5.1.4—The concept of efficiency to solve problems

Rearrange η=E_useful/E_input to find an unknown energy or power, then calculate the non-useful share as input minus useful output.

Convert percentages to decimals before substituting and label the direction of the calculation. Check that the answer is physically plausible.

At 80% efficiency, a 2.0 kW useful output requires 2.5 kW input; 0.5 kW is dissipated.

Do not multiply by 100 twice, and do not use dissipated energy as the denominator unless the question defines it as input.