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GUJCET · Physics · Alternating Current
The output of a stepdown transformer is measured to be 24 V , When connected to 12 watt light bulb. The value of peak current is ______________ A.
- A 1.41 A
- B 0.71 A
- C 2
- D 2.83
Answer & Solution
Correct Answer
(B) 0.71 A
Step-by-step Solution
Detailed explanation
(B) 0.71 A
output power \(\mathrm{P}_{\mathrm{s}}=\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}\)
\(
\begin{array}{ll}
\therefore & \mathrm{I}_{\mathrm{s}}=\frac{\mathrm{P}_{\mathrm{s}}}{\mathrm{~V}_{\mathrm{s}}} \\
\therefore & \mathrm{I}_{\mathrm{s}}=\frac{12}{24} \\
\therefore & \mathrm{I}_{\mathrm{s}}=0.5 \mathrm{~A} \\
\text { peak current } & \mathrm{I}_{\mathrm{m}}=\sqrt{2} \mathrm{I}_{\mathrm{s}} \\
& \mathrm{I}_{\mathrm{m}}=\sqrt{2} \times 0.5 \\
& \mathrm{I}_{\mathrm{m}}=\frac{1}{\sqrt{2}} \mathrm{~A} \\
& \mathrm{I}_{\mathrm{m}}=0.71 \mathrm{~A}
\end{array}
\)
output power \(\mathrm{P}_{\mathrm{s}}=\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{s}}\)
\(
\begin{array}{ll}
\therefore & \mathrm{I}_{\mathrm{s}}=\frac{\mathrm{P}_{\mathrm{s}}}{\mathrm{~V}_{\mathrm{s}}} \\
\therefore & \mathrm{I}_{\mathrm{s}}=\frac{12}{24} \\
\therefore & \mathrm{I}_{\mathrm{s}}=0.5 \mathrm{~A} \\
\text { peak current } & \mathrm{I}_{\mathrm{m}}=\sqrt{2} \mathrm{I}_{\mathrm{s}} \\
& \mathrm{I}_{\mathrm{m}}=\sqrt{2} \times 0.5 \\
& \mathrm{I}_{\mathrm{m}}=\frac{1}{\sqrt{2}} \mathrm{~A} \\
& \mathrm{I}_{\mathrm{m}}=0.71 \mathrm{~A}
\end{array}
\)
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