B1 , B2 and B3 are three identical bulbs connected as shown in Figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A. i) What happens to the glow of the other two bulbs when the bulb B1 gets fused? ii) What happens to the reading of A1 , A2 , A3 and A when the bulb B2 gets fused? iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Answer:

(i) When B1 is fused, there will be no influence on the glow of the other two bulbs, which will remain the same since bulb glowing is dependent on power, and the potential difference and resistance of the other two bulbs remain the same.

(ii) When there are parallel connections

Net resistance will be

1/R = 1/R1 + 1/R2 + 1/R3

Since resistance is the same so, R’ = R/3

As per ohm’s law

V = IR

R = 4.5Ω

Since B2 gets fused, so now only two bulbs B1 and B3 are in parallel

Net resistance in parallel 1/R’ =2/R

R’ = 4.5/2 Ω

I = V/R’

I = 2×4.5/4.5

I = 2A

Current will be distributed in both the bulbs as 1 A each.

(iii) Power dissipated when all three bulbs glow together

P = V × I

P= 4.5 × 3

P = 13.5 W

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