A small town with a demand of 800 kW of electric power

Solution:

Total electric power required, P = 800 kW = 800 × 103 W

Supply voltage, V = 220 V

Voltage at which electric plant is generating power, V' = 440 V

Distance between the town and power generating station, d = 15 km

Resistance of the two wire lines carrying power = 0.5 Ω/km

Total resistance of the wires, R = (15 + 15)0.5 = 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 is given as:

$I=\frac{P}{V_{1}}$

$=\frac{800 \times 10^{3}}{4000}=200 \mathrm{~A}$b

(a) Line power loss = I2R

= (200)2 × 15

= 600 × 103 W

= 600 kW

(b) Assuming that the power loss is negligible due to the leakage of the current:

Total power supplied by the plant = 800 kW + 600 kW

= 1400 kW

(c) Voltage drop in the power line = IR = 200 × 15 = 3000 V

Hence, total voltage transmitted from the plant = 3000 + 4000

= 7000 V

Also, the power generated is 440 V.

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