Correct Option: (D) 0.2 A
Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]
For $n$ identical cells connected in parallel, the equivalent emf ($E_{eq}$) and equivalent internal resistance ($r_{eq}$) are given by:
- $E_{eq} = E = 2\text{ V}$
- $r_{eq} = \frac{r}{n} = \frac{0.1}{5}\text{ }\Omega$
The current ($I$) flowing through the external resistor ($R$) is given by Ohm's Law:
$$I = \frac{E_{eq}}{R + r_{eq}}$$Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]
Given: $E = 2\text{ V}$, $r = 0.1\text{ }\Omega$, $n = 5$, $R = 9.98\text{ }\Omega$.
First, calculate the equivalent internal resistance:
$$r_{eq} = \frac{0.1}{5} = 0.02\text{ }\Omega$$Now, substitute the values into the current formula:
$$I = \frac{2}{9.98 + 0.02}$$ $$I = \frac{2}{10.00}$$Step 3 (Final Concluded Answer): [0.25 Mark]
$$I = 0.2\text{ A}$$The current flowing through the resistor is 0.2 A. Therefore, the correct option is (D).
💡 EL10 Concept Insight:
When you connect identical batteries in parallel, the total voltage stays the same as one battery, but the internal resistance drops significantly because there are more "paths" for the current to flow through, making the battery pack more efficient.