Q. In a transformer, the number of turns in the primary coil is $140$ and that in the secondary coil is $280$ . If the current in the primary coil is $4 \text{ A}$ , then that in the secondary coil is

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A

$4 \text{ A}$

B

$2 \text{ A}$

C

$6 \text{ A}$

D

$10 \text{ A}$

Solution:

$\textit{I}_{2} \textit{N}_{2} = \textit{I}_{1} \textit{N}_{1}$
$∴$ $\textit{I}_{2} = \frac{\textit{I}_{1} \textit{N}_{1}}{\textit{N}_{2}} = \frac{4 \times \text{140}}{\text{280}} = 2 \text{ A.}$

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