Two identical cells each of emf $$1.5$$ V are connected in parallel across a parallel combination of two resistors each of resistance $$20$$ $$\Omega$$. A voltmeter connected in the circuit measures $$1.2$$ V. The internal resistance of each cell is :

We are given: two identical cells each of emf $$E = 1.5$$ V with internal resistance $$r$$ connected in parallel, across a parallel combination of two $$20\ \Omega$$ resistors. The voltmeter reads $$V = 1.2$$ V.

Two $$20\ \Omega$$ resistors in parallel yield an equivalent external resistance of

$$ R_{ext} = \frac{20 \times 20}{20 + 20} = 10\ \Omega $$

Two identical cells in parallel have the same emf but combined internal resistance:

$$ r_{eq} = \frac{r}{2} $$ The equivalent emf remains $$E = 1.5$$ V.

The voltmeter reads the terminal voltage across the external resistance as $$V = 1.2\text{ V}$$, so the current is given by

$$ I = \frac{V}{R_{ext}} = \frac{1.2}{10} = 0.12 \text{ A} $$

Applying Kirchhoff's law, we have

$$ E = V + I \times r_{eq} $$ which gives

$$ 1.5 = 1.2 + 0.12 \times \frac{r}{2} $$. Simplifying,

$$ 0.3 = 0.06r $$ and hence

$$ r = \frac{0.3}{0.06} = 5\ \Omega $$

Therefore, the correct answer is Option C.