Electric light bulbs are normally connected in parallel across the 230V supply in your house.

Question

Assume you have 5 of them, each rated at 60W. In a re-wire they are mistakenly conneted in series. What would the power output of each lamp be now?

Really hope someone can help with this!

Answer 1

Now each bulb has (230/5) V across. Assuming resistance (230^2/60 W)

to be the same

power consumed will be

(230/5)^2 / (230^2/60)

= 1 X (60W/5^2)

= 2.4 W.

All the 5 together consume 12 W

But resistance changes (increases) with filament glow to complicates matters.

Answer 2

Easy enough, but I have to make one big false assumption: that the resistance of a light bulb does not vary with voltage. But in fact it does, by a large factor. So the answer does not have any relation to reality, and the power dissipation will be a lot higher.

P = E²/R

60 = 230²/R

R = 882 ohms

put them in series and the total R is 4408 ohms.

Power dissipated by that value of R is

P = E²/R

P = 230²/4408 = 12 watts

dividing that up  between 5 bulbs and we get

P = 12/5 = 2.4 watts each

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