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
The main fuse must carry the total current when every appliance is on, so add the powers and use \( P = VI \).
\( P_{\mathrm{bulbs}} = 20\times40 = 800\,\mathrm{W} \)
\( P_{\mathrm{bulbs}}' = 5\times100 = 500\,\mathrm{W} \)
\( P_{\mathrm{fans}} = 5\times80 = 400\,\mathrm{W} \)
\( P_{\mathrm{heater}} = 1000\,\mathrm{W} \)
\( P = 800 + 500 + 400 + 1000 = 2700\,\mathrm{W} \)
\( \displaystyle I = \frac{P}{V} = \frac{2700}{220} = 12.27\,\mathrm{A} \)