Physics

Grade-9

Easy

Question

# A technician examines different electrical devices to determine the one that is the most energy efficient. While conducting a test, he notes that one of these devices consumes 650 000 J of energy and loses 315 000 J at the same time. What is the energy efficiency of this device?

- 42.7%
- 51.5%
- 68.1%
- 174.6%

Hint:

- energy input = useful energy output + waste energy
- Efficiency (%) = (useful energy output ÷ energy input) × 100

## The correct answer is: 51.5%

- energy input = useful energy output + waste energy
- useful energy output=energy input- waste energy=650 000 J-315 000=335000J
- Efficiency (%) = (useful energy output ÷ energy input) × 100

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