Three identical bulbs are connected in parallel with a battery.The current drawn from the battery is 6A.If one of the bulbs fused, what will be the total current drawn from the battery.

Initially, for a parallel connection

 

1/Req = 1/R + 1/R + 1/R  (where R is the resistance of each bulb)

Req = R/3

I = 6A 

Hence V = IReq 

V = 6*R/3 = 2R

 

If one of the bulb is fused, then the current will still flow in the remaining 2 bulbs. The voltage will remain the same as the battery is same but the current will change. 

 

The new equivalent resistance would then be

 

1/Req = 1/R + 1/R i.e. Req = R/2

 

Hence I = V/Req = 2R/R/2 = 4 A