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The primary and secondary voltage of an ideal step-down transformer is $200 \mathrm{~V}$ and $25 \mathrm{~V}$ respectively. The secondary is connected to a device, which draws a current of $2 \mathrm{~A}$. The current in the primary is
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$250 \mathrm{~mA}$
For a lossless transformer, with primary coil $\left(I_P, V_P\right)$ and secondary coil $\left(I_S, V_S\right)$, there has to be conservation of energy per unit time:
$\begin{aligned} & I_P V_P=I_S V_S \\ & \Rightarrow \frac{I_P}{I_S}=\frac{V_S}{V_P} \\ & \therefore \frac{I_P}{2}=\frac{25}{500} \\ & \therefore I_P=\frac{1}{4} \mathrm{~A}=250 \mathrm{~mA}\end{aligned}$
$\begin{aligned} & I_P V_P=I_S V_S \\ & \Rightarrow \frac{I_P}{I_S}=\frac{V_S}{V_P} \\ & \therefore \frac{I_P}{2}=\frac{25}{500} \\ & \therefore I_P=\frac{1}{4} \mathrm{~A}=250 \mathrm{~mA}\end{aligned}$
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