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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
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450 V , 15 A
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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