JEE Main 2020 — Current Electricity Question with Solution
From: JEE Main 2020 (Online) 7th January Evening Slot
Question
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 :
Choose an option
Show full solutionCorrect option: B
Correct answer
B20 A
Step-by-step explanation
Total power is
= (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
= 4325 W
Current = = 19.66 A 20 A
= (15 × 45) + (15 × 100) + (15 × 10) + (2 × 1000)
= 4325 W
Current = = 19.66 A 20 A
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This is a previous-year question from JEE Main 2020, covering the Current Electricity chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.