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The ratio of secondary and primary turns of step-up transformer is $4: 1 .$ If a current of 4 A is applied to the primary, the induced current in secondary will be
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$1 \mathrm{~A}$
$\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}=\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{1}{4} \Rightarrow \mathrm{I}_{\mathrm{S}}=\frac{1}{4} \times 4=1 \mathrm{~A}$
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