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(C) For a step-down transformer, the number of turns in the primary coil $N_P$ is greater than the number of turns in the secondary coil $N_S$. Given:

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$40 \,A, 220 \,V$

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(C) For a step-down transformer, the number of turns in the primary coil $N_P$ is greater than the number of turns in the secondary coil $N_S$. Given: $N_P = 5000$, $N_S = 500$, $I_P = 4 \,A$, and $V_P = 2200 \,V$.Using the transformer ratio formula: $\frac{N_S}{N_P} = \frac{V_S}{V_P} = \frac{I_P}{I_S}$.First, calculate the secondary voltage $V_S$:$V_S = V_P 	imes \frac{N_S}{N_P} = 2200 	imes \frac{500}{5000} = 220 \,V$.Next, calculate the secondary current $I_S$ using the relation $\frac{V_S}{V_P} = \frac{I_P}{I_S}$:$I_S = I_P 	imes \frac{V_P}{V_S} = 4 	imes \frac{2200}{220} = 40 \,A$.Thus, the current is $40 \,A$ and the potential difference is $220 \,V$.

The coils of a step-down transformer have $500$ and $5000$ turns. In the primary coil, an $AC$ current of $4 \,A$ at $2200 \,V$ is sent. The value of the current and potential difference in the secondary coil is:

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