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