An accumulator is 90% efficient. If it gives out 2700J of energy while discharging, how much energy does it take in?

Analyze the question: An accumulator (a rechargeable battery) is 90% efficient. The energy it gives out (output) is 2700 J, and we need the energy it takes in (input).

Concept: Efficiency is the ratio of useful energy output to energy input:

Efficiency = (Energy output / Energy input) × 100%

Solution:

90 = (2700 / Input) × 100

Input = (2700 × 100) / 90

Input = 270000 / 90

Input = 3000 J

Check: 90% of 3000 J = 0.9 × 3000 = 2700 J ✓

Why not the other options?

• D (2430 J) is what you get if you wrongly multiply 2700 by 0.9. That would be finding 90% of the output, not the input.
• A and C do not satisfy the efficiency relationship when you check them: 90% of 3600 is 3240, and 90% of 3200 is 2880. Neither equals 2700.

Answer: The accumulator takes in 3000 J, which is option B.

Exam tip: Output is always smaller than input for real machines, so the answer must be bigger than 2700 J. That eliminates D immediately. Then simply divide the output by the efficiency as a decimal: 2700 ÷ 0.9 = 3000.