Given,

The power required, 𝑃 = 800 × 103 W

Voltage to be supplied, 𝑉 = 220 V

The voltage at power generating plant, 𝑉′ = 440 V

The total resistance of two wireline, 𝑅 = 2 × 15 km × 0.5 Ω per km = 15 Ω

A step-down transformer of rating 4000 − 220 V is used in the sub-station

Input Voltage, 𝑉1 = 4000 V

Output Voltage, 𝑉2 = 220 V

rms current in the wire lines can be given by,

𝐼$I=\frac{P}{V_1}=\frac{800\times {10}^3}{4000}=200A$

(a) Line power loss in the form of heat,

$P^{\mathrm{\prime }}=I^{2}R={200}^2\times 15=600kW$

(b) As there is no power loss due to leakage current. Total power,

$P_T=P+P^{\mathrm{\prime }}=800kW+600kW=1400kW$

(c) Drop across the power line = 𝐼𝐼𝐼𝐼 = 200 × 15 = 3000 Ω

Total voltage transmitted by the plant = 3000 + 4000 = 7000 V

The voltage at power generating plant, 𝑉𝑉′ = 440 V

Hence, the rating of the step-up transformer at the power plant is 440 V −
7000 V.