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(A) For a $100\%$ efficient transformer,the input power is equal to the output power.$P_{in} = P_{out}$$V_p I_p = V_s I_s$From the transformer ratio,w

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$1$

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(A) For a $100\%$ efficient transformer,the input power is equal to the output power.$P_{in} = P_{out}$$V_p I_p = V_s I_s$From the transformer ratio,we know that $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$.Given:Number of turns in primary,$N_p = 100$Number of turns in secondary,$N_s = 25$Current in secondary,$I_s = 4\, A$Using the relation $\frac{I_p}{I_s} = \frac{N_s}{N_p}$:$\frac{I_p}{4} = \frac{25}{100}$$\frac{I_p}{4} = \frac{1}{4}$$I_p = 1\, A$Therefore,the current in the primary coil is $1\, A$.

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

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