Three bulbs of 40w, 60w, 100w are connected in series across the mains. Across which potential difference is highest? Across which is lowest?-plz answer fast
Answers
Answer:
Resistance of 100Wbulb=200
2
/100=400ohms
Resistance of 60Wbulb=200
2
/60=666.67ohms
Resistance of 40Wbulb=200
2
/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=0.0967
2
x1000=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.
Take this,
Resistance of 100W bulb = 200^2/100 = 400 ohms
Resistance of 60W bulb = 200^2/60 = 666.67 ohms
Resistance of 40W bulb = 200^2/40 = 1000 ohms
Therefore, total resistance in series = (400+666.67+1000) = 2066.67 ohms
Current in the circuit = 200/2066.67 = 0.0967 A
Therefore, actual power consumed by “40W” bulb = 0.0967^2 x 1000 = 9.35 W (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.
N.B. Filament resistance also depends on temperature of the filament. Hence actual current and actual power consumed would be even lower than calculated above.
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