Output voltage = 220 ± 1%
1% of 220V = 2.2v
The resistance of bulb R = V2/P = (220)2/100 = 484Ω
(a) For minimum power consumed V1 = 220 – 1% = 220 – 2.2 = 217.8
∴ i = V1/R = 217.8/484 = 0.45A
Power consumed = i × V1 = 0.45 × 217.8 = 98.01W
(b) for maximum power consumed V2 = 220 + 1% = 220 + 2.2 = 222.2
∴ i = V2/R = 222.2/484 = 0.459
Power consumed = i × V2 = 0.459 × 222.2 = 102W