In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A ) of the building should be

Question

In a marriage hall, there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. If the voltage of the electric main is 220 V, then the minimum fuse capacity (in A ) of the building should be (A) 20 A (B) 15A (C) 10A (D) 25A

Tardigrade Admin · Accepted Answer

Total power is given by, sum of power of all the electrical components,
i.e.

P_{total} = P_{1} + P_{2} + P_{3} + P_{4}

, where

P_{1}, P_{2}, P_{3}, P_{4}

represents power of

45 W

bulb, power of

100 W

bulb, power of small fan and power of heater respectively.

\Rightarrow P_{total} = (15 \times 45) + (15 \times 100) + (15 \times 10) + (2 \times 1000) = 4325 W

So current is

= \frac{4325}{220} = 19.66 A \approx 20 A

Q. In a marriage hall, there are $15$ bulbs of $45 W,$ $15$ bulbs of $100 W,$ $15$ small fans of $10 W$ and $2$ heaters of $1 kW .$ If the voltage of the electric main is $220 V,$ then the minimum fuse capacity (in $A$ ) of the building should be

$20 A$

$15 A$

$10 A$

$25 A$

Q. In a marriage hall, there are 15 bulbs of 45W, 15 bulbs of 100W, 15 small fans of 10W and 2 heaters of 1kW. If the voltage of the electric main is 220V, then the minimum fuse capacity (in A ) of the building should be

20A

15A

10A

25A

Q. In a marriage hall, there are $15$ bulbs of $45 W,$ $15$ bulbs of $100 W,$ $15$ small fans of $10 W$ and $2$ heaters of $1 kW .$ If the voltage of the electric main is $220 V,$ then the minimum fuse capacity (in $A$ ) of the building should be

$20 A$

$15 A$

$10 A$

$25 A$