A cell of internal resistance 2Ω supplies current to a 6Ω resistor. The efficiency of the cell is
- 12.0%
- 25.0%
- 33.3%
-
75.0%
The correct answer is: D
Explanation
Internal resistance determine the maximum current that can be supplied
Efficiency = \(\frac{R}{R+ r}\) × 100
= \(\frac{6}{8}\) × 100
= 75%
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