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]
