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A heater rated to dissipate power P is run from a fixed supply. If its resistance R is halved (voltage unchanged), the power it dissipates becomes
AP/4
BP/2
C2P
D4P
Answer & Solution
Correct answer: C. 2P
1. At fixed voltage V, use $P = V^2/R$.
2. So at constant V, power is inversely proportional to R.
3. Halving R doubles the power: new power $= 2P$.
4. Using $P = I^2 R$ here is a trap because I changes when R changes; P/4 and P/2 misapply that form.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.12_
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