Trinomial Expressions in Application to the Calculation and Study of Unbalanced Three-Phase Circuits

R. A. Voronov

Professor, Doctor of Technical Sciences

In calculating the distribution of currents and voltages in unbalanced three-phase circuits, considerable simplifications are obtained by using the techniques of the method of symmetrical components. In the author’s work “Calculation of the Distribution of Currents and Voltages in Symmetrical Three-Phase Circuits,” the application of this method was examined in the most general case—for circuits both having a neutral conductor, i.e., capable of carrying zero-sequence current components, and for circuits without one. In the latter case it is possible to use a special form of the general mathematical expressions for the components, which makes it possible to simplify the solution of a number of different questions that present considerable difficulties under the usual methods of solution. This part of the work is presented in the present article.

Since all voltages and currents enter the equations exclusively in complex form, the dots over the letters have been omitted throughout. For a number of reasons, the work introduced certain notations and indices different from those adopted in other books and articles. These notations are fully retained in the present article as well, although for circuits without zero-sequence components one could have dispensed with them.

Symmetrical Components of Currents and Voltages

According to the method of symmetrical components, any system of three voltages \(U_a\), \(U_b\), and \(U_c\) may be represented through the positive-sequence component \(U_I\), the negative-sequence component \(U_{II}\), and the zero-sequence component \(U_0\), taken for one of the phases. If phase \(a\) is adopted as the reference phase, then

\[ \begin{aligned} U_a &= U_I + U_{II} + U_0;\\ U_b &= a^2 U_I + a U_{II} + U_0;\\ U_c &= a U_I + a^2 U_{II} + U_0, \end{aligned} \tag{1} \]

where \(a\) and \(a^2\) are operators rotating the vectors through angles of \(120^\circ\) and \(240^\circ\),

\[ a = e^{j\frac{2}{3}\pi} = -\frac{1}{2} + j\frac{\sqrt{3}}{2}; \qquad a^2 = e^{j\frac{4}{3}\pi} = -\frac{1}{2} - j\frac{\sqrt{3}}{2}. \tag{2} \]

In the same way, the three currents \(I_a\), \(I_b\), and \(I_c\) may be represented through the components \(I_I\), \(I_{II}\), and \(I_0\).

In the general case of phase unbalance in a circuit, each of the component currents causes the appearance of each of the component voltages, so that the dependence between them may be represented in the form of the relations

\[ \begin{aligned} U_{\mathrm{I}}&=I_{\mathrm{I}}\cdot Z'_0+I_{\mathrm{II}}\cdot Z_{\mathrm{II}}+I_0\cdot Z'_{\mathrm{I}},\\ U_{\mathrm{II}}&=I_{\mathrm{I}}\cdot Z_{\mathrm{I}}+I_{\mathrm{II}}\cdot Z'_0+I_0\cdot Z_{\mathrm{II}},\\ U_0&=I_{\mathrm{I}}\cdot Z''_{\mathrm{II}}+I_{\mathrm{II}}\cdot Z'_{\mathrm{I}}+I_0\cdot Z_0. \end{aligned} \tag{3} \]

The nine coefficients entering into them, \(Z'_0, Z_{\mathrm{II}}\), etc., having the dimensions of resistances, can easily be determined from the parameters of the circuit. Since the equations by which they are determined are similar in their structure to the equations for determining the components of voltages and currents (see equations 9), they may be regarded as resistance components of the zero, positive, and negative sequences. In the general case all nine coefficients may be different; usually, however, some of them prove to be identical. Thus, for static circuits there will be the equalities

\[ Z'_0=Z''_0;\quad Z'_{\mathrm{I}}=Z''_{\mathrm{II}};\quad Z'_{\mathrm{II}}=Z''_{\mathrm{I}}. \]

The lower subscripts of the resistance components determine their sequence, while the upper ones have been chosen so that, together with the lower ones, they determine the sequence of the current components to which they correspond (i.e., with which they stand together in the equations). In this case the sum of the lower subscripts of the current and the resistance gives the sequence of the resulting voltage component (three is to be counted as zero, and four as one).

Solving the system of equations (3) with respect to the current components, we obtain a new system of equations

\[ \begin{aligned} I_{\mathrm{I}}&=U_{\mathrm{I}}\cdot Y'_0+U_{\mathrm{II}}\cdot Y_{\mathrm{II}}+U_0\cdot Y'_{\mathrm{I}},\\ I_{\mathrm{II}}&=U_{\mathrm{I}}\cdot Y_{\mathrm{I}}+U_{\mathrm{II}}\cdot Y''_0+U_0\cdot Y'_{\mathrm{II}},\\ I_0&=U_{\mathrm{I}}\cdot Y''_{\mathrm{II}}+U_{\mathrm{II}}\cdot Y'_{\mathrm{I}}+U_0\cdot Y_0. \end{aligned} \tag{4} \]

The nine coefficients entering into them, \(Y'_0, Y_{\mathrm{II}}\), etc., will have the dimensions of conductances and may be regarded as certain conductance components of the circuit of the zero, positive, and negative sequences, their indices being chosen in the same way as for the resistance components. In most cases these conductance components cannot be determined directly from the circuit parameters, but are obtained only by solving the system of equations (3).

With symmetry of the phases of a circuit, the positive- and negative-sequence components, both for resistances and for conductances, become zero, and only the zero-sequence components \(Z'_0, Z''_0, Z_0\) and \(Y'_0, Y''_0, Y_0\) remain. In this case the well-known proposition on the independence of the separate sequences from one another is obtained; that is, in symmetrical circuits the components of the positive-sequence currents produce voltage components also only of the positive sequence, etc.

When calculating short-circuit currents for symmetrical circuits, the components \(Z'_0, Z''_0\), and \(Z_0\) represent the total resistances for currents of the positive, negative, and zero sequences (usually in this case they are denoted by \(Z_{\mathrm{I}}, Z_{\mathrm{II}}\), and \(Z_0\)).

Without dwelling on the general properties of all these components and their uses for calculations of three-phase circuits, let us proceed directly to the consideration of circuits without a neutral conductor, widely used in all electrical installations.

