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Asked in GSEB Board March 2019 · Rated bulbs and appliances
Given: each bulb is rated 220 V, 100 W; supply 220 V.
Resistance of each bulb: R = (V²)/P = (220²)/(100) = 484 Ω.
Parallel: each bulb gets the full 220 V and gives its rated 100 W, so the total is 200 W.
Series: total resistance 2R = 968 Ω, so P = (V²)/(2R) = (220²)/(968) = 50 W.
So the total powers are 200 W (parallel) and 50 W (series), in that order.
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