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 wires, 𝑅 = 2 × 15 km × 0.5 Ω per km = 15 Ω

A step-down transformer of rating 40000 − 220 V is used in the sub-station.

Input Voltage, 𝑉1 = 40000 V

Output Voltage, 𝑉2 = 220 V

rms current in the wire lines can be given by,

$I=\frac{P}{V_1}=\frac{800\times {10}^3}{40000}=20A$

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

$P^{\mathrm{\prime }}=I^{2}R={20}^2\times 15=6kW$

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

$P_T=P+P^{\mathrm{\prime }}=800kW+6kW=806kW$

(c) Drop across the power line = IR = 20 × 15 = 300 V

Total voltage transmitted by the plant = 300 + 40000 = 40300 V

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

Hence, the rating of the step-up transformer at the power plant is 440 V −
40300 V.

Percentage loss during transmission = $\frac{6}{806}\times 100=0.74\mathrm{\%}$

Percentage loss in the previous exercise = $\frac{600}{1400}\times 100=43\mathrm{\%}$

The percentage of power loss greatly reduced by high voltage
transmission.