Question

30 bulbs are connected in series 1 bulb fused. remining 29 bulbs are connected in series. What is the brightness of 29 bulbs

Answer 1

when all 30 bulbs are connected in series, the current flowing through each bulb is the same, but potential different across the series connection is different. So, when one bulb get fused the remaining bulbs does not glow, because the fused bulb offer a large resistance to the flow of current or it has infinite resistance.

Answer 2

Answer:

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 .

Explanation: