Three light bulbs of 40W, 60W and 100W are connected in series with 220 volt source. Which of the bulbs will grow brightest?
A
60W
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B
40W
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C
100W
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D
All with same brightness
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Solution
The correct option is B 40W
Resistance of 100Wbulb=2002/100=400ohms
Resistance of 60Wbulb=2002/60=666.67ohms
Resistance of 40Wbulb=2002/40=1000ohms
Therefore, total resistance in series = (400+666.67+1000)=2066.67ohms
Current in the circuit =200/2066.67=0.0967A
Therefore, actual power consumed by “40W”bulb=0.09672x1000=9.35W (much lesser than any of the original)
The 40W bulb will grow the brightest as the current is constant in all three and it has the maximum resistance. But it would consume much less than 40W as the bulbs are connected in series, and voltage would be divided across all three filaments depending upon resistances.