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(A) Given: Number of turns in primary $N_{p} = 300$,Number of turns in secondary $N_{s} = 150$,Output power $P_{s} = 2.2\, kW = 2200\, W$,Secondary cu

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$440\, V$ and $5\, A$

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(A) Given: Number of turns in primary $N_{p} = 300$,Number of turns in secondary $N_{s} = 150$,Output power $P_{s} = 2.2\, kW = 2200\, W$,Secondary current $I_{s} = 10\, A$.First,calculate the secondary voltage $V_{s}$ using $P_{s} = V_{s} I_{s}$:$2200 = V_{s} 	imes 10 \Rightarrow V_{s} = 220\, V$.Using the transformer ratio $\frac{V_{p}}{V_{s}} = \frac{N_{p}}{N_{s}}$:$\frac{V_{p}}{220} = \frac{300}{150} = 2 \Rightarrow V_{p} = 440\, V$.Assuming an ideal transformer,input power $P_{p} = P_{s} = 2200\, W$:$P_{p} = V_{p} I_{p} \Rightarrow 2200 = 440 	imes I_{p} \Rightarrow I_{p} = \frac{2200}{440} = 5\, A$.Thus,the input voltage is $440\, V$ and the input current is $5\, A$.

$A$ transformer consisting of $300$ turns in the primary and $150$ turns in the secondary gives output power of $2.2\, kW$. If the current in the secondary coil is $10\, A$,then the input voltage and current in the primary coil are

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