JEE PYQ: Current Electricity Question 28
Question 28 - 2020 (07 Jan 2020 Shift 2)
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:
(1) 10 A
(2) 25 A
(3) 15 A
(4) 20 A
Show Answer
Answer: (4)
Solution
Total power $P = 15 \times 45 + 15 \times 100 + 15 \times 10 + 2 \times 1000 = 675 + 1500 + 150 + 2000 = 4325$ W. But considering $15 \times 155 + 2000$: $I_{\text{min}} = \frac{P}{V} = \frac{15 \times 155 + 2000}{220} \approx 19.66 \approx 20$ A.
BETA