Circuit without a Neutral Conductor

In the absence in the circuit of zero-sequence current components (for example, in a circuit without a neutral conductor), all calculations are considerably simplified. Equations (3) take the form

\[ \begin{aligned} U_I &= I_I \cdot Z'_0 + I_{II}\cdot Z_{II};\\ U_{II} &= I_I \cdot Z_I + I_{II}\cdot Z''_0;\\ U_0 &= I_I \cdot Z''_{II} + I_{II}\cdot Z'_I. \end{aligned} \tag{5} \]

From the first two equations we obtain, for \(I_I\) and \(I_{II}\), the expressions

\[ I_I = U_I \cdot \frac{Z''_0}{Z'_0 \cdot Z''_0 - Z_I \cdot Z_{II}} + U_{II} \cdot \frac{-Z_{II}}{Z'_0 \cdot Z''_0 - Z_I \cdot Z_{II}}; \]

\[ I_{II} = U_I \cdot \frac{-Z_I}{Z'_0 \cdot Z''_0 - Z_I \cdot Z_{II}} + U_{II} \cdot \frac{Z'_0}{Z'_0 \cdot Z''_0 - Z_I \cdot Z_{II}}. \]

Introducing notation for the admittance components, we find

\[ \begin{aligned} I_I &= U_I \cdot Y'_0 + U_{II}\cdot Y_{II};\\ I_{II} &= U_I \cdot Y_I + U_{II}\cdot Y''_0, \end{aligned} \tag{6} \]

where

\[ Y'_0=\frac{Z''_0}{Z'_0\cdot Z''_0-Z_I\cdot Z_{II}}; \qquad Y''_0=\frac{Z'_0}{Z'_0\cdot Z''_0-Z_I\cdot Z_{II}}; \]

\[ Y_I=\frac{-Z_I}{Z'_0\cdot Z''_0-Z_I\cdot Z_{II}}; \qquad Y_{II}=\frac{-Z_{II}}{Z'_0\cdot Z''_0-Z_I\cdot Z_{II}}. \tag{7} \]

Solving equations (6) with respect to the voltage components, we have

\[ Z'_0=\frac{Y''_0}{Y'_0\cdot Y''_0-Y_I\cdot Y_{II}}; \qquad Z''_0=\frac{Y'_0}{Y'_0\cdot Y''_0-Y_I\cdot Y_{II}}; \]

\[ Z_I=\frac{-Y_I}{Y'_0\cdot Y''_0-Y_I\cdot Y_{II}}; \qquad Z_{II}=\frac{-Y_{II}}{Y'_0\cdot Y''_0-Y_I\cdot Y_{II}}. \tag{8} \]

Since in such a circuit the zero-sequence component of the applied voltage has no influence on the currents flowing, and the component \(I_0\) is equal to zero, the admittance components \(Y_0\), \(Y'_I\), \(Y''_I\), \(Y'_{II}\), and \(Y''_{II}\) must be equal to zero. All resistance components will exist, with the exception of \(Z_0\), which becomes infinite. In this case \(Z'_I\) and \(Z'_{II}\) play no role, while \(Z''_I\) and \(Z''_{II}\) participate only in determining the component \(U_0\) for the load, according to the last of equations (5).

Finding the Distribution of Currents and Voltages

To find the distribution of currents and voltages in a given circuit, the following procedure may be outlined.

1. All elements of the individual sections of the circuit must be expressed in components of impedances and admittances.

2. In a series connection of several sections of the circuit, the corresponding impedance components are summed, which may be expressed symbolically as

\[ Z_k = Z_1 + Z_2 + \cdots = \sum Z_n. \]

1. When branches are present, the corresponding components of the branch admittances are summed

\[ Y_k = Y_1 + Y_2 + \ldots = \sum Y_n . \]

1. By a gradual transition from impedances to admittances and back, the values of the components of the total impedance and total admittance of the circuit are found for the point of application of the known voltage (or e.m.f.), or else for the point of passage of the known current.

2. From the values found and the known voltage, the components of the total current of the given part of the circuit are found

\[ I = U \cdot Y = U \cdot Z^{-1}. \]

1. In the reverse manner, the components of voltages and currents at various points of the circuit are found, using the previously obtained components of the impedances and admittances of these parts of the circuit

\[ U_p = I_p \cdot Z_p; \qquad I_q = U_q \cdot Y_q \]

and so on.

1. From the components of currents and voltages found, the currents and voltages in the phases and the powers for individual sections of the circuit and for the entire circuit as a whole are determined.

Determination of the Components of Impedances

In a circuit without a neutral wire, in order to carry out the calculations it is necessary to have the values of only four components of impedance \(Z_0'\), \(Z_0''\), \(Z_1\), and \(Z_{II}\). These components can be found from consideration of the ordinary equations of voltage drop.

For the static circuit shown in the diagram of Fig. 1, for each of the phases one may write expressions for the voltage drop

\[ U_a = I_a \cdot (r_a + j \cdot L_a \cdot \omega) + I_b \cdot j \cdot M_{ab} \cdot \omega + I_c \cdot j \cdot M_{ac} \cdot \omega; \]

\[ U_b = I_b \cdot (r_b + j \cdot L_b \cdot \omega) + I_a \cdot j \cdot M_{ab} \cdot \omega + I_c \cdot j \cdot M_{bc} \cdot \omega; \]

\[ U_c = I_c \cdot (r_c + j \cdot L_c \cdot \omega) + I_a \cdot j \cdot M_{ac} \cdot \omega + I_b \cdot j \cdot M_{bc} \cdot \omega. \]

Fig. 1

Introducing notation for the phase impedances

\[ Z_a = r_a + j \cdot L_a \cdot \omega; \qquad Z_b = r_b + j \cdot L_b \cdot \omega; \qquad Z_c = r_c + j \cdot L_c \cdot \omega \]

and for the impedances of mutual inductances

\[ j \cdot M_{bc} \cdot \omega = H_A; \qquad j \cdot M_{ac} \cdot \omega = H_B; \qquad j \cdot M_{ab} \cdot \omega = H_C \]

(the designations by \(H\) have been introduced for convenience, since their usual designations by \(x\) or \(Z\) require the introduction of double subscripts), we obtain

\[ U_a = I_a \cdot Z_a + I_b \cdot H_C + I_c \cdot H_B; \]

\[ U_b = I_a \cdot H_C + I_b \cdot Z_b + I_c \cdot H_A; \]

\[ U_c = I_a \cdot H_B + I_b \cdot H_A + I_c \cdot Z_c. \]

