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JEE Main 2020 (7 Jan, Shift 2) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from JEE Main 2020 (7 Jan, Shift 2)

Q39JEE Main 2020 · 7 JAN · SHIFT 2Medium

Electrical Power & Heating Effect

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:

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Answer: (b)

  1. Total power =15(45)+15(100)+15(10)+2(1000)=675+1500+150+2000=4325= 15(45) + 15(100) + 15(10) + 2(1000) = 675 + 1500 + 150 + 2000 = 4325 W.
  2. Current I=PV=4325220≈19.7I = \frac{P}{V} = \frac{4325}{220} \approx 19.7 A.
  3. Fuse must be rated just above this: 20 A. So the answer is (b).
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