B1 B2 and B3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.
(i) Since B1, B2 and B3 are in parallel, the potential difference across each of them will remain same. So when the bulb Bx gets fused,B1, B2 and B3 have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.
(ii) Resistance of the parallel combination when all the three bulbs are glowing
1/RP = 1/R + 1/R + 1/R = 3/R
RP = R/3
Ammeter 'A' reads 3A current
So, V = IRP
4.5 = 3 x R/3
R = 4.5
So, resistance of each bulb = 4.5
Now when bulb B2 gets fused, the equivalent resistance of parallel combination of B1 and B3 is
As 1/RP = 1/R + 1/R = 2/R (Bulbs are identical)
RP' = R/2
Ammeter 'A' reads now, I' = V/R'P
I' = 4.5/R/2
= 4.5 x 2/4.5 = 2 A
Since resistance of each arm is same and p.d. is also same, current divides them equally. So lA current will pass through each bulb B1 and By
Therefore, ammeter A1 and A3 reads lA current while A2 will read zero and ammeter A read 2A current.
(iii) In parallel, total power consumed
Peq =P1 + P2 + P3
So, when all the three bulbs glow together
Peq = P + P + P (As P1 = P2 = P3 = P)
= 3P = 3 x V x I
= 3 x 4.5 x I = 13.5 W (Current through each bulb = 1A)