The fuse of the circuit will blow and the circuit will break.
First, calculate the current drawn by the oven: \[ P = VI \quad \Longrightarrow \quad I = \frac{P}{V} \] Given: $P = 3000$ W,\; $V = 220$ V \[ I = \frac{3000}{220} \approx 13.64 \text{ A} \] So, the oven will draw about 13.6 A, which is higher than the circuit’s current rating of 10 A. The fuse wire in the circuit is designed to melt when the current exceeds 10 A to protect the circuit from overheating or fire hazards. Therefore, the fuse will blow and the circuit will open.