In a large building, there are 20 bulbs of 40 W each, 5 bulbs of 100 W each, 5 fans of 80 W each, and 1 heater of 1 kW. The voltage supply is 220 V. The minimum capacity of the main fuse of the building should be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 12.27 A
Given:
20 bulbs of 40 W → $20\times40=800\ \text{W}$
5 bulbs of 100 W → $5\times100=500\ \text{W}$
5 fans of 80 W → $5\times80=400\ \text{W}$
Heater → $1000\ \text{W}$
Total power:
$P_{\text{total}}=800+500+400+1000=2700\ \text{W}$
Supply voltage: $V=220\ \text{V}$
Required current:
$I=\frac{P_{\text{total}}}{V}=\frac{2700}{220}\approx12.27\ \text{A}$