Advantages of AC in long distance power transmission:
Electric power is produced in a t large scale at electric power stations with the help of AC generators. These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places.
Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission. But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (i2R) in the transmission lines which are hundreds of kilometer long.
This power loss can be tackled either by reducing current i or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases the cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.
Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent.
At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss.
At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.
Illustration:
An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.
Case I:
P = 2 MW; R == 40 Ω; V = 10 kV Power,
Power, P = VI
Power loss = Heat produced = I2 R = (200)2 × 40 = 1.6 × 106 W
% of power loss = \(\frac{1.6 \times 10^6}{2\times 10^6}\) x100% = 0.8 x 100% = 80%
Case II:
P = 2 MW; R == 40 Ω; V = 100 kV
Power loss = Heat produced = I2 R = (20)2 × 40 = 0.016 × 106 W
% of power loss = \(\frac{0.6 \times 10^6}{2 \times 10^6}\) x 100% = 0.008 x 100% = 0.8%