Determining the components of the voltages and remembering that \(I_0\) is equal to zero, we find

\[ U_I = I_I \cdot \left[ \frac{Z_a + Z_b + Z_c}{3} - \frac{H_A + H_B + H_C}{3} \right] + \]

\[ + I_{II} \cdot \left[ \frac{Z_a + a^2 \cdot Z_b + a \cdot Z_c}{3} + 2 \cdot \frac{H_A + a^2 \cdot H_B + aH_C}{3} \right]; \]

\[ U_{\mathrm{II}}=I_{\mathrm{I}}\cdot\left[\frac{Z_a+a\cdot Z_b+a^2\cdot Z_c}{3} +2\cdot\frac{H_A+a\cdot H_B+a^2\cdot H_C}{3}\right]+ \]

\[ +I_{\mathrm{II}}\cdot\left[\frac{Z_a+Z_b+Z_c}{3} -\frac{H_A+H_B+H_C}{3}\right]; \]

\[ U_0=I_{\mathrm{I}}\cdot\left[\frac{Z_a+a^2\cdot Z_b+a\cdot Z_c}{3} -\frac{H_A+a^2\cdot H_B+a\cdot H_C}{3}\right]+ \]

\[ +I_{\mathrm{II}}\cdot\left[\frac{Z_a+a\cdot Z_b+a^2\cdot Z_c}{3} -\frac{H_A+a\cdot H_B+a^2\cdot H_C}{3}\right]. \]

Comparing these expressions with (5), we find the components of the resistances

\[ Z_0'=Z_0''=\frac{Z_a+Z_b+Z_c}{3}-\frac{H_A+H_B+H_C}{3}; \]

\[ Z_{\mathrm{I}}=\frac{Z_a+a\cdot Z_b+a^2 Z_c}{3} +2\cdot\frac{H_A+a\cdot H_B+a^2\cdot H_C}{3}; \]

\[ Z_{\mathrm{II}}=\frac{Z_a+a^2\cdot Z_b+a\cdot Z_c}{3} +2\cdot\frac{H_A+a^2\cdot H_B+a\cdot H_C}{3}; \tag{9} \]

\[ Z_{\mathrm{I}}'=\frac{Z_a+a\cdot Z_b+a^2\cdot Z_c}{3} -\frac{H_A+a\cdot H_B+a^2\cdot H_C}{3}; \]

\[ Z_{\mathrm{II}}'=\frac{Z_a+a^2\cdot Z_b+a\cdot Z_c}{3} -\frac{H_A+a^2\cdot H_B+a\cdot H_C}{3}. \]

In the general case, when zero-sequence currents pass, these expressions become more complicated because of the introduction of mutual inductances between the phases and the neutral conductor, and because of the resistance of the latter.

The components of the conductances usually have to be determined through these components of the resistances by equations (7), but in some cases they have to be sought directly from the parameters of the circuit, since the components of the resistances go to infinity. Such a case will occur, for example, when a resistance is connected between two phases, i.e. when the third phase is disconnected. If we put \(Z_b=\infty\), then equations (7) give an indeterminacy; resolving it, we find

\[ Y_0'=Y_0''=\frac{1}{Z_a+Z_c-2H_B};\qquad Y_{\mathrm{I}}=-a\cdot Y_0;\qquad Y_{\mathrm{II}}=-a^2\cdot Y_0. \]

Since in this case the resistances \(Z_a\) and \(Z_c\) are connected in series, they may be regarded as one

\[ Z_a+Z_c-2H_B=Z_B, \]

which gives for the components of the conductance the expression

\[ Y_0'=Y_0''=\frac{1}{Z_B}=Y_B;\qquad Y_{\mathrm{I}}=-a\cdot Y_B;\qquad Y_{\mathrm{II}}=-a^2Y_B. \tag{9a} \]

When phase C is disconnected, i.e. when resistance \(Z_c\) is connected between phases a and b, in expressions (9a) the factors \(a\) and \(a^2\) interchange places; and when phase a is disconnected, i.e. when resistance \(Z_A\) is connected between phases b and c, the factors \(a\) and \(a^2\) are replaced by units.

When impedances are connected in a delta, such a connection should first be replaced by an equivalent star, and only then should the component impedances be found from equations (9). For rotating machines (usually with symmetrical phases), the impedances \(Z'_0\) and \(Z''_0\) will not be equal to one another and are determined by the usual methods set forth in courses on electrical machines.

Equations (9) cannot be used to determine the component impedances for circuits with distributed parameters, since in this case it is not possible to find separately the self-inductances and mutual inductances of each of the phases. In this case the components may be determined as follows.

In the absence of zero-sequence currents, we have the voltage drops in the conductors of the individual phases

\[ U_a=-\frac{1}{2}\left(j\omega L_{AB} I_b+j\omega L_{AC} I_c\right); \]

\[ U_b=-\frac{1}{2}\left(j\omega L_{BC} I_c+j\omega L_{AB} I_a\right); \]

\[ U_c=-\frac{1}{2}\left(j\omega L_{CA} I_a+j\omega L_{CB} I_b\right), \]

where

\[ L_{AB}=\frac{\mu}{\pi}\cdot l\left(\ln \frac{d_{AB}}{R}+0.25\right);\quad L_{BC}=\frac{\mu}{\pi}\cdot l\left(\ln \frac{d_{BC}}{R}+0.25\right); \]

\[ L_{AC}=\frac{\mu}{\pi}\cdot l\left(\ln \frac{d_{AC}}{R}+0.25\right). \]

Assuming the presence only of positive-sequence current components,

\[ I_b=a^2 I_a;\quad I_c=a I_a, \]

and resolving the voltage drops into components, we have

\[ U_I=\frac{1}{6}j\omega\left(L_{AB}+L_{AC}+L_{BC}\right)I_I; \]

\[ U_{II}=-\frac{1}{3}j\omega\left(a^2 L_{AB}+aL_{AC}+L_{BC}\right)I_I. \]

Adding the active resistances of the conductors, we obtain the values of the component impedances

\[ Z'_0=\frac{1}{3}(r_a+r_b+r_c)+\frac{1}{6}j\omega\left(L_{AB}+L_{AC}+L_{BC}\right); \]

\[ Z_1=\frac{1}{3}(r_a+a r_b+a^2 r_c)-\frac{1}{3}j\omega\left(a^2 L_{AB}+aL_{AC}+L_{BC}\right). \]

In the same way, taking in the phases only negative-sequence current components,

\[ I_b=a I_a;\quad I_c=a^2 I_b, \]

we obtain the following values for the components:

\[ Z''_0=\frac{1}{3}(r_a+r_b+r_c)+\frac{1}{6}j\omega\left(L_{AB}+L_{AC}+L_{BC}\right); \]

\[ Z_{II}=\frac{1}{3}(r_a+a^2 r_b+a r_c)+r_c+\frac{1}{3}j\omega\left(aL_{AB}+a^2L_{AC}+L_{BC}\right). \]

