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A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is

1. 25 J
2. 50 J
3. 200 J
4. 400 J

Correct answer: 400 J

Solution

The resistance of the original coil is R = V²/P = (220)²/100 = 484 Ω. When the coil is cut in half, each half has resistance R/2 = 242 Ω. When connected in parallel, the equivalent resistance is (1/Req) = (1/242) + (1/242), giving Req = 121 Ω. The power is now P = V²/Req = (220)²/121 = 400 W, so the energy liberated per second is 400 J.

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