Chapter 31 - Fundamentals of Circuits - Exercises and Problems - Page 917: 37

$P_{actual} = 96.64W$

Resistance of bulb $R = \frac{V^2}{P} = \frac{(120)^2}{100}\Omega$ $R = 144\Omega$ equivalent resistance of corroded and bulb $R_{eq} =144\Omega + 5\Omega =149\Omega$ Actual power dissipated $P_{actual} =\frac{V^2}{R_{eq}}= \frac{(120)^2}{149} W$ $P_{actual} = 96.64W$