Discussion
What will be the power consumed by a resistor of \(1\ \mathrm{k\Omega}\) carrying a current of 20 mA?
Answer: B
The power dissipated in a resistor is \(P = I^2 R\). The current of 20 mA is \(0.02\) A and the resistance is \(1000\ \Omega\).
\[ \begin{aligned} P = (0.02)^2 \times 1000 \\ &= 4 \times 10^{-4} \times 10^3 = 0.4\ \text{W} \end{aligned} \]
Check with \(V = IR = 20\) V, then \(P = VI = 20 \times 0.02 = 0.4\) W. The value 4 W would come from taking the current as 63 mA, and 0.04 W or 0.004 W from slips of a power of ten when converting milliamperes. The answer is 0.4 W, option B.
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