and B is

V = E − Ir