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(D) In an ideal transformer, the input power is equal to the output power, so $V_p I_p = V_s I_s$.
For a transformer, the relationship between voltag

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(D) In an ideal transformer, the input power is equal to the output power, so $V_p I_p = V_s I_s$.
For a transformer, the relationship between voltage, current, and number of turns is given by $\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$.
Given values are $N_p = 100$, $N_s = 200$, and $I_s = 5 	ext{ A}$.
Using the relation $\frac{I_p}{I_s} = \frac{N_s}{N_p}$, we get $I_p = I_s 	imes \frac{N_s}{N_p}$.
Substituting the values: $I_p = 5 	imes \frac{200}{100} = 5 	imes 2 = 10 	ext{ A}$.
Therefore, the input current is $10 	ext{ A}$.

In an ideal step-up transformer, the number of turns in the primary coil and secondary coil are $100$ and $200$ respectively. If the output current is found to be $5 \text{ A}$, then the input current will be . . . . . . . (in $\text{ A}$)

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