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(C) For an ideal transformer, the input power equals the output power, meaning $P_{in} = P_{out}$.Since $P = V 	imes I$, we have $V_{p}I_{p} = V_{s}I

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(C) For an ideal transformer, the input power equals the output power, meaning $P_{in} = P_{out}$.Since $P = V 	imes I$, we have $V_{p}I_{p} = V_{s}I_{s}$, which implies $\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$.In a step-down transformer, the secondary voltage $V_{s}$ is less than the primary voltage $V_{p}$ $(V_{s} According to the relation $\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$, if $V_{s}  I_{p}$.Therefore, the current increases in the secondary coil.

In an ideal step-down transformer, out of the following quantities, which quantity increases in the secondary coil?

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