KCET · Physics · Mathematics in Physics
A step down transformer has \( 50 \) turns on secondary and \( 1000 \) turns on primary winding. If a
transformer is connected to \( 220 \mathrm{~V} 1 \mathrm{AA} \).C. source, what is output current of the transformer?
- A \( \frac{1}{20} A \)
- B \( 20 \mathrm{~A} \)
- C \( 100 \mathrm{~A} \)
- D \( 2 \mathrm{~A} \)
Answer & Solution
Correct Answer
(B) \( 20 \mathrm{~A} \)
Step-by-step Solution
Detailed explanation
Given, number of turns on secondary, \(N_{s}=50 ;\) number of turns on primary, \(N_{P}=1000 ;\) source voltage \(=220 \mathrm{~V} ;\)
current \(=1 \mathrm{~A}\).
We know \(\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}=\frac{I_{p}}{I_{s}}\)
Therefore \(\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}} \Rightarrow I_{s}=\frac{N_{p}}{N_{s}} I_{p}\)
Substituting the values, we get
\(I_{s}=\frac{1000}{50} \times 1=20 \mathrm{~A}\)
Thus, output current of transformer \(=20 \mathrm{~A}\)
current \(=1 \mathrm{~A}\).
We know \(\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}=\frac{I_{p}}{I_{s}}\)
Therefore \(\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}} \Rightarrow I_{s}=\frac{N_{p}}{N_{s}} I_{p}\)
Substituting the values, we get
\(I_{s}=\frac{1000}{50} \times 1=20 \mathrm{~A}\)
Thus, output current of transformer \(=20 \mathrm{~A}\)
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