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

We have an ideal transformer with primary voltage $$V_p = 8$$ kV $$= 8000$$ V, secondary voltage $$V_s = 160$$ V, and a load power of $$P = 80$$ kW $$= 80000$$ W. The load is purely resistive and operates at unity power factor.

For an ideal transformer, the input power equals the output power. The current in the secondary coil is $$I_s = \frac{P}{V_s} = \frac{80000}{160} = 500$$ A. The load resistance in the secondary circuit is therefore $$R_s = \frac{V_s}{I_s} = \frac{160}{500} = 0.32 \ \Omega$$. We can verify this using $$R_s = \frac{V_s^2}{P} = \frac{(160)^2}{80000} = \frac{25600}{80000} = 0.32 \ \Omega$$.

Now, the current in the primary coil is $$I_p = \frac{P}{V_p} = \frac{80000}{8000} = 10$$ A. The equivalent load resistance as seen from the primary side is $$R_p = \frac{V_p}{I_p} = \frac{8000}{10} = 800 \ \Omega$$. Alternatively, using $$R_p = \frac{V_p^2}{P} = \frac{(8000)^2}{80000} = \frac{64 \times 10^6}{80000} = 800 \ \Omega$$.

We can also verify using the impedance transformation relation: $$R_p = R_s \times \left(\frac{N_p}{N_s}\right)^2$$. The turns ratio is $$\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{8000}{160} = 50$$. So $$R_p = 0.32 \times 50^2 = 0.32 \times 2500 = 800 \ \Omega$$, which is consistent.

So the loads in the primary and secondary circuits are 800 $$\Omega$$ and 0.32 $$\Omega$$ respectively.

Hence, the correct answer is Option C.