## College Physics (4th Edition)

$P_0 = V_0^2/R_0 = 100.0 ~W$ $V = 0.900~V_0$ $P = V^2/R = (0.900~V_0)^2/R_0 = 0.810 ~V_0^2/R_0$ $P = 0.810 ~P_0 = 81.0 ~W$ The lightbulb draws 81.0 watts of power during the brownout.