Do the same exercise as above with the replacement of the earlier transformer

Solution:

The rating of a step-down transformer is 40000 V−220 V.

Input voltage, V1 = 40000 V

Output voltage, V2 = 220 V

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

Source potential, V = 220 V

Voltage at which the electric plant generates 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 wire lines, R = (15 + 15)0.5 = 15 Ω

P = V1I

Rms current in the wire line is given as:

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

$=\frac{800 \times 10^{3}}{40000}=20 \mathrm{~A}$

(a) Line power loss = I2R

= (20)2 × 15

= 6 kW

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

Hence, power supplied by the plant = 800 kW + 6kW = 806 kW

(c) Voltage drop in the power line = IR = 20 × 15 = 300 V

Hence, voltage that is transmitted by the power plant

= 300 + 40000 = 40300 V

The power is being generated in the plant at 440 V.

Hence, the rating of the step-up transformer needed at the plant is

440 V − 40300 V.

Hence, power loss during transmission $=\frac{600}{1400} \times 100=42.8 \%$

In the previous exercise, the power loss due to the same reason is $\frac{6}{806} \times 100=0.744 \%$. Since the power loss is less for a high voltage transmission, high voltage transmissions are preferred for this purpose.