Compensation of Phase Asymmetry in a Three-Phase Circuit

R. A. Voronov

Professor, Doctor of Technical Sciences

In a number of cases, the individual phases of a three-phase circuit operate under unequal conditions, as a result of which asymmetry arises in the currents and voltages; in some cases this asymmetry may reach considerable magnitudes, exceeding those permitted by the rules of technical operation. To compensate this asymmetry, additional resistances may be inserted both in series in the individual phases and between these phases.

In the present article the general case of compensation of such phase asymmetry is considered for three-phase circuits without a neutral conductor, i.e., for those cases in which the main circuit may be regarded as a six-terminal network, and for it the properties of ternary expressions can be used in full measure¹.

An exact solution of the given problem in many cases proves impossible, and the determination of the elements of a single common additional six-terminal network serving for such complete compensation is very difficult. In this connection it is more convenient and simpler to carry out the calculation of compensation not at once for all elements of the circuit, but in parts, as indicated below.

The six-terminal network of the main circuit is characterized by the coefficients \(A\), \(B\), \(C\), and \(D\) of the matrix

\[ \begin{bmatrix} U_1\\ I_1 \end{bmatrix} = \begin{bmatrix} A; B\\ C; D \end{bmatrix} \cdot \begin{bmatrix} U_2\\ I_2 \end{bmatrix}, \]

which may be represented in the form of ternary expressions

\[ \begin{aligned} A &= A_0 + \alpha A_{\mathrm{I}} + \beta A_{\mathrm{II}};\\ B &= B_0 + \alpha B_{\mathrm{I}} + \beta B_{\mathrm{II}};\\ C &= C_0 + \alpha C_{\mathrm{I}} + \beta C_{\mathrm{II}};\\ D &= D_0 + \alpha D_{\mathrm{I}} + \beta D_{\mathrm{II}}. \end{aligned} \tag{1} \]

The general circuit after compensation has been carried out is likewise characterized by similar coefficients, and the task of compensation is to eliminate in the latter the terms with the factors \(\alpha\) and \(\beta\), i.e., the components that cause phase asymmetry. The smaller these coefficients are, the more complete and better

there will be compensation. If only zero-sequence components \(A_0, B_0, C_0\), and \(D_0\) are present, the compensation will be complete and the overall circuit will be equivalent to a circuit with symmetrical phases.

The connection of additional resistances (six-terminal networks) may be made either at the generator end of the circuit or at the load end; and if it is desired also to preserve the symmetry of the phase-equivalent circuit (i.e., to preserve the equality \(A = D\)), the connection must be made simultaneously at both ends. Parallel connection of the main and additional compensating six-terminal networks is also possible in some cases, but in practice it is not used because of the resulting increases in currents and powers.

Suppose that the additional compensating six-terminal network is connected at the generator end of the circuit and consists of series resistances in the phases, without interphase conductances. Then for it the matrix will have the form

\[ \begin{bmatrix} 1; K\\ 0; 1 \end{bmatrix}, \]

where

\[ K = K_0 + \alpha \cdot K_{\mathrm{I}} + \beta \cdot K_{\mathrm{II}} \]

has the dimension of resistance. The matrix for the entire circuit will be equal to the product

\[ \begin{bmatrix} 1; K\\ 0; 1 \end{bmatrix} \cdot \begin{bmatrix} A; B\\ C; D \end{bmatrix} = \begin{bmatrix} A + K \cdot C;\quad B + K \cdot D\\ C;\quad D \end{bmatrix}. \]

From this expression we see that the coefficients \(C\) and \(D\) remain unchanged under such a connection, and compensation can be carried out only for the coefficients \(A\) and \(B\). Since the voltages depend on them, it follows that compensation will take place only for the voltages, and not for the currents.

