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

Explanation:

Given

Power of light bulbs is 45 W

and voltage applied is 240 V

Allowable current is 50 A

and $P=\frac{V^2}{R}$

$R=\frac{V^2}{P}$

$R=\frac{240\times 240}{45}=1280 \Omega$

and wire will trip if resistance drop below

$R_{total}=\frac{240}{50}=4.8 \Omega$

Therefore $R_{total}=\frac{R}{n}$

$n=\frac{R}{R_{total}}=\frac{1280}{4.8}=266.667 \approx 267$