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Asked in GUJCET 2009 · Rated bulbs and appliances
Given: each bulb is 250 V, 100 W, so R=(V²)/P=(250²)/(100)=625 Ω.
Series: the total resistance is 2R=1250 Ω across 250 V.
Pₛ=(250²)/(1250)=50 W.
Parallel: each bulb has its rated 250 V across it and gives 100 W.
Pₚ=2×100=200 W.
So the total powers are 50 W and 200 W.
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