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
- 0.5 Ω
- 1/3 Ω
- 1/4 Ω
- 1 Ω
Answer
(2) 1/3 Ω
Solution
= I(R + r)
2(2 + r) = 0.5(9 + r)
4 + 2r = 4.5 + 0.5r
1.5r = 0.5
r = 13
Answer: (2)