Thus, instead of equations (9), for determining the component impedances we have the expressions

\[ Z_0' = Z_0'' = \frac{1}{3}(r_a+r_b+r_c) +\frac{1}{2} j\omega \frac{\mu}{\pi} l \left( \ln \frac{\sqrt[3]{d_{AB}d_{AC}d_{BC}}}{R}+0.25 \right) \]

\[ Z_{\mathrm{I}}= \frac{1}{3}(r_a+a r_b+a^2 r_c) -\frac{1}{3} j\omega \frac{\mu}{\pi} l \left( a^2\ln\frac{d_{AB}}{R} +a\ln\frac{d_{AC}}{R} +\ln\frac{d_{BC}}{R} \right); \tag{10} \]

\[ Z_{\mathrm{II}}= \frac{1}{3}(r_a+a^2 r_b+a r_c) -\frac{1}{3} j\omega \frac{\mu}{\omega} l \left( a\ln\frac{d_{AB}}{R} +a^2\ln\frac{d_{AC}}{R} +\ln\frac{d_{BC}}{R} \right). \]

Transformations of Circuits without a Neutral Wire

As was indicated above, in solving various problems one has to combine series sections of a circuit, find equivalent values for parallel sections, etc., which usually has the aim of reducing the given circuit to a simpler form and then finding for it the values of the currents in its individual parts. By an inverse recalculation of these values, the distribution of currents in the original complex circuit is found.

A similar situation will also occur when it is necessary to extend certain properties of simple circuits to complex circuits. Thus, for example, the circle diagram for a simple circuit (for a quadripole) can be extended to any three-phase circuit while preserving all its properties.

In solving problems of this kind, one has to deal with the transformation of circuits in general form, which is greatly simplified by introducing one common notation for all component impedances or admittances. Such a notation may be a matrix form of the type of the equations

\[ \begin{bmatrix} U_{\mathrm{I}}\\ U_{\mathrm{II}} \end{bmatrix} = \begin{bmatrix} Z_0' & Z_{\mathrm{II}}\\ Z_{\mathrm{I}} & Z_0'' \end{bmatrix} \cdot \begin{bmatrix} I_{\mathrm{I}}\\ I_{\mathrm{II}} \end{bmatrix} \]

or, in general form,

\[ \mathrm{U}=Z\cdot \mathrm{I}. \]

For circuits without a neutral wire, having no paths for the passage of zero-sequence currents, instead of the matrix form, for the joint expression of component impedances or admittances one may propose a simpler form, very convenient in various transformations. In its structure this form represents trinomial hypercomplex numbers with complex elements. The consideration of the properties of these trinomial expressions and their use for three-phase circuits is the principal task of the present work.

Trinomials of Impedances and Admittances

In a three-phase circuit without a neutral wire, in the absence of components of zero-sequence currents, of all the component impedances only four are of significance: \(Z_0'\), \(Z_0''\), \(Z_{\mathrm{I}}\), and \(Z_{\mathrm{II}}\). Since for stationar-

circuits that have no rotating parts (machines), the components \(Z'_0\) and \(Z''_0\) are equal to one another; therefore, denoting them without superscript indices, we obtain only three distinct quantities \(Z_0, Z_{\mathrm{I}}\), and \(Z_{\mathrm{II}}\).

These three values, determining the total impedance \(Z\) of the circuit and its asymmetry, can be represented in the form of a trinomial

\[ Z=Z_0+\alpha Z_{\mathrm{I}}+\beta Z_{\mathrm{II}}, \tag{11} \]

where \(\alpha\) and \(\beta\) are conditional factors determining the order of these components. These factors, together with the unity of the first term, represent, as it were, unit vectors determining the entire impedance as a vector of a certain three-dimensional space.

For the same circuit there will be only four components of admittance \(Y'\), \(Y'_0\), \(Y_{\mathrm{I}}\), and \(Y_{\mathrm{II}}\), of which the first two are usually equal to one another. Denoting them without superscript indices, we may write the admittance in the form

\[ Y=Y_0+\alpha Y_{\mathrm{I}}+\beta Y_{\mathrm{II}}. \tag{12} \]

Since these components of admittance refer to the same part of the circuit as the impedance components of equation (11), relations according to equations (7) must exist between them:

\[ Y_0=\frac{Z_0}{Z_0^2-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}};\quad Y_{\mathrm{I}}=\frac{-Z_{\mathrm{I}}}{Z_0^2-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}};\quad Y_{\mathrm{II}}=\frac{-Z_{\mathrm{II}}}{Z_0^2-Z_{\mathrm{I}}Z_{\mathrm{II}}}. \tag{13} \]

or, according to equations (8),

\[ Z_0=\frac{Y_0}{Y_0^2-Y_{\mathrm{I}}\cdot Y_{\mathrm{II}}};\quad Z_{\mathrm{I}}=\frac{-Y_{\mathrm{I}}}{Y_0^2-Y_{\mathrm{I}}\cdot Y_{\mathrm{II}}};\quad Z_{\mathrm{II}}=\frac{-Y_{\mathrm{II}}}{Y_0^2-Y_{\mathrm{I}}\cdot Y_{\mathrm{II}}}. \tag{14} \]

Substituting expressions (13) into expression (12), we obtain

\[ Y=\frac{Z_0-\alpha Z_{\mathrm{I}}-\beta Z_{\mathrm{II}}}{Z_0^2-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}}. \tag{15} \]

The numerator represents a trinomial of impedances in which the signs of the second and third terms have been changed to their opposites. We shall call such a trinomial conjugate and introduce for it the notation

\[ \widehat{Z}=Z_0-\alpha Z_{\mathrm{I}}-\beta Z_{\mathrm{II}}. \tag{16} \]

If we assume that

\[ Y=Z^{-1}=\frac{1}{Z}, \]

then expression (15) takes the form

\[ \frac{1}{Z}=\frac{\widehat{Z}}{Z_0^2-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}}, \]

or

\[ Z_0^2-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}=Z\cdot \widehat{Z}. \]

Thus, the expression appearing in the denominator of (13) and (15) represents the value of the product of two conjugate trinomials. We introduce for it the notation in square brackets

\[ [Z]=Z\cdot \widehat{Z}=Z_0^2-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}. \tag{17} \]

Analogous expressions can also be obtained for admittances.

