In a transformer,the number of turns in the p | vedclass

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(B) For an ideal transformer,the relationship between the number of turns $(N)$ and the current $(I)$ is given by the inverse ratio: $\frac{N_1}{N_2}

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$2$

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(B) For an ideal transformer,the relationship between the number of turns $(N)$ and the current $(I)$ is given by the inverse ratio: $\frac{N_1}{N_2} = \frac{I_2}{I_1}$.Given:Number of turns in primary coil,$N_1 = 140$.Number of turns in secondary coil,$N_2 = 280$.Current in primary coil,$I_1 = 4 \, A$.Substituting these values into the formula:$\frac{140}{280} = \frac{I_2}{4}$.$\frac{1}{2} = \frac{I_2}{4}$.$I_2 = \frac{4}{2} = 2 \, A$.Therefore,the current in the secondary coil is $2 \, A$.

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 $4 \, A$,then the current in the secondary coil is.....$A$.

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