KCET · Physics · Alternating Current
A step-up transformer operates on a \( 230 \mathrm{~V} \) line and a load current of \( 2 \mathrm{~A} \). The ratio of primary and secondary windings is \( 1: 25 \). Then the current in the primary is
- A \( 25 \) A
- B \( 50 \mathrm{~A} \)
- C \( 15 \mathrm{~A} \)
- D \( 12.5 \mathrm{~A} \)
Answer & Solution
Correct Answer
(B) \( 50 \mathrm{~A} \)
Step-by-step Solution
Detailed explanation
Transformer operates on \( 230 \mathrm{~V} \) line; load current \( =2 \mathrm{~A} \)
Ratio of primary and secondary windings, \( \frac{N_{P}}{N_{S}}=\frac{1}{25} \)
Since, \( \frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}} \)
\( \Rightarrow I_{p}=\frac{N_{s}}{N_{p}} \times I_{s} \) \( \Rightarrow I_{p}=\frac{25}{1} \times 2=50 \)
Therefore, current in primary \( =50 \mathrm{~A} \)
Ratio of primary and secondary windings, \( \frac{N_{P}}{N_{S}}=\frac{1}{25} \)
Since, \( \frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}} \)
\( \Rightarrow I_{p}=\frac{N_{s}}{N_{p}} \times I_{s} \) \( \Rightarrow I_{p}=\frac{25}{1} \times 2=50 \)
Therefore, current in primary \( =50 \mathrm{~A} \)
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