The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to a (220  x .8) volt source, then the actual power would be

P1= 2202R1  and P1= 220 × 0.82R2P2P1 =  220 × 0.822202  × R1R2⇒  P2P1 = 0.82  ×  R1R2Here R2 < R1(because voltage decreases from 220 V → 220 x 0.8 V. It means heat produced → decreases)So R1R2 > 1  ⇒  P2 0.82 P1 ⇒ P2 >  0.82 ×  100  WAlso P2P1 =  220 × 0.8i2220 i1,  Since  i2 < i1  we  expectSo  P2P1 <  0.8 ⇒  P2  <  100  ×  0.8Hence, the actual power would be between 100 x (0.8)2 W and (100 x 0.8) W