B₁, B₂ and B₃ are the three identical bulbs connected to a battery of steady e.m.f. with key K closed. What happens to the brightness of the bulbs B₁ and B₂ when the key is opened?

CONCEPT:

• The brightness of the bulb is directly proportional to the current passing through it.
• When the key is opened, the current through that wire does not pass.

Current (I) in a branch is given by the Ohm's law: 

\(I={V \over R}\)

where V is the voltage in that branch and R is the resistance of that branch.

EXPLANATION:

• Assuming that the 3 bulbs are identical, thus each of the bulbs offers a resistance of R Ω. Also let's assume the overall voltage across the ends of the battery be V volts.

• Hence the current through the bulb B₁ decreases from 2V/3R to V/2R.
• So the bulb B₁ will decrease brightness.
• The bulb B₂ increases from V/3R to V/2R.
• So the bulb B₂ will decrease brightness.
• So the correct answer is option 1.