JEE Main2014PhysicsCurrent ElectricityActual
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:
Options
- A8 A
- B10 A
- C12 A
- D14 A
Correct answer
C. 12 A
Step-by-step solution
The power generated by each electrical device is calculated below Item Number Power Consumed 40   W bulb 15 40 × 15 = 600   Watt 100   W bulb 5 100 × 5 = 500   Watt 80   W fan 5 80 × 5 = 400   Watt 1000   W heater 1 1000   Watt The total power consumed = 2500   Watt So the minimum current capacity i = P V = 2500 220 = 125 11 = 11 . 36 ≅ 12   A