STANDARD Physics Power and Energy MCQ Question
A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel, to the same source of 200 V. The energy liberated per second in the new combination is
100 J
200 J
300 J
400 J
Correct Answer
Detailed Explanation
The resistance of the original heater coil is 400 Ω. When cut into two identical parts, each part has a resistance of 200 Ω. When connected in parallel, the equivalent resistance is 100 Ω. The power dissipated is calculated using P = V²/R, which results in 400 J per second.
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