A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns giving the output power at 230 V. If the current in the primary coil of the transformer is 5 A and its efficiency is 90%, the output current would be:

We have the primary (input) voltage $$V_p = 2300\ \text{V}$$ and the current in the primary coil $$I_p = 5\ \text{A}$$.

The electrical power delivered to the primary side of a transformer is given by the product of voltage and current, that is the formula

$$P_{\text{in}} = V_p \, I_p.$$

Substituting the given values,

$$P_{\text{in}} = 2300 \times 5 = 11500\ \text{W}.$$

The transformer is not ideal; its efficiency is stated to be $$\eta = 90\% = 0.90.$$ By definition of efficiency,

$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}.$$

Re-arranging to obtain the output power,

$$P_{\text{out}} = \eta \, P_{\text{in}}.$$

Substituting the known quantities,

$$P_{\text{out}} = 0.90 \times 11500 = 10350\ \text{W}.$$

The secondary (output) voltage is given as $$V_s = 230\ \text{V}.$$ The power leaving the secondary is also the product of this voltage and the secondary current $$I_s$$, so

$$P_{\text{out}} = V_s \, I_s.$$

Solving for the unknown current $$I_s$$, we write

$$I_s = \frac{P_{\text{out}}}{V_s}.$$

Substituting the calculated output power and the given secondary voltage,

$$I_s = \frac{10350}{230} = 45\ \text{A}.$$

Hence, the correct answer is Option A.