Question

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be

Answer 1

Step 1: Given data:

15 Bulbs Of $40W=15\times 40=600$

5 Bulbs Of $100W=5\times 100=500$

5 Fans Of $80W=5\times 80=400$

1 Heater Of $1kW=1*1000=1000W$

The voltage of the electric mains is $220V$

Step 2: Formula used:

Power is the rate at which an activity or work is completed in the shortest amount of time is referred to as power.

$P=V\times I$ [ where P=power, V=voltage, I=current ]

Step3: Calculating the current

Total Power= $600W+500W+400W+1000W=2500W$

$$\begin{gathered} P=V\times I \\ I=\frac{P}{V} \\ I=\frac{2500}{220} \\ I=\frac{125}{11} \\ I=11.36\cong 12A \end{gathered}$$

Therefore, the minimum capacity of the main fuse of the building will be 12 Ampere.