If in this expression (17) we substitute the values of the trinomials (11) and (16) and perform their algebraic multiplication, we obtain

\[ Z_0^2 - Z_1 Z_{\mathrm{II}} = \left(Z_0+\alpha Z_1+\beta Z_{\mathrm{II}}\right) \left(Z_0-\alpha Z_1-\beta Z_{\mathrm{II}}\right) = \]

\[ = Z_0^2-2\alpha\beta Z_1 Z_{\mathrm{II}} -\alpha^2 Z_1^2-\beta^2 Z_{\mathrm{II}}^2 . \]

This equality is satisfied if the coefficients \(\alpha\) and \(\beta\) (i.e., the unit vectors) are assigned the properties¹

\[ 2\alpha\beta=1;\qquad \alpha^2=0;\qquad \beta^2=0. \tag{18} \]

Below it will be shown that in most cases there is no need to resort to these relations, since they enter automatically together with relations of the type of products (17).

The principal application of conjugate trinomials follows from equation (15), which may be represented in the form

\[ Y=\frac{\widehat Z}{[Z]}. \tag{19} \]

This expression serves for the transition from impedances to admittances. For the inverse transition we have the analogous expression

\[ Z=\frac{\widehat Y}{[Y]}. \tag{20} \]

It is easy to see that the following relations also hold:

\[ [Z]\,[Y]=1;\qquad \widehat Y=\frac{Z}{[Z]},\qquad \widehat Z=\frac{Y}{[Y]}. \tag{21} \]

Basic relations for trinomials

For convenience in subsequent transformations we introduce the following notation:

\[ [Z]_{A+B+C+\ldots} = \left(Z_A+Z_B+Z_C+\ldots\right) \left(\widehat Z_A+\widehat Z_B+\widehat Z_C+\ldots\right) = \]

\[ = [Z_A+Z_B+Z_C+\ldots] \tag{22} \]

\[ [Z]_{AB}=\frac{1}{2}\left(Z_A Z_B+Z_B Z_A\right). \tag{23} \]

The last expression, after substituting the trinomials and multiplying them, taking into account relations (18), gives

\[ [Z]_{AB} = Z_{0A}Z_{0B} + \frac{1}{2}\left(Z_{IA}Z_{IIB}+Z_{IIA}Z_{IB}\right), \tag{24} \]

and equation (22), after opening the brackets and grouping the terms, gives

\[ [Z]_{A+B+C+\ldots} = [Z]_A+[Z]_B+[Z]_C+\ldots +2[Z]_{AB}+2[Z]_{AC}+2[Z]_{BC}+\ldots . \tag{25} \]

If there are only two trinomials, as a sum or a difference, then this expression takes the form

\[ [Z]_{A\pm B}=[Z]_A+[Z]_B\pm 2[Z]_{AB}. \tag{26} \]

In addition, it is easy to verify the existence of the relations

\[ [Z]_{A(B+C)}=[Z]_{AB}+[Z]_{AC};\quad [Z]_{AA}=[Z]_{A}. \tag{27} \]

Exactly the same relations are also obtained for admittances. For the transition from resistances to admittances and back, besides the fundamental equation (21), which may be written as

\[ [Z]_{A}\cdot [Y]_{A}=1, \tag{28} \]

the following can also be obtained:

\[ [Z]_{AB}= \frac{Z_A\cdot \widehat{Z}_B+Z_B\cdot \widehat{Z}_A}{2} = \frac{\widehat{Z}_A\cdot \widehat{Z}_B+\widehat{Z}_B\cdot \widehat{Z}_A} {2[Z]_A\cdot [Y]_A\cdot [Z]_B\cdot [Y]_B} = \]

\[ = \frac{\widehat{Y}_A\cdot Y_B+\widehat{Y}_B\cdot Y_A} {2[Y]_A\cdot [Y]_B} = -\frac{[Y]_{AB}}{[Y]_A\cdot [Y]_B}, \]

or

\[ [Z]_{AB}=-\frac{[Y]_{AB}}{[Y]_A\cdot [Y]_B}; \quad [Y]_{AB}=-\frac{[Z]_{AB}}{[Z]_A\cdot [Z]_B}. \tag{29} \]

In a similar way,

\[ [Z]_{A+B}=[Z]_A+[Z]_B+2[Z]_{AB} = \frac{1}{[Y]_A}+\frac{1}{[Y]_B} +2\frac{[Y]_{AB}}{[Y]_A\cdot [Y]_B} = \]

\[ = \frac{[Y]_B+[Y]_A+2[Y]_{AB}}{[Y]_A\cdot [Y]_B}, \]

or

\[ [Z]_{A+B}=\frac{[Y]_{A+B}}{[Y]_A\cdot [Y]_B}; \quad [Y]_{A+B}=\frac{[Z]_{A+B}}{[Z]_A\cdot [Z]_B}. \tag{30} \]

These fundamental relations make it possible to carry out transformations in equations, as will be indicated below.

Equivalent circuits

The simplest use of trinomials will be the finding of expressions for the resistance of two parallel branches, under the condition that there are no paths for the zero-sequence component of the current. If one of the branches has the resistance

\[ Z_A=Z_{0A}+\alpha\cdot Z_{IA}+\beta\cdot Z_{IIA}, \]

and the second

\[ Z_B=Z_{0B}+\alpha\cdot Z_{IB}+\beta\cdot Z_{IIB}, \]

then their admittances are determined as

\[ Y_A=\frac{\widehat{Z}_A}{[Z]_A}; \quad Y_B=\frac{\widehat{Z}_B}{[Z]_B}. \]

The total admittance of both branches will be equal to the sum

\[ Y=Y_A+Y_B=\frac{\widehat{Z}_A}{[Z]_A}+\frac{\widehat{Z}_B}{[Z]_B}. \]

and the total resistance will be equal to

\[ Z=\frac{1}{Y}=\frac{1}{Y_A+Y_B}=\frac{\hat Y_A+\hat Y_B}{[Y]_{A+B}}. \]

Using relations (21) and (30), we have

\[ Z= \frac{\dfrac{Z_A}{[Z]_A}+\dfrac{Z_B}{[Z]_B}} {\dfrac{[Z]_{A+B}}{[Z]_A\cdot [Z]_B}} = \frac{Z_A\cdot [Z]_B\cdot [Z]_A}{[Z]_{A+B}+Z_B}. \]

The same relation can also be obtained in another way. It is known that the equivalent resistance of two parallel branches may be expressed as

\[ Z=\frac{Z_A\cdot Z_B}{Z_A+Z_B}. \]

