Kevin Nash
Last Activity: 10 Years ago
Sol. Output voltage = 220 ± 1% 1% of 220 V = 2.2 v
The resistance of bulb R = V^2/P = (220)^2/100 = 484 Ω
(a) For minimum power consumed V base 1 = 2220 – 1% = 220 – 2.2 = 217.8
∴ i = V base 1/R = 217.8/484 = 0.45 A
Power consumed = i * V base 1 = 0.45 * 217.8 = 98.01 W
(b) for maximum power consumed V base 2 = 220 + 1% = 220 + 2.2 = 222.2
∴ i = V base 2/R = 222.2/484 = 0.459
Power consumed = i * V base 2 = 0.459 * 222.2 = 102 W
