This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
The power rating of 106 kW for a kettle is extremely high and unrealistic for a household appliance. It is highly probable that there is a typo in the question, and the intended power rating was 1.06 kW (1060 W). I will proceed with this assumption to provide a realistic fuse value for a kettle. Step 1: Calculate the operating current (I). The formula for power is P = V × I, where P is power, V is voltage, and I is current. Given: Power (P) = 1.06 kW = 1060 W (assuming a typo in the question) Voltage (V) = 220 V I = P / V I = 1060 W / 220 V I = 4.818 A Step 2: Determine the most appropriate fuse value. A fuse should be rated slightly higher than the normal operating current to prevent it from blowing unnecessarily, but low enough to protect the appliance from overcurrent. Common standard fuse ratings include 1A, 2A, 3A, 5A, 10A, and 13A. Since the calculated operating current is 4.818 A, the next standard fuse rating above this value is 5 A. The most appropriate value of fuse for a kettle rated 220 V, 1.06 kW (1060 W) is a 5 A fuse. That's 2 down. 3 left today — send the next one.