We consider ROES with a closed circuit, therefore, we must also take into account the power transmitted by the conduction current through the conductor section. It will consist of the power of the Joule losses in the conductor and the power circulating in the closed circuit. Here, we neglect the losses due to ionization and attenuation in the line because of their smallness.

Since we believe that the second wire has a relatively low resistance, since part of the current flows through the ground, then here we will take into account only Joule losses on the active resistance of the central core of the power line cable. If we enter the active resistance of this conductor, as \(R_l \), then the power of active losses can be found as follows: \[P_J = I ^ 2 R_l \qquad (1.21) \] Recall that the current, in the general case, is described as follows: \(I = I_0 \mathrm {e} ^ {i \omega t} \). Substituting here the complex values ??of the current and finding the real part of the power, we finally get the power of the Joule losses: \[P_J = I_0 ^ 2 R_l \cos (2 \omega t) \qquad (1.22) \] The power circulating in a closed circuit is also found in the classical way: \[P _ {\rho} = U_0 I_0 \mathrm {e} ^ {i \omega t} \qquad (1.23) \] Selecting only the real part, we get: \[P _ {\rho} = U_0 I_0 \cos (2 \omega t) \qquad (1.24) \] In the case of an open circuit, for example, when the Avramenko plug is located at the receiving end, these powers will be zero.

Let's draw up a general equation in the form of a power balance. To do this, take the obtained formulas (1.13, 1.20, 1.22) and substitute in (1.6): \[P_S - \omega C U_0 ^ 2 \sin (2 \omega t) - \omega L I_0 ^ 2 \sin (2 \omega t) = 0 \qquad (1.25) \] From here we can get the power that is transmitted through a given power line by an electromagnetic wave: \[P_S = \omega C U_0 ^ 2 \sin (2 \omega t) + \omega L I_0 ^ 2 \sin (2 \omega t) \qquad (1.26) \] In the most general case, to find the total power transmitted through the transmission line, it is necessary to add the power \(P_J \) from (1.22) \(P _ {\rho} \) from (1.24). Then we finally get: \[P _ {\Sigma} = \left [\omega C U_0 ^ 2 + \omega L I_0 ^ 2 \right] \sin (2 \omega t) + \left [U_0 I_0 - I_0 ^ 2 R_l \right] \cos (2 \omega t) \qquad (1.27) \] From the obtained expression, it can be seen that if the frequency is equal to zero, then the formula will go into the classical one, for calculating the direct current. Moving on to the effective (rms) values ??of current and voltage, we get: \[P _ {\Sigma} = \sqrt {\left [\omega (CU ^ 2 + LI ^ 2) \right] ^ 2 + \left [UI - I ^ 2 R_l \right] ^ 2} \qquad (1.28) \] Let's, for clarity, write the same formula in a different form: \[P _ {\Sigma} = \sqrt {P_D ^ 2 + P ^ 2} \\P_D = \omega (CU ^ 2 + LI ^ 2), \quad P = UI - I ^ 2 R_l \qquad (1.29) \] If we compare this formula with the very first (1.1), we can see an obvious analogy: the total current, like the total power, is formed due to two components. It would be better to formulate the result obtained as follows: the total energy transmitted through the power transmission line consists of the energy carried by the displacement current and the energy carried by the conduction current. But in practice it is more convenient to deal with the cardinalities, which were obtained in (1.29). Therefore, in what follows we will call the corresponding powers as follows: displacement power - \(P_D \), and conduction power - \(P \) (our classical power). In this case, the power of Joule losses in the transmission line is \(P_J = I ^ 2 R_l \).

Interestingly, unlike the conduction current, an electric circuit with a displacement current cannot be calculated using Ohm's and Kirchhoff's laws and requires a completely new mathematical apparatus.

Another conclusion from the obtained result can be the fact that in the case of alternating current and a closed circuit, two currents always flow through the power line : conduction current - along the cross-section of the conductor and displacement current - along the conductor, but not in it. Many contemporaries-researchers spoke about the second property of the current, for example, D. Smith. With an open circuit, only one current flows through the conductor - the displacement current.

If we assume that the resulting formula works not only for coaxial transmission lines, but also for its other types - and this will be shown further in the examples - it will be interesting to evaluate the operation of classical power grids on its basis.

Let us estimate what percentage of the total transmission power is occupied by the offset power for 110 kV transmission lines, per one kilometer of the line. If we take the capacitance between the wires of such a transmission line - 4 nF / km [9], then at a frequency of 50 Hz the displacement power will be 15 kW, which, with a total power of such a line of 30 MW, will be 0.05%. With the maximum length of such a transmission line between substations - 80 km, the share of the bias power will be only 4%. In the 220 V power grid, this figure will be orders of magnitude less. Therefore, the bias power for classical power grids is not taken into account or transferred to the reactive category. If we increase the frequency of energy transmission, for example, to 20 kHz, then for a 110 kV transmission line, the fraction of the bias power will increase to 20% and can, in principle, completely replace the power transmitted by conduction currents. At the same time, the cross-section of the wires of such a line can become several times smaller, and in the material of the conductor, you can completely get away from expensive components.

From formula (1.25) it is obvious that the efficiency of the transmission line will be estimated based on the total power without Joule losses and power, taking into account them: \[\eta_l = {\sqrt {P_D ^ 2 + P ^ 2 \over P_D ^ 2 + (P + P_J) ^ 2}} \qquad (2.1) \] The overall efficiency of the entire installation will consist of the efficiency of the transmitting and receiving parts \(\eta_ {tx}, \eta_ {rx} \), and the efficiency of the line: \[\eta = \eta_ {tx} \eta_l \eta_ {rx} \qquad (2.2) \] Further, we will receive the transfer characteristic, and hence the resonant frequency, and proceed to calculate specific ROES.