Office cable cross-section

In the office will be connected: computer 800 W, kettle 2.5 kW, lighting 650 W, oven 1.9 kW; electric stove 3.3 kW; air conditioning 6100 W. As the smallest cross-section of the cables, we have to bring them to this room. Recalculate for Copper and Aluminum wires.

Final Answer:

S =  6.216 mm²

Step-by-step explanation:

$$\begin{gathered} P_1=800\text{W} \\ {P}_2=2.5\text{kW}\to \text{W}=2.5\cdot 1000\text{W}=2500\text{W} \\ {P}_3=650\text{W} \\ {P}_4=1.9\text{kW}\to \text{W}=1.9\cdot 1000\text{W}=1900\text{W} \\ {P}_5=3.3\text{kW}\to \text{W}=3.3\cdot 1000\text{W}=3300\text{W} \\ {P}_6=6100\text{W} \\ P=P_1+P_2+P_3+P_4+P_5+P_6=800+2500+650+1900+3300+6100=15250\text{W} \\ U=230\text{V} \\ P=UI \\ I=P/U=15250/230=\frac{1525}{23}\doteq 66.3043\text{A} \\ {S}_{16}=1.5{\text{mm}}^2 \\ n=I/16=66.3043/16=\frac{1525}{368}\doteq 4.144 \\ S=n\cdot S_{16}=4.144\cdot 1.5=6.216{\text{mm}}^2 \end{gathered}$$

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