A cell of internal resistance 2 π supplies current to a 6 π resistor. The efficiency of the cell is
12.0%
25.0%
33.3%
75.0%
Correct answer is C
Internal resistance determine the maximum current that can be supplied
Efficiency = \(\frac{r}{R}\) × 100
= \(\frac{2}{6}\) × 100
= 33.3%