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A step-down transformer has turns ratio 20: 1 . If $8 \mathrm{~V}$ is applied across 0.4 ohm secondary coil, then the primary current will
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$\begin{aligned} & \mathrm{V}_{\mathrm{s}}=8 \mathrm{~V}, \mathrm{R}_{\mathrm{s}}=0.4 \Omega \\ & \mathrm{I}_{\mathrm{s}}=\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{s}}}=\frac{8}{0.4}=20 \mathrm{~A} \\ & \frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_{\mathrm{s}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}=\frac{1}{20} \\ & \therefore \mathrm{I}_{\mathrm{p}}=\frac{\mathrm{I}_{\mathrm{s}}}{20}=\frac{20}{20}=1 \mathrm{~A}\end{aligned}$
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