MHT CET · Physics · Alternating Current
A transformer having efficiency of \(90 \%\) is working on \(200 \mathrm{~V}\) and \(3 \mathrm{~kW}\) power supply. If the current is the secondary coil is \(6 \mathrm{~A}\), the voltage across the secondary coil and the current in the primary coil respectively are
- A 450 V , 12 A
- B 600 V , 15 A
- C 300 V , 15 V
- D 450 V , 15 A
Answer & Solution
Correct Answer
(D) 450 V , 15 A
Step-by-step Solution
Detailed explanation
Primary current, \(I_{\mathrm{P}}=\frac{P}{V}=\frac{3000}{200}=15 \mathrm{~A}\)
Output power \(=90 \%\) of \(3000 \mathrm{~W}=2700 \mathrm{~W}\)
Therefore, voltage across the secondary coil, \(V_{\mathrm{s}}=\frac{2700}{6}=450 \mathrm{~V}\)
Output power \(=90 \%\) of \(3000 \mathrm{~W}=2700 \mathrm{~W}\)
Therefore, voltage across the secondary coil, \(V_{\mathrm{s}}=\frac{2700}{6}=450 \mathrm{~V}\)
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