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Physics

Fuses

In a large bu...

Question

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

12 A

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14 A

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8 A

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10 A

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Solution

The correct option is A 12 A
Item No. Power
40 W bulb 15 600 Watt
100 W bulb 5 500 Watt
80 W fan 5 400 Watt
1000 W heater 1 1000 Watt

Total Wattage $=$ 2500 Watt

So current capacity $i = \frac{P}{V} = \frac{2500}{220} = \frac{125}{11} = 11.36 ≅ 12 A m p .$

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Similar questions

Q. In a large building, there are

15

bulbs of

40 W

5

bulbs of

100 W

5

fans of

80 W

and

1

heater of

1 kW

. The voltage of the electric mains is

220 V

. The minimum capacity of the main fuse of the building will be:

Q. In a large building , there are

15

bulbs of

40 W

5

bulbs of

100 W

5

fans of

80 W

and

1

heater of

1 KW

. If the voltage of main electric supply is

220 V

then the minimum current capacity of the main fuse will be:

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

Q. In a building there are

15

bulbs of

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bulbs of

100 W, 15

small fans of

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and

2

heaters of

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. The voltage of electric main is

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. The minimum fuse capacity (rated value) of the building will be:

Fig. 9.12 shows three bulbs A, B and C each of rating 100 W, 220 V connected to the mains of 220 V. Answer the following:

(a) How is the bulb A connected with the mains? At what voltage does it glow?

(b) How are the bulbs B and C connected with the mains? At what voltage does the bulb B glow?

(d) How is the glow of bulbs B and C affected if bulb A gets fused?

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Item	No.	Power
40 W bulb	15	600 Watt
100 W bulb	5	500 Watt
80 W fan	5	400 Watt
1000 W heater	1	1000 Watt

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

The correct option is A 12 AItem No.Power40 W bulb15600 Watt100 W bulb5500 Watt80 W fan5400 Watt1000 W heater11000 WattTotal Wattage = 2500 WattSo current capacity i=PV=2500220=12511=11.36≅12Amp.

The correct option is A 12 A
Item No. Power
40 W bulb 15 600 Watt
100 W bulb 5 500 Watt
80 W fan 5 400 Watt
1000 W heater 1 1000 Watt

Total Wattage $=$ 2500 Watt

So current capacity $i = \frac{P}{V} = \frac{2500}{220} = \frac{125}{11} = 11.36 ≅ 12 A m p .$