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A current of 2A flows through a 2Ω resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9Ω resistor. The internal resistance of the battery is
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- 0.5 Ω
- 1/3 Ω
- 1/4 Ω
- 1 Ω
Correct Option: B
Let the internal resistance of the battery be r. Then the current flowing through the circuit is given by
| i = | E | |
| R + r |
In first case,
| 2 = | E | ...(i) |
| 2 + r |
In second case,
| 0.5 = | E | ...(i) |
| 9 + r |
From (1) & (2), 4 + 2r = 4.5 + 0.5 r
⇒1.5 r = 0.5 ⇒ r = 1/3 Ω.