For such compensation it is necessary to choose the value \(K\) so that in the expressions \((A + K \cdot C)\) and \((B + K \cdot D)\) the terms with \(\alpha\) and \(\beta\) vanish. After substituting the trinomials, this condition reduces to the relations

\[ \begin{aligned} \alpha \cdot (A_{\mathrm{I}} + K_0 \cdot C_{\mathrm{I}} + K_{\mathrm{I}} \cdot C_0) &= 0;\\ \beta \cdot (A_{\mathrm{II}} + K_0 \cdot C_{\mathrm{II}} + K_{\mathrm{II}} \cdot C_0) &= 0;\\ \alpha \cdot (B_{\mathrm{I}} + K_0 \cdot D_{\mathrm{I}} + K_{\mathrm{I}} \cdot D_0) &= 0;\\ \beta \cdot (B_{\mathrm{II}} + K_0 \cdot D_{\mathrm{II}} + K_{\mathrm{II}} \cdot D_0) &= 0. \end{aligned} \tag{3} \]

From these relations we find

\[ \frac{A_{\mathrm{I}} + K_{\mathrm{I}} \cdot C_0}{C_{\mathrm{I}}} = \frac{A_{\mathrm{II}} + K_{\mathrm{II}} \cdot C_0}{C_{\mathrm{II}}} = \frac{B_{\mathrm{I}} + K_{\mathrm{I}} \cdot D_0}{D_{\mathrm{I}}} = \frac{B_{\mathrm{II}} + K_{\mathrm{II}} \cdot D_0}{D_{\mathrm{II}}} = - K_0. \tag{4} \]

Taking together the first and third terms, and then the second and fourth, we obtain expressions for determining the values of the components \(K_{\mathrm{I}}\) and \(K_{\mathrm{II}}\):

\[ K_{\mathrm{I}} = \frac{A_{\mathrm{I}} \cdot D_{\mathrm{I}} - B_{\mathrm{I}} \cdot C_{\mathrm{I}}} {C_{\mathrm{I}} \cdot D_0 - D_{\mathrm{I}} \cdot C_0}; \qquad K_{\mathrm{II}} = \frac{A_{\mathrm{II}} \cdot D_{\mathrm{II}} - B_{\mathrm{II}} \cdot C_{\mathrm{II}}} {C_{\mathrm{II}} \cdot D_0 - D_{\mathrm{II}} \cdot C_0}. \tag{5} \]

If these values are now substituted into equation (4), we obtain two expressions for determining \(K_0\):

\[ K_0 = \frac{B_{\mathrm{I}} \cdot C_0 - A_{\mathrm{I}} \cdot D_0} {C_{\mathrm{I}} \cdot D_0 - D_{\mathrm{I}} \cdot C_0}; \qquad K_0 = \frac{B_{\mathrm{II}} \cdot C_0 - A_{\mathrm{II}} \cdot D_0} {C_{\mathrm{II}} \cdot D_0 - D_{\mathrm{II}} \cdot C_0}. \tag{6} \]

In the general case these two expressions give different results, which indicates the impossibility of jointly solving the four equations (3), which have only three unknowns. Consequently, in this case the value \(K\) cannot be chosen so as simultaneously to compensate completely the asymmetry in both terms \(A\) and \(B\). As will be shown below, this is usually impossible for other reasons as well.

The asymmetry in the coefficients \(A\) and \(D\) is in most cases very small, and in a first approximation their components \(A_{\mathrm{I}}, A_{\mathrm{II}}, D_{\mathrm{I}}\), and \(D_{\mathrm{II}}\) may be set equal to zero. Then equations (5) and (6) are simplified to

\[ K_{\mathrm{I}}=-\frac{B_{\mathrm{I}}}{D_0};\qquad K_{\mathrm{II}}=-\frac{B_{\mathrm{II}}}{D_0};\qquad K_0=\frac{B_{\mathrm{I}}}{C_{\mathrm{I}}}\cdot \frac{C_0}{D_0};\qquad K_0=\frac{B_{\mathrm{II}}}{C_{\mathrm{II}}}\cdot \frac{C_0}{D_0}. \tag{7} \]

From the basic property of the coefficients of four-terminal networks,

\[ \begin{bmatrix} A; & B\\ C; & D \end{bmatrix} = A\cdot D-B\cdot C=1, \]

there follow, in the general case, the relations

\[ A_0\cdot D_0+\frac{1}{2}(A_{\mathrm{I}}\cdot D_{\mathrm{II}}+A_{\mathrm{II}}\cdot D_{\mathrm{I}}) = 1+B_0\cdot C_0+\frac{1}{2}(B_{\mathrm{I}}\cdot C_{\mathrm{II}}+B_{\mathrm{II}}\cdot C_{\mathrm{I}}); \]

\[ A_{\mathrm{I}}\cdot D_0+D_{\mathrm{I}}\cdot A_0 = B_{\mathrm{I}}\cdot C_0+C_{\mathrm{I}}\cdot B_0; \]

\[ A_{\mathrm{II}}\cdot D_0+D_{\mathrm{II}}\cdot A_0 = B_{\mathrm{II}}\cdot C_0+C_{\mathrm{II}}\cdot B_0, \]

which, under the assumptions indicated above, are simplified to

\[ B_{\mathrm{I}}\cdot C_0+C_{\mathrm{I}}\cdot B_0=0; \]

\[ B_{\mathrm{II}}\cdot C_0+C_{\mathrm{II}}\cdot B_0=0. \]

