Let,the resistance of each of the three lamp is #r#,
Now, lamp #X# and #Y# are in parallel combination,so their net resistance is #(r*r)/(r+r) =r/2#
They are again in series with lamp #Z#, so net resistance of the circuit is #r+ r/2 =(3r)/2#
If the voltage of the battery is #V# ,then current flowing through the circuit is #V/((3r)/2)=(2V)/(3r)#
So,current flowing through lamp #Z# is #(2V)/(3r)#
Now,potential drop across lamp #X# and #Y# is the same as they are in parallel combination, So the value is #(V - (2V)/(3r)*r)=V/3#
So,current flowing through lamp #Y# is #(V/3) /r =V/(3r)#
So,initially, more current is flowing through lamp #Z# than #X#,so it was glowing more.
when #X# is no more, lamp #Z# and #Y# comes in series combination,so net resistance is #r+r=2r# hence,current flowing through each of them is #V/(2r)#
So, this value of current is higher w.r.t initial value of current flowing through lamp #Y# so it becomes glower, but for #Z# this value has decreased than previous,so its brightness decreases.
