KCET · Physics · Magnetic Effects of Current
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 \( \sqrt{2} A \)
- C \( \frac{1}{2 \sqrt{2}} A \)
- D \( \frac{1}{4} A \)
Answer & Solution
Correct Answer
(C) \( \frac{1}{2 \sqrt{2}} A \)
Step-by-step Solution
Detailed explanation
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 \(I_{S} \sqrt{2}=0.25 \times \sqrt{2}=\frac{1}{2 \sqrt{2}} A\)
\(\Rightarrow\) 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 \(I_{S} \sqrt{2}=0.25 \times \sqrt{2}=\frac{1}{2 \sqrt{2}} A\)
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