Multiplying both the numerator and the denominator by the conjugate trinomial of the denominator, we obtain directly

\[ Z= \frac{Z_A\cdot Z_B\cdot(\hat Z+\hat Z_B)} {(Z_A+Z_B)\cdot(\hat Z+\hat Z_B)} = \frac{Z_B\cdot [Z]_A+Z_A\cdot [Z]_B}{[Z]_{A+B}}. \]

Thus, for two parallel branches,

\[ Z=\frac{Z_A\cdot Z_B}{Z_A+Z_B} = \frac{Z_A\cdot [Z]_B+Z_B\cdot [Z]_A}{[Z]_{A+B}}. \tag{31} \]

If this equation is expanded, then for the component \(Z_0\) we shall have the expression

\[ Z_0= \frac{ Z_{0A}\cdot\left(Z_{0B}^{2}-Z_{IB}\cdot Z_{IIB}\right) + Z_{0B}\cdot\left(Z_{0B}^{2}-Z_{IA}\cdot Z_{IIA}\right) } { \left(Z_{0A}+Z_{0B}\right)^2 - \left(Z_{IA}+Z_{IB}\right)\cdot\left(Z_{IIA}+Z_{IIB}\right) }, \]

to obtain which without the aid of trinomials would be considerably more difficult. Similar expressions are obtained for the remaining components as well.

In the case of series connection of two sections of a circuit having conductances \(Y_A\) and \(Y_B\), a similar equation will give the total conductance of the entire circuit

\[ Y=\frac{Y_A\cdot [Y]_B+Y_B\cdot [Y]_A}{[Y]_{A+B}}. \tag{32} \]

Let us now suppose that three branches, having resistances \(Z_A\), \(Z_B\), and \(Z_C\), form a triangle (Fig. 2a), which it is desirable to transform into an equivalent circuit connected in star (Fig. 2b). The need for such a transformation may easily arise when solving problems connected with a complex form of circuit.

Fig. 2

In the absence of paths for zero-sequence currents, we may write, in general form,

\[ Z_a=\frac{Z_B\cdot Z_C}{Z_A+Z_B+Z_C}, \]

with cyclic replacement of the indices for the other branches of the star. Multiplying both the numerator and the denominator by the conjugate trinomial of the denominator, we obtain

\[ Z= \frac{Z_B\cdot Z_C\cdot(\hat Z_A+\hat Z_B+\hat Z_C)} {[Z]_{A+B+C}} = \frac{ \hat Z_A\cdot Z_B\cdot Z_C + \hat Z_B\cdot Z_B\cdot Z_C + Z_B\cdot Z_C\cdot\hat Z_C } {[Z]_{A+B+C}}. \]

or

\[ Z_a=\frac{\widehat{Z}_A\cdot Z_B\cdot Z_C+Z_C\cdot [Z]_B+Z_B\cdot [Z]_C}{[Z]_{A+B+C}}. \]

It remains to transform the term \(\widehat{Z}_A\cdot Z_B\cdot Z_C\), reducing it to such a form as to eliminate the product of three terms. For this purpose we add and subtract identical terms, representing them in the form

\[ \widehat{Z}_A\cdot Z_B\cdot Z_C= \left(\frac{1}{2}\cdot \widehat{Z}_A\cdot Z_B\cdot Z_C+ \frac{1}{2}\cdot \widehat{Z}_A\cdot Z_B\cdot Z_C\right)+ \]

\[ +\left(\frac{1}{2}Z_A\cdot Z_B\cdot \widehat{Z}_C- \frac{1}{2}\cdot Z_A\cdot Z_B\cdot \widehat{Z}_C\right)+ \left(\frac{1}{2}Z_A\cdot \widehat{Z}_B\cdot Z_C- \frac{1}{2}Z_A\cdot \widehat{Z}_B\cdot Z_C\right). \]

Combining the first term with the third, the second with the fifth, and the fourth with the sixth, we obtain

\[ \widehat{Z}_A\cdot Z_B\cdot Z_C = Z_B\cdot\frac{\widehat{Z}_A\cdot Z_C+Z_A\cdot\widehat{Z}_C}{2} + Z_C\cdot\frac{\widehat{Z}_A\cdot Z_B+Z_A\cdot\widehat{Z}_B}{2} - Z_A\cdot\frac{\widehat{Z}_B\cdot Z_C+Z_B\cdot\widehat{Z}_C}{2}, \]

or, according to (23),

\[ \widehat{Z}_A\cdot Z_B\cdot Z_C = Z_B\cdot [Z]_{AC}+Z_C\cdot [Z]_{AB}-Z_A\cdot [Z]_{BC}, \tag{33} \]

which gives, for the sought resistance, the final expression

\[ Z_a= \frac{ Z_B\cdot\left([Z]_C+[Z]_{AC}\right) + Z_C\cdot\left([Z]_B+[Z]_{AB}\right) - Z_A\cdot [Z]_{BC} }{ [Z]_{A+B+C} }. \tag{34} \]

By cyclic substitution of indices we obtain expressions for the resistances \(Z_b\) and \(Z_c\) of the other two branches.

Let us now clarify the question of the general case of an expression of the type

\[ Z=\frac{Z_m\cdot Z_n}{Z_k}. \]

Multiplying both the numerator and the denominator by the conjugate trinomial of the denominator, we obtain

\[ Z=\frac{\widehat{Z}_k\cdot Z_m\cdot Z_n}{Z_k\cdot \widehat{Z}_k}. \]

Since the numerator is similar to equation (33), the final expression will have the form

\[ Z=\frac{ Z_m\cdot [Z]_{nk}+Z_n\cdot [Z]_{mk}-Z_k\cdot [Z]_{mn} }{ [Z]_k }. \tag{35} \]

If in this equation we replace \(m\) by \(A\), \(n\) by \(B\), and \(k\) by \(A+B+C\), we obtain equation (34).

Equation (35) also makes it possible to solve the problem of the inverse transition from star to delta. In simple circuits, for this purpose one uses the equation

\[ Z_A=\frac{Z_a\cdot Z_b+Z_a\cdot Z_c+Z_b\cdot Z_c}{Z_a} = Z_b+Z_c+\frac{Z_b\cdot Z_c}{Z_a}. \]

According to the other variant, having two sides in each triangle, it is also possible to compose trinomials and obtain directly

\[ Z_A = Z_B + Z_C + \frac{Z_B\cdot[Z]_{ac} + Z_C\cdot[Z]_{ab} - Z_a\cdot[Z]_{bc}}{[Z]_a}. \]

or, reducing to a common denominator,

\[ Z_A = \frac{ Z_b\cdot\bigl([Z]_a + [Z]_{ac}\bigr) + Z_c\cdot\bigl([Z]_a + [Z]_{ab}\bigr) - Z_a[Z]_{bc} }{ [Z]_a }, \tag{36} \]

i.e., an equation of the same type as (34).

