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A servo voltage stabiliser restricts the voltage output to 220V ± 1%. If an electric bulb rated at 220V, 100W is connected to it, what will be the minimum and maximum power consumed by it?

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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

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