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Asked in RS Academy GUJCET booklet · Rated bulbs and appliances
Given: each bulb is rated 220 V, 100 W; the supply is 220 V.
Idea: get each bulb's resistance from its rating, then use P = (V²)/(R_eq) for each arrangement.
R = (V²)/P = (220²)/(100) = 484 Ω.
Series: R_eq = 968 Ω, so P = (220²)/(968) = 50 W.
Parallel: R_eq = 242 Ω, so P = (220²)/(242) = 200 W.
So the total power is 50 W in series and 200 W in parallel.
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