In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be :
Solution
Total power is <br><br>= (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
<br><br>= 4325 W
<br><br>$\therefore$ Current = ${{4325} \over {220}}$ = 19.66 A $\simeq$ 20 A
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Electrical Power
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