Question

The output of a step-down transformer is measured to be 24 V when connected to a 12 watt light bulb. The value

of the peak current is

• A. $\frac{1}{\sqrt{2}}A$
• B. $\sqrt{2}A$
• C. $\text{2 A}$
• D. $2\sqrt{2}\text{ A}$

Answer 1

Answer: (A)

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

Answer 2

: Here, $V_{s} = 24V,P_{s} = 12W$

$I_{s} = \frac{P_{s}}{V_{s}} = \frac{12}{24} = 0.5A$

$I_{m} = \sqrt{2}I_{s} = \sqrt{2} \times 0.5 = \frac{1}{\sqrt{2}}A$