Can anyone answer and explain? Thanks.

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1 Answer
Feb 25, 2018

C.C.

Explanation:

Let,the resistance of each of the three lamp is rr,

Now, lamp XX and YY are in parallel combination,so their net resistance is (r*r)/(r+r) =r/2rrr+r=r2

They are again in series with lamp ZZ, so net resistance of the circuit is r+ r/2 =(3r)/2r+r2=3r2

If the voltage of the battery is VV ,then current flowing through the circuit is V/((3r)/2)=(2V)/(3r)V3r2=2V3r

So,current flowing through lamp ZZ is (2V)/(3r)2V3r

Now,potential drop across lamp XX and YY is the same as they are in parallel combination, So the value is (V - (2V)/(3r)*r)=V/3(V2V3rr)=V3

So,current flowing through lamp YY is (V/3) /r =V/(3r)V3r=V3r

So,initially, more current is flowing through lamp ZZ than XX,so it was glowing more.

when XX is no more, lamp ZZ and YY comes in series combination,so net resistance is r+r=2rr+r=2r hence,current flowing through each of them is V/(2r)V2r

So, this value of current is higher w.r.t initial value of current flowing through lamp YY so it becomes glower, but for ZZ this value has decreased than previous,so its brightness decreases.