In a step-up transformer,if the ratio of turn | vedclass

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(A) For an ideal transformer,the power input equals the power output,which implies $V_P I_P = V_S I_S$.Also,the transformer ratio is given by $\frac{N

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$20$

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(A) For an ideal transformer,the power input equals the power output,which implies $V_P I_P = V_S I_S$.Also,the transformer ratio is given by $\frac{N_P}{N_S} = \frac{V_P}{V_S} = \frac{I_S}{I_P}$.Given the ratio of turns $\frac{N_P}{N_S} = \frac{1}{10}$ and the load current (secondary current) $I_S = 2 \, A$.Using the relation $\frac{N_P}{N_S} = \frac{I_S}{I_P}$,we get $I_P = \frac{N_S}{N_P} 	imes I_S$.Substituting the values: $I_P = \frac{10}{1} 	imes 2 \, A = 20 \, A$.Therefore,the current in the primary coil is $20 \, A$.

In a step-up transformer,if the ratio of turns of primary to secondary is $1:10$ and the primary voltage is $230 \, V$. If the load current is $2 \, A$,then the current in the primary is.......$A$.

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