Question

Three identical bulbs are connected in parallel with the battery the current drawn from the battery is 6 ampere if one of the bulb get fused what will be the total cureent drawn from the battery

Answer 1

4 ampere of curent will be drawn

Answer 2

Answer:

Let the resistance of three bulb be 'R' each.

Total current drawn from the battery is I= 6A

Let the potential difference accross the circuit be V.

Total resistance of the circuit will be "RT".

using law of parallel addition of resistances we get:

1/RT = 1/R + 1/R + 1/R

RT = R/3

Hence using Ohms law we get:

V = I x RT

V = 6 x R/3

V = 2 R ----------------------------------eq 1

Now if one of the bulbs get fused  the new total resistance of the circuit will be RT' , due to the two bulbs left.

Thus going the same way as above:

RT = 1/R + 1/R

RT = 2/R --------------------------------eq 2

Now using ohms law again with new value of total resistance along with the potential difference calculated we can get the current drawn by the circuit:

(now from eq 1 and eq 2)

I = 2R/R/2

I = 4 A