MHT CET · Physics · Alternating Current
A transformer having efficiency \(90 \%\) is working on 200 V and 3 kw power supply. If the current in the secondary coil is 6 A , the voltage across the secondary coil and the current in the primary coil are respectively
- A \(300 \mathrm{~V}, 15 \mathrm{~A}\)
- B \(450 \mathrm{~V}, 15 \mathrm{~A}\)
- C \(450 \mathrm{~V}, 13 \cdot 5 \mathrm{~A}\)
- D \(600 \mathrm{~V}, 15 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(450 \mathrm{~V}, 15 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\text { Power output }=3 \times \frac{90}{100}=2.7 \mathrm{~kW}\)
\(I_s=6 \mathrm{~A}\)
\(\therefore V_S=\frac{2.7 \times 10^3}{6}=450 \mathrm{~V} \text { and } I_p=\) \(\frac{3 \times 10^3}{200}=15 \mathrm{~A}\)
\(I_s=6 \mathrm{~A}\)
\(\therefore V_S=\frac{2.7 \times 10^3}{6}=450 \mathrm{~V} \text { and } I_p=\) \(\frac{3 \times 10^3}{200}=15 \mathrm{~A}\)
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