## Appendix C  
For the case of the circuit without a compensator, the consumed power was determined as follows:  
\[\mathrm{P_{1}}=\mathrm{U_{d}}\cdot\mathrm{I_{d}} \tag{10}\]  
where \(\mathrm{I_{d}}\) is the value of the consumed current and \(\mathrm{U_{d}}\) is the value of the inverter supply voltage. As mentioned above, \(\mathrm{U_{d}}=2645\) V.  
For the cases of circuits with compensators, the consumed power was determined based on the following considerations. The power part of the drive with a compensator consists of the power part without a compensator and the power part of the compensator itself, which is additional inverter, transformer, and RC-high frequency filters (Figure 1). To simplify the modeling on the comprehensive model (Figure 6), the inverter was replaced by three controlled voltage sources. It follows that using the determined value of the consumed current \(\mathrm{I_{d}}\) for cases 2-6, it is possible to calculate the power consumed by the power part ofthe drive \(\mathrm{P_{1}}^{\prime}\) based on Equation (A36) without considering the power consumed by the inverter of the compensator \(\mathrm{P_{Cl}}\), i.e.,:  
\[\mathrm{P_{1}^{\prime}}=\mathrm{P_{1}}=\mathrm{U_{d}}\cdot\mathrm{I_{d}}\] (A37)  
Power losses in TV transformer were determined based on its energy efficiency. The full rated power of the transformer should be slightly greater than the full rated power of the traction motor. The full rated power of the traction motor is as follows:  
\[\mathrm{S_{nom}}=\frac{\mathrm{P_{nom}}\cdot 100}{\cos\varphi\cdot\eta}=\frac{1,200,000\cdot 100}{0.88\cdot 95.5}=1,428,000,\mathrm{V}\cdot\mathrm{A}\] (A38)  
where \(\mathrm{P_{nom}}\) is the rated power of the traction motor (Table 1), \(\cos\varphi\) is the power factor of the traction motor (Table 1), and \(\eta\) is the efficiency factor of the traction motor (Table 1). A three-phase transformer with a full power of 1600 \(\mathrm{V}\cdot\mathrm{A}\) was selected from the standard series [43]. The losses in such a transformer should not exceed 1%. In further calculations of the power consumed by the compensator, the losses in the transformer will be considered as 1% of the power consumed by the traction motor, i.e.,:  
\[\mathrm{P_{VT}}=0.01\cdot\mathrm{P_{1}^{\prime}}\] (A39)  
To simplify the calculations, the transformation coefficient TV is taken equal to 1 (\(\mathrm{K_{TV}}=1\)). The power consumed by the inverter is equal to the power losses in the inverter. The power losses in the inverter can be divided into static and switching losses. Static losses are power losses caused by the flow of current through the inverter elements. Dynamic losses depend on the parasitic values of the capacitance and inductance of the p-n junctions and the rate of increase in the current and voltage in the inverter elements. In this paper, dynamic losses are neglected, since their magnitude is much smaller than static losses and switching losses [44]. To account for the static power consumed by the power part of the compensator, the following considerations were considered:  
* The power part of the compensator is powered by a direct current link, the voltage which is equal to \(\mathrm{U_{d}}\). * The power circuit of the compensator is assembled according to the bridge circuit made on IGBT modules of the same series as the autonomous voltage inverter (parameters of the IGBT transistor and the reverse diode are given in Table 2). The current equal to the phase current of the induction motor will flow through the open transistor. Since the drive has symmetrical windings and symmetrical inverter arms, then \(\mathrm{I_{sa}}=\mathrm{I_{sb}}=\mathrm{I_{sc}}=\mathrm{I_{s}}\). When implementing the algorithm, a mode with a pulse duty cycle was selected (50% of the time in the pulse period is occupied by the pulse duration, and 50% by the pause).