If this equation is expressed in admittances, interchanging uppercase and lowercase subscripts,

\[ Y_a = \frac{ Y_B\cdot\bigl([Y]_A + [Y]_{AC}\bigr) + Y_C\cdot\bigl([Y]_A + [Y]_{AB}\bigr) - Y_A[Y]_{BC} }{ [Y]_A }, \tag{37} \]

then it will serve for the transition from a delta to a star. If this replacement is made in equation (34),

\[ Y_A = \frac{ Y_b\cdot\bigl([Y]_c + [Y]_{ac}\bigr) + Y_c\cdot\bigl([Y]_b + [Y]_{ab}\bigr) - Y_a[Y]_{bc} }{ [Y]_{a+b+c} }, \tag{38} \]

then we obtain an expression serving for the transition from a star to a delta.

If, for some part of the circuit, in each phase there is a Π- or T-shaped equivalent circuit, then they may be considered as a delta and a star (Fig. 3a and 3b). In this case, in the first of them the points \(q\) and \(p\) (the neutrals for the positive- and negative-sequence components) will be of the same potential, i.e., they may be regarded as one point. It follows from this that, for transition from one circuit to the other, equations (34), (36), (37), and (38) may be used. Similar relations may also be obtained for other cases of circuit transformation.

Fig. 3.

Expressions for current, voltage, and power

In some cases it proves convenient to introduce, for current and voltage, expressions analogous to trinomials of impedances and admittances. Since in this case the zero-sequence components \(U_0\) and \(I_0\) do not participate in the calculations, the expressions take the form of binomials:

\[ U = \alpha\cdot U_1 + \beta\cdot U_2;\qquad I = \alpha\cdot I_1 + \beta\cdot I_2. \tag{39} \]

In multiplications according to equations of the Ohm’s-law type,

\[ U = Z \times I;\qquad I = Y \times U, \]

this may be performed by the ordinary algebraic method, but, for the unit vectors \(\alpha\) and \(\beta\), one must adopt different properties than was done above for operations only with impedances and admittances. Introducing the relations

\[ \alpha \times \alpha = \beta;\qquad \beta \times \beta = \alpha;\qquad \alpha \times \beta = 0, \tag{40} \]

we obtain

\[ \alpha \cdot U_{\mathrm{I}}+\beta \cdot U_{\mathrm{II}} = \left(Z_0+\alpha \cdot Z_{\mathrm{I}}+\beta \cdot Z_{\mathrm{II}}\right) \times \left(\alpha \cdot I_{\mathrm{I}}+\beta \cdot I_{\mathrm{II}}\right) = \alpha \cdot \left(I_{\mathrm{I}}\cdot Z_0+I_{\mathrm{II}}\cdot Z_{\mathrm{II}}\right)+ \]

\[ +\beta \cdot \left(I_{\mathrm{I}}\cdot Z_{\mathrm{I}}+I_{\mathrm{II}}\cdot Z_0\right); \]

\[ \alpha \cdot I_{\mathrm{I}}+\beta \cdot I_{\mathrm{II}} = \left(Y_0+\alpha \cdot Y_{\mathrm{I}}+\beta \cdot Y_{\mathrm{II}}\right) \times \left(\alpha \cdot U_{\mathrm{I}}+\beta \cdot U_{\mathrm{II}}\right) = \alpha \cdot \left(U_{\mathrm{I}}\cdot Y_0+U_{\mathrm{II}}\cdot Y_{\mathrm{II}}\right)+ \]

\[ +\beta \cdot \left(U_{\mathrm{I}}\cdot Y_{\mathrm{I}}+U_{\mathrm{II}}\cdot Y_0\right), \]

which fully corresponds to relations (3) and (4).

In order that the properties of the coefficients \(\alpha\) and \(\beta\) in these two cases of multiplication not be confused, in the last equations a cross multiplication sign has been introduced instead of the dot, although even without this it is almost impossible to make a mistake.

Dot and cross multiplication in these two cases correspond to two different kinds of multiplication in vector algebra, in the method of complex vectors, and in applications of tensor analysis. In all these cases two different types of properties are ascribed to unit vectors under different operations with them.

If the expression for the current is multiplied by the expression for the conjugate voltages

\[ \left(\alpha \cdot I_{\mathrm{I}}+\beta \cdot I_{\mathrm{II}}\right) \times \left(\alpha \cdot \widehat{U}_{\mathrm{I}}+\beta \cdot \widehat{U}_{\mathrm{II}}\right), \]

then, taking into account relations (40), we obtain

\[ \left(\alpha \cdot I_{\mathrm{I}}+\beta \cdot I_{\mathrm{II}}\right) \times \left(\alpha \cdot \widehat{U}_{\mathrm{I}}+\beta \cdot \widehat{U}_{\mathrm{II}}\right) = \beta \cdot \widehat{U}_{\mathrm{I}}\cdot I_{\mathrm{I}} + \alpha \cdot \widehat{U}_{\mathrm{II}}\cdot I_{\mathrm{II}}. \]

If in this expression the sequence coefficients, which play no role for the power, are discarded, then the power of one phase is obtained. Thus, for

\[ \widehat{U}=\alpha \cdot \widehat{U}_{\mathrm{I}}+\beta \cdot \widehat{U}_{\mathrm{II}}; \qquad I=\alpha \cdot I_{\mathrm{I}}+\beta \cdot I_{\mathrm{II}}, \]

the value of the power is found directly from

\[ P=3\cdot \left(I\times \widehat{U}\right) \tag{41} \]

with the sequence coefficients discarded in the final result.

EXAMPLE 1. A three-phase circuit is given according to the diagram of Fig. 4. The resistances of the separate phases are equal to:

1) delta—\(Z_{A1}=Z_{B1}=Z_{C1}=12-j\cdot 9\,\Omega\);

2) star—\(Z_{a2}=6+j\cdot 0\,\Omega;\quad Z_{b2}=Z_{c2}=3+j\cdot 6\,\Omega\)

and, in addition, an inductive coupling between phases \(b\) and \(c\)

\[ H_{A2}=-j\cdot 3\,\Omega; \]

3) series part (wires)—\(Z_{a3}=Z_{b3}=Z_{c3}=2+j\cdot 1\,\Omega\).

The network voltages are symmetrical, 220 volts per phase, i.e. line voltages of 380 volts. Find the distribution of currents.

