JEE Main 2025 · 4 April · evening · Single correct · +4 / −1
There are ' n ' number of identical electric bulbs, each is designed to draw a power p independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is :
- (a) np
- (b) p
- (c) p/n²
- (d) p/n
Answer and solution
Answer (d) p/n
Solution
Physics sir · 3 steps
Each bulb keeps its own resistance, so find R from p first and let the series sum do the rest.
- Resistance of one bulb. On its own across the mains voltage V, a bulb draws p = V²/R. So each bulb has resistance R = V²/p, and this stays the same however the bulbs are connected.
- Series resistance. In series the resistances add, so n identical bulbs give Rₜₒₜₐₗ = nR. The full mains voltage V now acts across this larger resistance.
Pₜₒₜₐₗ = V²/nR = 1/n × V²/R = p/n
- So. The combination draws p/n, which is option (d). Each bulb then gets only p/n², but that is one bulb's share, not the total.
The idea to keep. At fixed supply voltage, P = V²/R. Adding loads in series raises R and lowers the total power. Adding them in parallel lowers R and raises it.
Answer from the NTA final answer key. Solution by PrepWiser.