Physics – 4.4 Electrical safety | e-Consult

To calculate the minimum current rating of the fuse, we can use Joule's Law, which relates power (P), current (I), resistance (R), and time (t):

P = I²R

We know the resistance of the lamp (R = 240Ω) and the voltage of the mains supply (V = 120V). We can rearrange the formula to solve for the current (I):

I = √(P/R)

First, we need to calculate the power (P) dissipated by the lamp:

P = V² / R = (120V)² / 240Ω = 14400 / 240 = 60W

Now, we can use Joule's Law to find the current:

I = √(P/R) = √(60W / 240Ω) = √(0.25 A²) = 0.5 A

The fuse must be rated slightly higher than the current flowing through the circuit to prevent nuisance tripping. A standard fuse rating slightly above 0.5A would be appropriate. Therefore, a 1A fuse would be a suitable choice. It's always better to choose a fuse with a slightly higher rating than the calculated current to avoid unnecessary interruptions to the circuit. A 1A fuse provides adequate protection for the lamp and the wiring in this scenario.

Table of Fuse Ratings (Example):