30 bulbs were connected in series and one was fused the other 29 bulbs were joint in series to the same supply then what will be the light in the room

471 Views Updated: 19 Apr 2018
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Ans : Increase

Explanation :- Let power and voltage of each bulb are P and V respectively.

Then, resistance of each bulb is R = V²/P 

Case 1:- When all the bulbs are connected in series then, equivalent resistance of circuit is given by Req = R + R + R + ..... Upto 30 terms 

Req = 30R = 30V²/P 

current passing through each bulb is i₁ = E/Req = E/30V²/P = PE/30V² -----(1)

Here E is the potential difference of circuit.

Case2:- Now, a bulb is fused. And 29 bulbs are connected in circuit.

So, equivalent resistance of 29 bulbs in series combination, 

Req = R + R + R + .... Upto 29 terms 

Req ' = 29R = 29V²/P 

Current passing each bulb is i₂ = E/29V²/P = PE/29V² -----(2/

What we observed ? From case 1 and case 2 i₂ > i₁ 

e.g., current flow through each bulb is higher than initial. 

Means power out put = i²R will be higher 

Hence, room's light increased.

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