A 15 A circuit at 220 V has a refrigerator of 200 W, an iron of 1100 W and a geyser of 1800 W, all…

Step 1. (a) $P = 220\ \text{V} \times 15\ \text{A} = 3300\ \text{W}$.

Step 2. (b) $200\ \text{W} + 1100\ \text{W} + 1800\ \text{W} = 3100\ \text{W}$, so $I = \dfrac{3100\ \text{W}}{220\ \text{V}} = 14.1\ \text{A}$: safe, but close to the limit.

Step 3. (c) $3500\ \text{W}$, so $I = \dfrac{3500\ \text{W}}{220\ \text{V}} = 15.9\ \text{A}$: overloaded, and the fuse melts.