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(A) For a $100\%$ efficient transformer, the input power equals the output power.$P_{in} = P_{out}$$V_{p} I_{p} = V_{s} I_{s}$We know that for an idea

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(A) For a $100\%$ efficient transformer, the input power equals the output power.$P_{in} = P_{out}$$V_{p} I_{p} = V_{s} I_{s}$We know that for an ideal transformer, the ratio of currents is inversely proportional to the ratio of turns:$\frac{I_{p}}{I_{s}} = \frac{N_{s}}{N_{p}}$Given:$N_{p} = 100$$N_{s} = 25$$I_{s} = 4 	ext{ A}$Substituting the values:$\frac{I_{p}}{4} = \frac{25}{100}$$\frac{I_{p}}{4} = \frac{1}{4}$$I_{p} = 1 	ext{ A}$Therefore, the current in the primary coil is $1 	ext{ A}$.

$A$ $100\%$ efficient transformer has $100$ turns in the primary and $25$ turns in its secondary coil. If the current in the secondary coil is $4 \text{ A}$, then the current in the primary coil is ... $ \text{A}$

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