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Question

The output of a step-down transformer is measured to be 48 V when connected to a 12 W bulb. The value of the peak current is?

$\frac{1}{\sqrt{2}} \text{ A}$

$\sqrt{2} \text{ A}$

$\frac{1}{2\sqrt{2}} \text{ A}$

$\frac{1}{4\sqrt{4}} \text{ A}$

Solution

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