AIIMS 2018 Physics Ohm's Law and Circuits MCQ Question
The current in the bulb is, I = \frac{P}{V} = \frac{4.5}{1.5} = 3 Å. Current in 1 Ω resistance wire = \frac{1.5}{1} = 1.5 Å. Total current through the battery, I = 3 + 1.5 = 4.5 Å. Emf of the battery, E = V + Ir = 1.⁵⁺ (4.5 × 2.67) = 13.5 V.
Correct Answer
Detailed Explanation
The calculation of the emf of the battery is based on the total current flowing through the circuit and the internal resistance. The derived emf, E = V + Ir, where V is the voltage across the load and Ir accounts for the voltage drop across the internal resistance, leads to E = 1.5 V + (4.5 A × 2.67 Ω) = 13.5 V, confirming that option D is accurate. Other options are not applicable or do not provide relevant information regarding the emf calculation, making them incorrect. Understanding the relationship between current, voltage, and resistance is crucial for analyzing electrical circuits effectively.
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