WAEC Physics 2023 Paper

Concept:The current drawn by the kettle is calculated from its power rating, and the fuse value must be slightly higher than this normal operating current.Explanation:The kettle is rated at $P = 1.06\,\text{kW} = 1060\,\text{W}$ and connected to a supply of $V = 220\,\text{V}$.Using the electrical power formula $P = VI$, we can find the current drawn.Rearrange to get $I = \frac{P}{V}$.Substitute the values: $I = \frac{1060}{220}$.This gives $I \approx 4.82\,\text{A}$.A fuse of exactly $4\,\text{A}$ would melt during normal operation because the kettle draws more than $4\,\text{A}$.A fuse of $6\,\text{A}$ would work, but it is less protective because it allows a larger overload before blowing.The most appropriate fuse rating is the one just above $4.82\,\text{A}$, which is $5\,\text{A}$.Answer:C. 5A