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.

Asked by Jai Gupta | 18th May, 2013, 09:37: AM

Expert Answer:

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

Answered by  | 21st May, 2013, 05:34: AM

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