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Asked in CBSE AIPMT 2003 · Rated bulbs and appliances
Each bulb: R=(220²)/(100)=484 Ω.
Series: 2R, so Pₛ=(220²)/(2R)=(100)/2=50 W.
Parallel: R/2, so Pₚ=(220²)/(R/2)=2×100=200 W.
Answer: 50 watt, 200 watt.
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