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 \(I_1 = \frac{E_1 - V}{r_1}\)
Now, the second cell is also connected between the points A and B. So,
or \(I_2 = \frac{E_2 - V}{r_2}\)
Thus, substituting for I1 and I2
\(I = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2}\)
or \(I\left(\frac{E_1}{r_1} + \frac{E_2}{r_2}\right) - V\left(\frac 1{r_1} + \frac 1{r_2}\right)\)
\(V = \left(\frac{E_1r_2 + E_2 r_1}{r_1 + r_2}\right) - I\left(\frac{r_1r_2}{r_1 + r_2}\right)\) ......(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),
(i) \(E= \frac{E_1r_2 + E_2 r_1}{r_1 + r_2}\)
This is equivalent e.m.f of the combination
(ii) \(r = \frac{r_1r_2}{r_1 + r_2}\)
This is equivalent resistance of the combination.
(iii) The potential difference between the point A and B is
V = E − Ir