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A $12\,\text{V}$ battery drives a current of $2\,\text{A}$ through a resistor. The POWER dissipated is
A$6\,\text{W}$
B$10\,\text{W}$
C$24\,\text{W}$
D$144\,\text{W}$
Answer & Solution
Correct answer: C. $24\,\text{W}$
1. POWER: $P = VI$.
2. $P = 12 \times 2 = 24\,\text{W}$.
3. Equivalents: $P = I^2 R$ (using $V=IR$) and $P = V^2/R$. All give 24 W here.
4. Option A is $V/I$. Option B is unrelated. Option D is $V^2$.
_Source: Tony Kuphaldt, "Lessons in Electric Circuits — DC", Vol I, Ch 2 (Power)._
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