Three identical bulbs, B1, B2 and B3 connected. Current of 3A is recorded by the ammeter when the bulbs glow.

Three identical bulbs, B1, B2 and B3 connected. A current of 3A is recorded by the ammeter when ​the bulbs glow. What happens when:

To the glow of the other two bulbs when one of them gets fused?

Assuming that B₁ is fused, there supposedly will be no impact on the other two bulbs. The two bulbs will remain the same since they are dependent on the power. This happens because the potential difference and resistance of the other two bulbs remain the same, and do not get affected by an external fuse of another bulb.

To the reading of A1, A2, A3 and A when the bulb B2 gets fused? 

Considering that there are parallel connections, the net resistance should be:

1/R = 1/R₁ + 1/R₂ + 1/R₃

Now, since the resistance remains same, thus, R’ = R/3

Hence, according to Ohm’s Law,

Thus,

R = 4.5Ω

Assuming that B₂ is fused, two bulbs, B₁ and B₃, remain parallel

Therefore, the net resistance in parallel combination is = 1/R’ = 2/R

⇒ R’ = 4.5/2 Ω

⇒ I = V/R’

⇒ I = 2 × 4.5 / 4.5

⇒ I = 2A

The distribution of current in both the bulbs will be1 A each.

How much power is dissipated in the circuit when all the three bulbs glow together?

The amount of power that is dissipated when all the three bulbs glow together:

P = V × I

Therefore, after replacing the values, we get,

P= 4.5 × 3

⇒ P = 13.5 W

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