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(i) There will be no effect on glow of other two bulbs and will remain same when B1 gets fused because glowing of bulb depends on power and the potential difference and resistance remains same of other two bulbs.

(ii) When there are parallel connections:

Net resistance will be 1/R = 1/R1+1/R2+1/R3

Since resistance is same so, R’ = R/3

Applying ohm’s law V= IR

R =4.5Ω

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

Therefore net resistance in parallel 1/R’ =2/R

R’ = 4.5/2 Ω

I = V/R’ = 2×4.5/4.5

I = 2A

So, 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 = 13.5 W

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