The output of a step-down transformer is measured to be $24 V$ when connected to a $12 w a t t$ light bulb. The value of the peak current is

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

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\sqrt{2} A

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2 A

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2 \sqrt{2} A

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Solution

The correct option is A $\frac{1}{\sqrt{2}} A$
Here, $V_{s} = 24 V, P_{s} = 12 W$
$I_{s} = \frac{P_{s}}{V_{s}} = \frac{12}{24} = 0.5 A;$
$I_{m} = \sqrt{2} I_{s} = \sqrt{2} \times 0.5 = \frac{1}{\sqrt{2}} A$

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