Q159 / 180 · KCET 2017 · Physics

The output of a step down transformer is measured to be \( 48 \mathrm{~V} \) when connected to a \( 12 \mathrm{~W} \)
bulb. The value of peak current is

  • A
    ( \( \frac{1}{\sqrt{2}} \mathrm{~A} \)
  • B
    \( \sqrt{2} A \)
  • Correct answer
    C
    \( \frac{1}{2 \sqrt{2}} A \)
  • D
    \( \frac{1}{4} A \)

Solution

Given, output of step down transformer is $48 \mathrm{~V}$ $\Rightarrow$ peak secondary voltage $V_{s}=48 \mathrm{v}$ Also, power at secondary $P_{s}=12 W$ Now, current at secondary, $I_{s}=\frac{P_{s}}{V_{s}}=\frac{12}{48}=0.25 \mathrm{~A}$ Thus, value of peak current is $I_{S} \sqrt{2}=0.25 \times \sqrt{2}=\frac{1}{2 \sqrt{2}} A$

Hence, the correct answer is option: 'C'.

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