UPSC NDA I 2024 General Ability Paper

Concept:Parallel batteries combine to produce an equivalent voltage and internal resistance, which determines the current through an external resistor.Explanation:Two batteries with voltages $V_1 = 1\,\mathrm{V}$ and $V_2 = 5\,\mathrm{V}$ are connected in parallel.Each has an internal resistance of $0.1\,\Omega$.The equivalent voltage is:$E_{\text{eq}} = \frac{\frac{V_1}{r_1} + \frac{V_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{1}{0.1} + \frac{5}{0.1}}{\frac{1}{0.1} + \frac{1}{0.1}} = \frac{10 + 50}{10 + 10} = \frac{60}{20} = 3\,\mathrm{V}.$The equivalent internal resistance is:$\frac{1}{r_{\text{eq}}} = \frac{1}{0.1} + \frac{1}{0.1} = 20, \quad \text{so } r_{\text{eq}} = 0.05\,\Omega.$Total resistance in the circuit is $R = 1000\,\Omega$ plus $r_{\text{eq}}$:$R_{\text{total}} = 1000 + 0.05 = 1000.05\,\Omega.$Using Ohm's law, the current through $I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{3}{1000.05} \approx 0.003\,\mathrm{A} = 3\,\mathrm{mA}.$ is:$I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{3}{1000.05} \approx 0.003\,\mathrm{A} = 3\,\mathrm{mA}.$Answer: C. 3.0 mA