1. Given data

• Primary voltage: $V_p = 2.3\,\text{kV} = 2300\,\text{V}$
• Secondary voltage: $V_s = 230\,\text{V}$
• Primary current: $I_p = 5\,\text{A}$
• Efficiency of transformer: $\eta = 90\% = 0.9$

We need the output current $I_s$.

2. Input power to transformer

The electrical input power is

$P_{\text{in}} = V_p I_p = 2300 \times 5 = 11500\,\text{W}$

3. Output power using efficiency

Efficiency is defined as

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

So,

$P_{\text{out}} = \eta P_{\text{in}} = 0.9 \times 11500 = 10350\,\text{W}$

4. Find secondary current

Output power is also

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

Thus,

$I_s = \frac{P_{\text{out}}}{V_s} = \frac{10350}{230} = 45\,\text{A}$

5. Final answer

$\boxed{45}$

The given number of primary turns and the copper winding information are not needed for this calculation.