Three similar light bulbs are connected to a constant voltage dc supply as shown in figure. Each bulb operates at normal brightness and the ammeter (of negligible resistance) registers a steady current. The filament of one of the bulbs  breaks. What happens to the ammeter reading and to the brightness of the remaining bulbs? 

Suppose V is the voltage of the supply and R is the resistance of each bulb. Now, $S_{1}nor{R}_p=R/3$ and current in ammeter is $I=V/R_p=3V/R,$ provided all three bulbs are working properly. If one bulb has broken down, then $R_p-\frac{R}{2}andI=2\frac{V}{R}$

Therefore, current decreases and since current through each bulb is V/R, the same as before, brightness of bulbs is not affected.