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Two identical bulbs of power P are connected in parallel. The third identical bulb of power...

Two identical bulbs of power P are connected in parallel. The third identical bulb of

power P is connected in series to this combination. What is the power dissipated in this

circuit in terms of P?

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Answer #1

In case of parallel connection P' = V2/(R/2) = 2V2/R =2P

In case of series connection P" = V2 / R

Total Power = V2/(R+R/2) =2V2/3R = 2P/3

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