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

Answer:

Explanation:

 Total power (P) consumed  

                            $=(15\times40)+(5\times100)+(5\times80)+(1\times1000)$

                          =2500W

 As we know,

 power i.e, P=VI

$\Rightarrow$              $I= \frac{2500}{220}A$

  $= \frac{125}{11}=11.3 A$

                 Minimum capacity should be 12 A