Since the coefficients \(C_0\) and \(B_0\) must necessarily have positive real parts of the complex quantities, these equalities can be satisfied only on condition that \(B_{\mathrm{I}}\) and \(C_{\mathrm{I}}\), and, respectively, \(B_{\mathrm{II}}\) and \(C_{\mathrm{II}}\), have different signs. In this case the complex quantity for \(K_0\) will in most cases have a negative real part (i.e., a negative active resistance) and, consequently, cannot be selected in practice. It is necessary to choose it with the smallest possible positive value, which does not make it possible to obtain full compensation.

Even in those cases where these values can be realized in practice, it is sometimes convenient to reject them and choose the smallest possible value, so as not to increase excessively the total resistance of the circuit.

Suppose that there is a four-terminal network with the coefficient matrix

\[ \begin{bmatrix} 1{,}099; & 10+(\alpha+\beta)\cdot 1\\ 0{,}01-(\alpha+\beta)\cdot 0{,}001; & 1 \end{bmatrix}. \]

Since in this case equations (7) are fully applicable, we determine from them the value of \(K\):

\[ K_{\mathrm{I}}=K_{\mathrm{II}}=-1;\qquad K_0=-10, \]

or in the form of a three-terminal network,

\[ K=-10-(\alpha+\beta)\cdot 1. \]

If these are active resistances, then it is impossible to select them. If they are pure reactive resistances (in this case all terms should be multiplied by \(j=\sqrt{-1}\)), then the circuit can be realized by combining inductances and capacitances. Choosing the minimal positive value for \(K_0\), we shall obtain compensation for the coefficient \(B\), but at the same time there will appear

new asymmetry in the coefficient \(A\). Thus, it is possible to insert resistances of 3 ohms in series in phases \(b\) and \(c\), which gives for \(K\) the value

\[ K=2-(\alpha+\beta)\cdot 1. \tag{8} \]

Carrying out the matrix multiplication, we obtain

\[ \begin{bmatrix} 1;\; 2-(\alpha+\beta)\cdot 1\\ 0;\; 1 \end{bmatrix} = \begin{bmatrix} 1{,}099;\; 10+(\alpha+\beta)\cdot 1\\ 0{,}01-(\alpha+\beta)\cdot 0{,}001;\; 1 \end{bmatrix} = \begin{bmatrix} 1{,}120-(\alpha+\beta)\cdot 0{,}012;\; 12\\ 0{,}01-(\alpha+\beta)\cdot 0{,}001;\; 1 \end{bmatrix}. \]

The appearance of asymmetry in the coefficient \(A\) is of no significance, since it can be corrected in a subsequent compensation.

To compensate the asymmetry in the currents, one may insert an additional six-terminal network consisting only of interphase conductances. The matrix for such a six-terminal network will have the form

\[ \begin{bmatrix} 1;\; 0\\ M;\; 1 \end{bmatrix}, \tag{9} \]

where the compensation will occur for the coefficients \(C\) and \(D\). Just as in the preceding case, we find the expressions

\[ M_{\mathrm{I}}=\frac{A_{\mathrm{I}}\cdot D_{\mathrm{I}}-B_{\mathrm{I}}\cdot C_{\mathrm{I}}}{B_{\mathrm{I}}\cdot A_0-A_{\mathrm{I}}\cdot B_0}; \qquad M_{\mathrm{II}}=\frac{A_{\mathrm{II}}\cdot D_{\mathrm{II}}-B_{\mathrm{II}}\cdot C_{\mathrm{II}}}{B_{\mathrm{II}}\cdot A_0-A_{\mathrm{II}}\cdot B_0}. \tag{10} \]

