Discussion
An alternating current having the peak value \(10\sqrt{2}A\) is used to heat a metal wire. To produce the same heating effect, the constant current required is
Answer: E
The constant current is 10A.
The heating effect of an alternating current equals that of a direct current equal to its rms value:
\[ \begin{aligned} I_{rms} &= \frac{I_{0}}{\sqrt{2}} \\ &= \frac{10\sqrt{2}}{\sqrt{2}} = 10\ \text{A} \end{aligned} \]
\(10\sqrt{2}\) A is the peak, which is reached only momentarily, so it would overheat the wire. 14 A is the peak rounded, and 5 A or 7 A halve the correct values.
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