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A step-down transformer has $N_{\mathrm{S}}$ to $N_{\mathrm{p}}$ ratio of $20: 1$. If $8 \mathrm{~V}$ are developed across $0.4 \Omega$ secondary, the primary current will be
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The correct answer is:
$1 \mathrm{~A}$
For a step-down transformer:
$\frac{N_{\mathrm{P}}}{N_{\mathrm{S}}}=\frac{20}{1}=\frac{V_p}{V_{\mathrm{s}}}=\frac{I_s}{I_p}$
$\begin{aligned} & \therefore I_s=\frac{8 \mathrm{~V}}{0.4 \mathrm{~A}}=20 \mathrm{~A} \\ & \Rightarrow I_p=I_s \times \frac{1}{20}=1 \mathrm{~A}\end{aligned}$
$\frac{N_{\mathrm{P}}}{N_{\mathrm{S}}}=\frac{20}{1}=\frac{V_p}{V_{\mathrm{s}}}=\frac{I_s}{I_p}$
$\begin{aligned} & \therefore I_s=\frac{8 \mathrm{~V}}{0.4 \mathrm{~A}}=20 \mathrm{~A} \\ & \Rightarrow I_p=I_s \times \frac{1}{20}=1 \mathrm{~A}\end{aligned}$
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