The turn ratio of a transformer is given as 2 | vedclass

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(C) For an ideal transformer,the relationship between the currents and the number of turns in the primary and secondary coils is given by the formula:

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$2$

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(C) For an ideal transformer,the relationship between the currents and the number of turns in the primary and secondary coils is given by the formula:$\frac{I_P}{I_S} = \frac{N_S}{N_P}$Given that the turn ratio is $\frac{N_P}{N_S} = \frac{2}{3}$,we have $\frac{N_S}{N_P} = \frac{3}{2}$.The current in the primary coil is $I_P = 3 \, A$.Substituting these values into the formula:$\frac{3}{I_S} = \frac{3}{2}$$3 	imes 2 = 3 	imes I_S$$6 = 3 	imes I_S$$I_S = \frac{6}{3} = 2 \, A$.Therefore,the current through the load resistance is $2 \, A$.

The turn ratio of a transformer is given as $2 : 3$. If the current through the primary coil is $3 \, A$,calculate the current through the load resistance (secondary coil). (in $, A$)

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