Question

What minimum number of 45 W lightbulbs must be connected in parallel to a single 240 V household circuit to trip a 50.0 A circuit breaker? ____lightbulbs

Answer 1

Answer:267

Step-by-step explanation:

Given

Power of light bulbs is 45 W

and voltage applied is 240 V

Allowable current is 50 A

and
`P=(V^2)/(R)`

`R=(V^2)/(P)`

`R=(240* 240)/(45)=1280 \Omega`

and wire will trip if resistance drop below

`R_(total)=(240)/(50)=4.8 \Omega`

Therefore
`R_(total)=(R)/(n)`

`n=(R)/(R_(total))=(1280)/(4.8)=266.667 \approx 267`