sianphilpott3365 sianphilpott3365
  • 02-09-2019
  • Physics
contestada

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

Respuesta :

nuuk nuuk
  • 05-09-2019

Answer:267

Explanation:

Given

Power of light bulbs is 45 W

and voltage applied is 240 V

Allowable current is 50 A

and [tex]P=\frac{V^2}{R}[/tex]

[tex]R=\frac{V^2}{P}[/tex]

[tex]R=\frac{240\times 240}{45}=1280 \Omega [/tex]

and wire will trip if resistance drop below

[tex]R_{total}=\frac{240}{50}=4.8 \Omega [/tex]

Therefore [tex]R_{total}=\frac{R}{n}[/tex]

[tex]n=\frac{R}{R_{total}}=\frac{1280}{4.8}=266.667 \approx 267[/tex]

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