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

Brightness of the bulb B₁ decreases and B₂ increases

Let EMF of the Source is E and resistance of each bulb is R.

When key is closed then Equivelent resistance is $R_{eq} = R + \frac{R}{2} = \frac{3R}{2}$

Total current is $I = \frac{E}{R_{eq}}= \frac{2E}{3R}$

Current through B₁ is $ \frac{2E}{3R}$ , current through B₂ and B₃ is $\frac{I}{2} =\frac{E}{3R}$

After Key is opened 

$R_{eq} = 2R$

Total current is $ I' = \frac{E}{2R}$ through both B₁ and B₂

Current through the bulb B₁ decreases and B₂ increases

Hence Brightness of the bulb B₁ decreases and B₂ increases.