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

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(B) For an ideal transformer,the power input equals the power output,which implies $V_{1}I_{1} = V_{2}I_{2}$.Since the voltage ratio is proportional t

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$2$

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(B) For an ideal transformer,the power input equals the power output,which implies $V_{1}I_{1} = V_{2}I_{2}$.Since the voltage ratio is proportional to the turns ratio,$\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$.Combining these,we get the relationship $\frac{N_{1}}{N_{2}} = \frac{I_{2}}{I_{1}}$.Given $N_{1} = 140$,$N_{2} = 280$,and $I_{1} = 4\,A$.Substituting the values: $\frac{140}{280} = \frac{I_{2}}{4}$.$\frac{1}{2} = \frac{I_{2}}{4}$.$I_{2} = \frac{4}{2} = 2\,A$.

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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