$A$ small town with a demand of $800 \; kW$ of electric power at $220 \; V$ is situated $15 \; km$ away from an electric plant generating power at $440 \; V$. The resistance of the two-wire line carrying power is $0.5 \; \Omega/km$. The town gets power from the line through a $4000-220 \; V$ step-down transformer at a sub-station in the town.
$(a)$ Estimate the line power loss in the form of heat.
$(b)$ How much power must the plant supply,assuming there is negligible power loss due to leakage?
$(c)$ Characterise the step-up transformer at the plant.

(N/A) Total electric power required,$P = 800 \; kW = 800 \times 10^3 \; W$.
Supply voltage at the town,$V = 220 \; V$.
Voltage at which the electric plant generates power,$V' = 440 \; V$.
Distance between the town and the power generating station,$d = 15 \; km$.
Resistance of the two-wire line $= 0.5 \; \Omega/km$.
Total resistance of the wires,$R = (15 + 15) \times 0.5 = 15 \; \Omega$.
$A$ step-down transformer of rating $4000-220 \; V$ is used in the sub-station.
Input voltage to the transformer,$V_1 = 4000 \; V$.
The current in the transmission line is $I = \frac{P}{V_1} = \frac{800 \times 10^3}{4000} = 200 \; A$.
$(a)$ Line power loss $= I^2 R = (200)^2 \times 15 = 40000 \times 15 = 600,000 \; W = 600 \; kW$.
$(b)$ Total power supplied by the plant $= \text{Power required} + \text{Power loss} = 800 \; kW + 600 \; kW = 1400 \; kW$.
$(c)$ Voltage drop in the power line $= I \times R = 200 \times 15 = 3000 \; V$.
Total voltage transmitted from the plant $= \text{Voltage at transformer input} + \text{Voltage drop} = 4000 \; V + 3000 \; V = 7000 \; V$.
Since the plant generates at $440 \; V$,the step-up transformer rating is $440 \; V - 7000 \; V$.