For \(M_0\) we shall likewise have two expressions

\[ M_0=\frac{C_{\mathrm{I}}\cdot B_0-D_{\mathrm{I}}\cdot A_0}{B_{\mathrm{I}}\cdot A_0-A_{\mathrm{I}}\cdot B_0}; \qquad M_0=\frac{C_{\mathrm{II}}\cdot B_0-D_{\mathrm{II}}\cdot A_0}{B_{\mathrm{II}}\cdot A_0-A_{\mathrm{II}}\cdot B_0}, \tag{11} \]

which may give different values because of the incompatibility of the solution of the four equations. For small values of the asymmetry in the coefficients \(A\) and \(D\), one may, with sufficient accuracy, take

\[ M_{\mathrm{I}}=-\frac{C_{\mathrm{I}}}{A_0}; \qquad M_{\mathrm{II}}=-\frac{C_{\mathrm{II}}}{A_0}; \qquad M_0=\frac{C_{\mathrm{I}}}{B_{\mathrm{I}}}\cdot\frac{B_0}{A_0}; \qquad M_0=\frac{C_{\mathrm{II}}}{B_{\mathrm{II}}}\cdot\frac{B_0}{A_0}. \tag{12} \]

When the six-terminal network (9) is inserted at the load end of the line, we shall have for the entire circuit the matrix

\[ \begin{bmatrix} A;\; B\\ C;\; D \end{bmatrix} \cdot \begin{bmatrix} 1;\; 0\\ M;\; 1 \end{bmatrix} = \begin{bmatrix} A+M\cdot B;\; B\\ C+M\cdot D;\; D \end{bmatrix}, \]

i.e., in this case compensation will occur for the coefficients \(A\) and \(C\). In doing so, we obtain

\[ M_{\mathrm{I}}=\frac{A_{\mathrm{I}}\cdot D_{\mathrm{I}}-B_{\mathrm{I}}\cdot C_{\mathrm{I}}}{B_{\mathrm{I}}\cdot D_0-D_{\mathrm{I}}\cdot B_0}; \qquad M_{\mathrm{II}}=\frac{A_{\mathrm{II}}\cdot D_{\mathrm{II}}-B_{\mathrm{II}}\cdot C_{\mathrm{II}}}{B_{\mathrm{II}}\cdot D_0-D_{\mathrm{II}}\cdot B_0}; \tag{13} \]

\[ M_0=\frac{C_{\mathrm{I}}\cdot B_0-A_{\mathrm{I}}\cdot D_0}{B_{\mathrm{I}}\cdot D_0-D_{\mathrm{I}}\cdot B_0}; \qquad M_0=\frac{C_{\mathrm{II}}\cdot B_0-A_{\mathrm{II}}\cdot D_0}{B_{\mathrm{II}}\cdot D_0-D_{\mathrm{II}}\cdot B_0}, \tag{14} \]

or, upon simplification,

\[ M_{\mathrm{I}}=-\frac{C_{\mathrm{I}}}{D_0}; \qquad M_{\mathrm{II}}=-\frac{C_{\mathrm{II}}}{D_0}; \qquad M_0=\frac{C_{\mathrm{I}}}{B_{\mathrm{I}}}\cdot\frac{B_0}{D_0}; \qquad M=\frac{C_{\mathrm{II}}}{B_{\mathrm{II}}}\cdot\frac{B_0}{D_0}. \tag{15} \]

This case is of interest because, in combination with the first case of compensation, it can yield complete compensation of the circuit. Continuing the numerical example given above, we find for this insertion by ...

according to equations (15), the values \(M_{\mathrm{I}}=M_{\mathrm{II}}=0.001\). This can be accomplished by connecting two admittances \(Y_{\mathrm{B}}=Y_{\mathrm{C}}=0.001\) between phases \(a\) and \(b\) and between \(a\) and \(c\). In this case we obtain

\[ M=0.002+(\alpha+\beta).0.001, \]

which gives, for the matrix of the entire circuit, the expression

\[ \begin{bmatrix} 0.12-(\alpha+\beta).0.012;\ 12\\ 0.01-(\alpha+\beta).0.001;\ 1 \end{bmatrix} \cdot \begin{bmatrix} 1;\ 0\\ 0.002+(\alpha+\beta).0.001;\ 1 \end{bmatrix} = \begin{bmatrix} 1.144;\ 12\\ 0.012;\ 1 \end{bmatrix}. \]

Thus, compensation of voltage asymmetry at one end of the circuit and compensation of current asymmetry at the other end make it possible to obtain complete compensation for the whole circuit as a whole.

