A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

Given, primary voltage $$V_p = 8\,\text{kV} = 8000\,\text{V}$$, secondary voltage $$V_s = 160\,\text{V}$$ and load power $$P = 80\,\text{kW} = 80000\,\text{W}$$.

Since the load is purely resistive, the power factor is 1.

For an ideal transformer, input power is equal to output power.

The secondary current is

$$I_s = \frac{P}{V_s} = \frac{80000}{160} = 500\,\text{A}$$

Therefore, the load resistance in the secondary circuit is

$$R_s = \frac{V_s}{I_s} = \frac{160}{500} = 0.32\,\Omega$$

The primary current is

$$I_p = \frac{P}{V_p} = \frac{80000}{8000} = 10\,\text{A}$$

Therefore, the equivalent load resistance as seen from the primary side is

$$R_p = \frac{V_p}{I_p} = \frac{8000}{10} = 800\,\Omega$$

We can also verify the result using the resistance transformation relation.

The turns ratio is

$$\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{8000}{160} = 50$$

Hence,

$$R_p = R_s\left(\frac{N_p}{N_s}\right)^2$$

$$R_p = 0.32 \times 50^2 = 0.32 \times 2500 = 800\,\Omega$$

Thus, the load resistance in the primary circuit is $$800\,\Omega$$ and in the secondary circuit is $$0.32\,\Omega$$.

Answer: Option C