An accumulator is 90% efficient. If it gives out 2700J of energy while discharging, how much energy does it take in?
A
3600J
B
3000J
C
3200J
D
2430J
Verified Correct Answer: B.
Physical Principles & Numerical Solution
Efficiency = \(\frac{\text{ Output energy}}{\text{input energy}}\) x 100
Efficiency = 90% = 0.9, Output energy = 2700J
Input energy = \(\frac{2700}{0.9}\) = 3000J
Source: WAEC / JAMB / NECO Classroom Archive (Record #70681)