The power transmission line transmits power of 500 kW at a voltage of 20 kV.

The power transmission line transmits power of 500 kW at a voltage of 20 kV. What is the efficiency of the line if its resistance is 5 ohms?

The efficiency η shows what part of the consumed (expended) power Ns is the useful power Ns, that is: η = Ns: Ns. The consumed power Nc = U² / R, we get η = (R ∙ Nп): U². From the condition of the problem, it is known that power Np = 500 kW = 500000 W is transmitted through the power transmission line of the transmission line under a voltage of U = 20 kV = 20,000 V with an electrical resistance of the line R = 5 Ohm. Substitute the values of physical quantities into the calculation formula and find the efficiency of the line:

η = (5 ∙ 500000): (20000) ²;

η = 0.00625 = 0.625%.

Answer: The efficiency of this line is 0.625%.