1. Given data

• Primary voltage: $V_p = 8\,\text{kV} = 8000\,\text{V}$
• Secondary voltage: $V_s = 160\,\text{V}$
• Load power: $P = 80\,\text{kW} = 80000\,\text{W}$
• Transformer is ideal and load is purely resistive, so power factor $=1$

For an ideal transformer, $P_p = P_s = 80000\,\text{W}$

2. Find primary-side load resistance

Since the load is resistive, $P = \frac{V^2}{R}$ So, $R_p = \frac{V_p^2}{P} = \frac{(8000)^2}{80000}$ $R_p = \frac{64\times 10^6}{8\times 10^4} = 800\,\Omega$

3. Find secondary-side load resistance

Similarly, $R_s = \frac{V_s^2}{P} = \frac{(160)^2}{80000}$ $R_s = \frac{25600}{80000} = 0.32\,\Omega$

4. Match with options

Primary and secondary loads are: $800\,\Omega \quad \text{and} \quad 0.32\,\Omega$

This matches Option C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So, the derived answer agrees with the stored correct answer.