Two cells of emf E and 2E with internal resistances r and 3r, respectively are connected in parallel. The emf and internal resistance of the combination is:

$\frac{5}{4}E,\frac{3}{4}r$

The correct answer is Option (3) → $\frac{5}{4}E,\frac{3}{4}r$

Given:

First cell: emf = $E_1 = E$, internal resistance = $r_1 = r$

Second cell: emf = $E_2 = 2E$, internal resistance = $r_2 = 3r$

For parallel connection:

Equivalent emf, $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$

Equivalent resistance, $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$

Substitute values:

$E_{eq} = \frac{E(3r) + 2E(r)}{r + 3r} = \frac{3Er + 2Er}{4r} = \frac{5E}{4}$

$r_{eq} = \frac{r(3r)}{r + 3r} = \frac{3r^2}{4r} = \frac{3r}{4}$

∴ Equivalent emf = $\frac{5E}{4}$, Equivalent internal resistance = $\frac{3r}{4}$