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

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Question

(B) Given: Current in primary coil $i_{p} = 4 \, A$,number of turns in primary coil $N_{p} = 140$,and number of turns in secondary coil $N_{s} = 280$.

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given: Current in primary coil $i_{p} = 4 \, A$,number of turns in primary coil $N_{p} = 140$,and number of turns in secondary coil $N_{s} = 280$.For an ideal transformer,the relationship between current and the number of turns is given by the formula:$\frac{i_{p}}{i_{s}} = \frac{N_{s}}{N_{p}}$Substituting the given values into the formula:$\frac{4}{i_{s}} = \frac{280}{140}$$\frac{4}{i_{s}} = 2$$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 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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