JEE MAIN - Physics (2020 - 7th January Evening Slot - No. 12)
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 :
15 A
20 A
25 A
10 A
Explanation
Total power is
= (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
= 4325 W
$$ \therefore $$ Current = $${{4325} \over {220}}$$ = 19.66 A $$ \simeq $$ 20 A
= (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
= 4325 W
$$ \therefore $$ Current = $${{4325} \over {220}}$$ = 19.66 A $$ \simeq $$ 20 A
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