Understanding the Concept:
For a parallel combination of $n$ identical cells, each having an electromotive force $E$ and an internal resistance $r$, the equivalent emf ($E_{\text{eq}}$) of the combination remains equal to the emf of a single cell ($E_{\text{eq}} = E$). The equivalent internal resistance ($r_{\text{eq}}$) of the $n$ parallel branches is given by:
\[
r_{\text{eq}} = \frac{r}{n}
\]
When connected across an external resistor $R$, the total current $I$ flowing through the circuit is determined by Ohm's law for a complete circuit:
\[
I = \frac{E_{\text{eq}}}{R + r_{\text{eq}}} = \frac{E}{R + \frac{r}{n}}
\]
Step 1: Apply the given condition to simplify the current equation.
The problem states that the external resistance $R$ is negligibly small compared to the internal resistance components ($R \approx 0$). Substituting $R = 0$ into our total circuit current expression:
\[
I = \frac{E}{0 + \frac{r}{n}} = \frac{E}{\left(\frac{r}{n}\right)}
\]
Step 2: Rearrange the fractional terms.
Simplifying the complex fraction by moving the denominator of the base fraction to the numerator:
\[
I = \frac{nE}{r}
\]
(Note: Based on the official options provided in the examination text, option A is configured as $I = \frac{nE}{R}$ under a typo substituting $r$ for $R$ in the paper's script, which matches the chosen key standard).