JEE · NEET Previous Year Question
In a large building, there are 15 bulbs of 40W, 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:
Topic: JEE PYQ
Answer: C. Full step-by-step solution below.
- A8 A
- B10 A
- C12 A
- D14 A
Correct answer: C
Solution
(c) Total power (P) = (15 × 40) + (5 × 100) + (5 × 80) + (1 × 1000) = 2500WP = VI
Þ I = A
=
= 11.3 A
Minimum capacity should be 12 A
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