Q. A current of 2 A flows through a 2 Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Ω resistor. The internal resistance of the battery is
A) 1 Ω
B) 0.5 Ω
C) 1/3 Ω
D) 1/4 Ω
Explanation
[Solution from source] The emf is the same in both cases: ε = 2(2 + r) = 0.5(9 + r). So 4 + 2r = 4.5 + 0.5r, giving 1.5r = 0.5 and r = 1/3 Ω (ε = 4.67 V). Checking 1/3 Ω: 2 × 2.333 = 4.67 and 0.5 × 9.333 = 4.67, consistent. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.