If you have an engine that supplies 5 hp but is only 85% efficient, what's the power input?

\(\text { Efficiency }=\frac{\text { Power }_{\text {out }}}{\text { Power }_{\text {in }}} \times 100\)
\(E=\frac{P_{o}}{P_{\mathbf{i}}} \times 100\)
So
\( {P_i} = \frac{{{P_o}}}{{\frac{E}{{100}}}} \) 

\({P_i} = \frac{5}{{\frac{{85}}{{100}}}} = \frac{5}{{0.85}} = 5.88hp\)