## Electrical Engineering: Principles & Applications (6th Edition)

$R=100\Omega$ 19% reduction in power
$P=\frac{V^2}R$ so $R=\frac{V^2}{P}$ $R=\frac{(100V)^2}{100W} = 100\Omega$ $P=\frac{(90V)^2}{100\Omega}=81W, 100W-81W=19W\div100W=19\%$