Question

30 bulbs are connected in series one bulb gets fused remaining 29 bulbs are joined in series and connected to the same supply the light in the room will be

Answer 1

Answer :- option (a) is correct

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.

Answer 2

The correct option is option a.

Explanation to the 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 current.

Means power output = i²R will be higher

Hence, room's light increased.