Question

$\boxed\bf{Question}$

Three bulbs of 40 W 60 W and 100 W are arranged in series across the mains, Across which the potential difference is highest? Which bulb has lowest potential difference? ​

Answer 1

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.

Answer 2

Formula of Power = V × I

In Series the Current across all the components remain Same ' ɪ '

• Potential Difference across,

40w bulb = 40w / I

• Potential Difference across,

60w bulb = 60 / I

• Potential Difference across,

100w bulb = 100 / I

Here, Potential Difference across 100 w bulb is Maximum & minimum across 40 w bulb.

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