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 that the ratio of the number of turns is inversely proportional to

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$2$

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(B) For an ideal transformer,the power input equals the power output,which implies that the ratio of the number of turns is inversely proportional to the ratio of the currents.The relationship is given by: $\frac{N_p}{N_s} = \frac{i_s}{i_p}$Given:$N_p = 140$$N_s = 280$$i_p = 4 \ A$Substituting the values into the formula:$\frac{140}{280} = \frac{i_s}{4}$$\frac{1}{2} = \frac{i_s}{4}$$i_s = \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 (in $A$):

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In a transformer,the number of turns in the primary coil is $140$ and 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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