JAMB Physics 2024 Paper

Concept:Efficiency compares useful output energy to the total input energy, expressed as a percentage.Explanation:The formula for efficiency is:$$\text{Efficiency} = \frac{\text{Output energy}}{\text{Input energy}} \times 100\%$$The accumulator is $90\%$ efficient, so in fraction form it is $0.9$.The output energy given out is $2700\, \text{J}$.Substitute into the formula:$$90\% = \frac{2700\, \text{J}}{\text{Input energy}} \times 100\%$$Convert $90\%$ to decimal and remove the percentage factor:$$0.9 = \frac{2700\, \text{J}}{\text{Input energy}}$$Make input energy the subject:$$\text{Input energy} = \frac{2700\, \text{J}}{0.9}$$$$\text{Input energy} = 3000\, \text{J}$$Answer:The accumulator takes in $3000\, \text{J}$ of energy, so the correct option is B. $3000\, \text{J}$.