If, finally, a six-terminal network with matrix (2) is connected at the load end of the circuit, we obtain the values

\[ K_{\mathrm{I}}=\frac{A_{\mathrm{I}}\cdot D_{\mathrm{I}}-B_{\mathrm{I}}\cdot C_{\mathrm{I}}}{C_{\mathrm{I}}\cdot A_0-A_{\mathrm{I}}\cdot C_0}; \quad K_{\mathrm{II}}=\frac{A_{\mathrm{II}}\cdot D_{\mathrm{II}}-B_{\mathrm{II}}\cdot C_{\mathrm{II}}}{C_{\mathrm{II}}\cdot A_0-A_{\mathrm{II}}\cdot C_0}; \tag{16} \]

\[ K_0=\frac{B_{\mathrm{I}}\cdot C_0-D_{\mathrm{I}}\cdot A_0}{C_{\mathrm{I}}\cdot A_0-A_{\mathrm{I}}\cdot C_0}; \quad K_0=\frac{B_{\mathrm{II}}\cdot C_0-D_{\mathrm{II}}\cdot A_0}{C_{\mathrm{II}}\cdot A_0-A_{\mathrm{II}}\cdot C_0}, \tag{17} \]

or, in simplified form,

\[ K_{\mathrm{I}}=-\frac{B_{\mathrm{I}}}{A_0}; \quad K_{\mathrm{II}}=-\frac{B_{\mathrm{II}}}{A_0}; \quad K_0=\frac{B_{\mathrm{I}}}{C_{\mathrm{I}}}\cdot\frac{C_0}{A_0}; \quad K_0=\frac{B_{\mathrm{II}}}{C_{\mathrm{II}}}\cdot\frac{C_0}{A_0}. \tag{18} \]

Simultaneous connection of the admittance six-terminal network (9) at the generator end of the circuit and of the resistance six-terminal network (2) at the load end can likewise provide complete compensation of phase asymmetry.

To preserve symmetry in the equivalent circuit, i.e., to satisfy the equality \(A=D\), simultaneous compensation should be carried out from both ends of the basic six-terminal network. In this case it is entirely permissible to determine the values \(K_{\mathrm{I}}, K_{\mathrm{II}}, M_{\mathrm{I}}\), and \(M_{\mathrm{II}}\) from relations (5), (10), (13), and (16), or from their simplified expressions (7), (12), (15), and (18), selecting \(K_0\) and \(M_0\) so as to make their magnitudes as small as possible, and then to insert at each end resistances corresponding to \(\frac{1}{2}K\), and admittances corresponding to \(\frac{1}{2}M\).

As an example, let us consider compensation of phase asymmetry of a three-phase line with one grounded phase, used simultaneously as a protective cable. The length of the line is 200 kilometers. The transmitted power is 68 megawatts at a voltage of 110 kilovolts and a power factor of 0.8. Having carried out the entire calculation of such a line, we obtain for it, when phase “a” is grounded, the coefficients of the six-terminal network

\[ A=D=(0.968+j.0.014)+(\alpha+\beta).(0.00165-j.0.00074); \]

\[ B=(41.6+j.93.7)-(\alpha+\beta).(8.0+j.18.0); \]

\[ C=(0.030+j.0.672).10^{-3}+(\alpha+\beta).(0.002+j.0.166).10^{-3}. \]

With symmetry of the load voltages and currents, this line, at full load, will have at the generator end a voltage asymmetry

\[ \frac{U_{\mathrm{II}_1}}{U_{\mathrm{I}_1}}\cdot 100 = 8.7\% \]

and a current asymmetry

\[ \frac{I_{\mathrm{II}_1}}{I_{\mathrm{I}_1}}\cdot 100 = 2.44\%, \]

which are still within the permissible norm.

Voltage compensation can be carried out by inserting a six-terminal network (2), for which

\[ K_{\mathrm{I}}=K_{\mathrm{II}}=\frac{8.0+j\,18.0}{2\cdot 0.968}=4.13+j\,9.3, \]

which can be accomplished by inserting in the grounded phase a resistance \(Z_a=24.68+j55.8\) ohms at either end of the circuit, or \(Z'_a=12.39+j27.9\) ohms at both ends. Compensation of current asymmetry can be carried out by inserting an additional six-terminal network (9), for which

\[ M_{\mathrm{I}}=M_{\mathrm{II}}=-\frac{(0.002+j\,0.166)\cdot 10^{-3}}{2\cdot 0.968} =-(1.03+j\,85.6)\cdot 10^{-6}, \]

which corresponds to inserting between the ungrounded phases a conductance \(Y_A=(2.06-j171.2)\cdot 10^{-6}\) siemens at either end of the line or, better, \(Y'_A=(1.03-j85.6)\cdot 10^{-6}\) siemens at both ends. Multiplying all the matrices in any combination, one can verify that the terms with coefficients \(\alpha\) and \(\beta\) will be very small and that the remaining asymmetry in the currents and voltages will be less than \(0.5\%\).

