An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Given: Power rating (P) = 2 kW = 2000 W, Voltage (V) = 220 V, Current rating of circuit = 5 A
Step 1: Calculate the current drawn by the electric oven. Using the formula: P = V × I Therefore, Current (I) = P/V = 2000 W ÷ 220 V = 9.09 A
Step 2: Compare the current drawn with the circuit rating. The current drawn by the oven (9.09 A) is greater than the current rating of the circuit (5 A).
Result: The circuit will get overloaded. Due to this overloading, the electric fuse will melt and break the circuit. This happens because the Joule heating produced by the high current causes the fuse to melt, thereby protecting the circuit and appliance from damage. The oven should be connected to a circuit with a higher current rating (like 15 A) meant for appliances with higher power ratings.
Explanation
This question tests the student's understanding of power, current, and the protective role of fuses in domestic circuits. The calculation shows that the oven draws approximately 9.09 A, which exceeds the 5 A circuit rating. The textbook context explains that a fuse prevents damage from overloading by melting when current exceeds the safe limit. The context also mentions that higher power appliances like geysers and air coolers require 15 A circuits, which would be appropriate for this 2 kW oven.
Solution Steps
Step 1: Convert power to watts: 2 kW = 2000 W
Step 2: Apply formula I = P/V to find current drawn
Step 3: Calculate I = 2000/220 = 9.09 A
Step 4: Compare 9.09 A with circuit rating of 5 A
Step 5: Conclude that overloading will occur and fuse will blow