Question

Three identical light bulbs, each of resistance 12 Ω, are connected in parallel to a 12 V battery. The current passing through each light bulb

Answer 1

Answer:3A

Explanation:total resistance of three bulbs

= 1/R1 + 1/R2 + 1/R3 = 1/Req

1/12 +1/12 +1/12 = 3/12= 1/4

Or Req = 4

I= V/R

Thus I= 12/4 = 3A