If the main circuit has no conductance between phases, then the compensation problem is greatly simplified, since coefficient \(C\) of the main six-terminal network becomes zero, and coefficients \(A\) and \(D\) (in the absence of transformers, or for per-unit values) are equal to unity. Coefficient \(B\) then corresponds to the sum of all resistances of the series elements of the circuit. Compensation of the asymmetry will occur when

\[ K_{\mathrm{I}}=-B_{\mathrm{I}};\qquad K_{\mathrm{II}}=-B_{\mathrm{II}}, \tag{19} \]

whereby the insertion of the additional six-terminal network (2) may be made at any point of the circuit.

As an example, let us take the supply of a three-phase arc circuit considered in the article by V. I. Vorob’ev and P. L. Kalantarov, “On balancing the three-phase system of currents in furnaces with an asymmetric current supply” (Elektrichestvo, No. 19, 1931).

When the supply conductors are arranged in one plane, the inductive coupling between the middle phase and the outer phases will be stronger than between the two outer phases. Taking the middle phase as the initial one (phase \(a\)), we have

\[ H_B=H_C;\qquad H_A<H_B. \quad \text{or} \quad M_{ac}=M_{ab};\qquad M_{bc}<M_{ac}, \]

where \(M_{ab}\), \(M_{ac}\), and \(M_{bc}\) are the mutual inductances of the phases.

Considering all the remaining elements of the circuit to be symmetrical, we obtain

\[ Z_{\mathrm{I}}=Z_{\mathrm{II}}=-\frac{2}{3}(H_B-H_A), \]

which also gives for the six-terminal network the coefficient values

\[ B_{\mathrm{I}}=B_{\mathrm{II}}=-\frac{2}{3}(H_B-H_A). \]

To compensate this asymmetry it is necessary to insert an additional six-terminal network which, according to relation (19), has

\[ K_{\mathrm{I}}=K_{\mathrm{II}}=\frac{2}{3}(H_B-H_A) =\frac{2}{3}\cdot j\omega\,(M_{ac}-M_{bc}), \]

which can be accomplished by inserting into the middle phase a reactive coil with impedance \(Z_k = r_k + j L_k \omega\), and two active resistances \(r_k\) into the remaining phases. In this case

\[ K_{\mathrm{I}} = K_{\mathrm{II}} = \frac{1}{3} j \cdot L_k \cdot \omega, \]

whence

\[ L_k = 2(M_{ac} - M_{bc}). \]

The same can be achieved by inserting into the outer phases two coils having strong inductive coupling with one another.

Since inserting coils into the low-voltage conductors is difficult because of the large currents flowing there, they are more often inserted into the high-voltage circuit ahead of the step-down transformer. If the latter has its windings connected according to the first group (star—star or delta—delta), then nothing changes in the referred quantities: the magnitude of the coils remains the same. If, however, the transformer has a connection according to the second group (star—delta or zigzag), then, when the windings are referred, the sign of the components of the impedances \(Z_{\mathrm{I}}\) and \(Z_{\mathrm{II}}\) changes to the opposite one, i.e., for this case it is necessary to have

\[ K_{\mathrm{I}} = K_{\mathrm{II}} = -\frac{2}{3}(H_B - H_A) = -\frac{2}{3} \cdot j \cdot \omega (M_{ac} - M_{bc}). \]

This can be accomplished by inserting into the outer phases two coils \(Z_k = r_k + j \cdot L_k \omega\) and an active resistance \(r_k\) into the middle phase, with, as in the first case, \(L_k = 2(M_{ac} - M_{bc})\).

Thus, compensation of asymmetry by inserting coils into the high-voltage circuit is possible for any transformer winding connection, and not only in the first case, as was noted in the article cited above.

1. R. A. Voronov, Ternary expressions as applied to the calculation and investigation of asymmetrical three-phase circuits. ↩