A 100 W bulb B 1 two 60 W bulb B 2 and B3 are connected to a 250 V source as shown in the figure- Now W 1- W 2 and W 3 are the output powers of the bulbsB 1- B 2 and B 3 respectively- ThenA- W 1- W 2- W 3B- W1-W2-w3C- W 1- W 2- W 3D- W 1- W 2- W 3

The correct option is C W1 lt; W2 lt; W3Voltage across B3 is greatest; hence B3 will show maximum brightness. In series combination of bulbs, the bulb of ...

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Standard XII

Physics

Self Induction

A 100 W bulb ...

Question

A 100 W bulb B₁two 60 W bulb B₂ and B₃ are connected to a 250 V source as shown in the figure. Now W₁, W₂ and W₃ are the output powers of the bulbs B₁, B₂ and B₃ respectively. Then

W₁> W₂ = W₃

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W₁> W₂> W₃

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W₁< W₂ = W₃

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W₁< W₂< W₃

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Solution

The correct option is D
W₁< W₂< W₃

Voltage across B₃ is greatest; hence B₃ will show maximum brightness. In series combination of bulbs, the bulb of lesser wattage will glow more bright.
Hence W₂> W₁.So, W₁< W₂< W₃.

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A 100 W bulb B1two 60 W bulb B2 and B3 are connected to a 250 V source as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively. Then

The correct option is D W1 < W2< W3Voltage across B3 is greatest; hence B3 will show maximum brightness. In series combination of bulbs, the bulb of lesser wattage will glow more bright. Hence W2> W1.So, W1 < W2< W3.

A 100 W bulb B₁two 60 W bulb B₂ and B₃ are connected to a 250 V source as shown in the figure. Now W₁, W₂ and W₃ are the output powers of the bulbs B₁, B₂ and B₃ respectively. Then

The correct option is D
W₁< W₂< W₃

Voltage across B₃ is greatest; hence B₃ will show maximum brightness. In series combination of bulbs, the bulb of lesser wattage will glow more bright.
Hence W₂> W₁.So, W₁< W₂< W₃.