Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
$\mathrm{B}_{1}, \mathrm{~B}_{2}$ and $\mathrm{B}_{3}$ are the three identical bulbs connected to a battery of steady emf with key $\mathrm{K}$ closed. What happens to the brightness of the bulbs $B_{1}$ and $B_{2}$ when the key is opened?


PhysicsCurrent ElectricityKCETKCET 2010
Options:
  • A Brightness of the bulb $\mathrm{B}_{1}$ increases and that of $\mathrm{B}_{2}$ decreases
  • B Brightness of the bulbs $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ increase
  • C Brightness of the bulb $\mathrm{B}_{1}$ decreases and $\mathrm{B}_{2}$ increases
  • D Brightness of the bulbs $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ decrease
Solution:
2410 Upvotes Verified Answer
The correct answer is: Brightness of the bulb $\mathrm{B}_{1}$ decreases and $\mathrm{B}_{2}$ increases
When key $\mathrm{K}$ is opened, bulb $\mathrm{B}_{2}$ will not draw any current from the source, so that terminal voltage of source increases. Hence, power consumed by bulb increases, so light of the bulb becomes more. The brightness of bulb $\mathrm{B}_{1}$ decreases.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.