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GUJCET · Physics · Alternating Current
A transformer with \(80 \%\) efficiency works at 4 kW and 100 V. If the secondary voltage is 240 V. Then the primary current is ____________ .
- A 0.4 A
- B 40 A
- C 10 A
- D 4 A
Answer & Solution
Correct Answer
(B) 40 A
Step-by-step Solution
Detailed explanation
(B) 40 A
Here phase difference is \(\phi=\frac{\pi}{2}\)
\(\Rightarrow\) power \(\mathrm{P}_{1}=\mathrm{V}_{1} \mathrm{I}_{1}\)
\(
\therefore \quad \begin{aligned}
\mathrm{I}_{1} & =\frac{\mathrm{P}_{1}}{\mathrm{~V}_{1}} \\
\mathrm{I}_{1} & =\frac{4 \times 10^{3}}{100} \\
\mathrm{I}_{1} & =40 \mathrm{~A}
\end{aligned}
\)
Here phase difference is \(\phi=\frac{\pi}{2}\)
\(\Rightarrow\) power \(\mathrm{P}_{1}=\mathrm{V}_{1} \mathrm{I}_{1}\)
\(
\therefore \quad \begin{aligned}
\mathrm{I}_{1} & =\frac{\mathrm{P}_{1}}{\mathrm{~V}_{1}} \\
\mathrm{I}_{1} & =\frac{4 \times 10^{3}}{100} \\
\mathrm{I}_{1} & =40 \mathrm{~A}
\end{aligned}
\)
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