The circuit has a fuse rating of 5 A, which means the total current drawn by all connected devices must not exceed 5 A to prevent the fuse from blowing. Each bulb is rated at 100 W at 220 V. The current drawn by a single bulb is calculated using the formula P = V I, so I = P/V. Current per bulb = 100 W / 220 V = 10/22 A = 5/11 A. When bulbs are connected in parallel, the voltage across each bulb is the same (220 V), and the total current drawn from the source is the sum of the currents drawn by each bulb. Let N be the number of bulbs connected. The total current is N (5/11) A. For safe operation, the total current must be less than or equal to the fuse rating: N (5/11) A ≤ 5 A. N 5 ≤ 5 11 N ≤ 11. Since the number of bulbs must be a whole number, the maximum number of bulbs that can be safely connected in parallel is 11.
A circuit has a fuse having a rating of 5 A. What is the maximum number of 100 W-220 V bulbs that can be safely connected in parallel in the circuit?
Asked in UPSC CDS 2 2018 · Topic: Electric Current (Electrical Power and Energy, Joule heating effect)
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