Solution. For the delta the resistances of the equivalent star will be equal to

\[ Z_{a1}=Z_{b1}=Z_{c1}=\frac{Z_{A1}}{3}=4-j\cdot 3\,\Omega, \]

Fig. 4

which gives, for the components of the resistances, the values

\[ Z'_{01}=Z''_{01}=4-j\cdot 3\ \Omega;\qquad Z_{\mathrm{I}1}=Z_{\mathrm{II}1}=0 \]

or, in trinomial form,

\[ Z_1=(4-j\cdot 3)+(\alpha+\beta)\cdot 0. \]

For the star, the components of the resistances will be equal to

\[ Z'_{02}=Z''_{02}=4+j\cdot 5\ \Omega; \]

\[ Z_{\mathrm{I}2}=Z_{\mathrm{II}2}=1-j\cdot 4\ \Omega \]

or, in trinomial form,

\[ Z_2=(4+j\cdot 5)+(\alpha+\beta)\cdot(1-j\cdot 4). \]

In the series part of the circuit the components will be

\[ Z'_{03}=Z''_{03}=2+j\cdot 1\ \Omega;\qquad Z_{\mathrm{I}3}=Z_{\mathrm{II}3}=0 \]

or, in trinomial form,

\[ Z_3=(2+j\cdot 1)+(\alpha+\beta)\cdot 0. \]

The corresponding trinomials of admittances will be

\[ Y_1=\frac{\widehat{Z}_1}{[Z]_1}=(0.16+j\cdot 0.12)+(\alpha+\beta)\cdot 0; \]

\[ Y_2=\frac{\widehat{Z}_2}{[Z]_2}=(0.1128-j\cdot 0.0693)+(\alpha+\beta)\cdot(0.0795+j\cdot 0.0308); \]

\[ Y_3=\frac{\widehat{Z}_3}{[Z]_3}=(0.40-j\cdot 0.20)+(\alpha+\beta)\cdot 0. \]

Let us find the admittance of the branch

\[ Y_4=Y_1+Y_2=(0.2728+j\cdot 0.0507)+(\alpha+\beta)\cdot(0.0795+j\cdot 0.0308). \]

From equations (17) we find the values

\[ [Y]_3=0.120-j\cdot 0.160; \]

\[ [Y]_4=0.0665+j\cdot 0.0228; \]

\[ [Y]_{3+4}=0.4250+j\cdot 0.2059, \]

which gives, by equation (32), the total admittance of the circuit

\[ Y=\frac{Y_3\cdot [Y]_4+Y_4\cdot [Y]_3}{[Y]_{3+4}} =(0.1758-j\cdot 0.0130)+(\alpha+\beta)\cdot(0.0359-j\cdot 0.038), \]

that is,

\[ Y'_0=Y''_0=0.1758-j\cdot 0.0130;\qquad Y_{\mathrm{I}}=Y_{\mathrm{II}}=0.0359-j\cdot 0.038. \]

Since the voltage has only the positive-sequence component, then

\[ U=\alpha\cdot(220+j\cdot 0)+\beta\cdot 0, \]

and the currents supplied from the network will be equal to

\[ I=U\times Y=\alpha\cdot(38.68-j\cdot 2.86)+\beta\cdot(7.90-j\cdot 8.36). \]

In the same way one can find the voltage at the branching point and the currents in the star and in the delta.

The resulting binomial expression for the currents makes it possible easily to determine the currents in the phases

\[ I_a=(38.68-j\,2.86)+(7.90-j\,8.36)=46.58-j\,11.22\ \mathrm{A}; \]

\[ I_b=a^2(38.68-j\,2.86)+a\cdot(7.90-j\,8.36)=-18.54-j\,21.05\ \mathrm{A}; \]

\[ I_c=a\cdot(38.68-j\,2.86)+a^2\cdot(7.90-j\,8.36)=-28.04+j\,32.27\ \mathrm{A}. \]

The power supplied from the network is found from equation (41)

\[ P=3\cdot(I\times \widehat{U})=3\cdot(38.68-j\cdot 2.86)\cdot 220=25500-j\cdot 1900\ \mathrm{W}. \]

Matrix Form of Expressions

The trinomial forms of expressions given above for the components of resistances and conductances can be used only for static circuits, since in this case both zero-sequence components are identical and may be replaced by one. If they differ, one may use the matrix form of the expressions mentioned above. For resistances and conductances, square matrices are obtained

\[ Z=\begin{bmatrix} Z'_0; & Z_{\mathrm{II}}\\ Z_{\mathrm{I}}; & Z''_0 \end{bmatrix}; \qquad Y=\begin{bmatrix} Y'_0; & Y_{\mathrm{II}}\\ Y_{\mathrm{I}}; & Y''_0 \end{bmatrix}, \tag{42} \]

and for currents and voltages—rectangular ones

\[ I=\begin{bmatrix} I_{\mathrm{I}}\\ I_{\mathrm{II}} \end{bmatrix}; \qquad U=\begin{bmatrix} U_{\mathrm{I}}\\ U_{\mathrm{II}} \end{bmatrix}. \tag{43} \]

For this form of matrices, the relations between currents and voltages of the Ohm’s-law type should be written in the form

\[ U=Z\cdot I;\qquad I=Y\cdot U, \tag{44} \]

where these products are not commutative and the factors cannot be interchanged. Performing all operations according to the rules of matrix algebra, one can obtain relations analogous to those given above for trinomials.

As an example, let us find the resistance of two parallel branches, which may be written in the form

\[ Z=(Z_A^{-1}+Z_B^{-1})^{-1}=(Y_A+Y_B)^{-1}. \]

Since

\[ Y_A=Z_A^{-1}= \frac{ \begin{bmatrix} Z''_{0A}; & -Z_{\mathrm{II}A}\\ -Z_{\mathrm{I}A}; & Z'_{0A} \end{bmatrix} }{D_A}; \qquad Y_B=Z_B^{-1}= \frac{ \begin{bmatrix} Z''_{0B}; & -Z_{\mathrm{II}B}\\ -Z_{\mathrm{I}B}; & Z'_{0B} \end{bmatrix} }{D_B}; \]

where \(D_A\) and \(D_B\) are determinants composed from the matrices

\[ D_A= \begin{vmatrix} Z'_{0A}; & Z_{\mathrm{II}A}\\ Z_{\mathrm{I}A}; & Z''_{0A} \end{vmatrix}; \qquad D_B= \begin{vmatrix} Z'_{0B}; & Z_{\mathrm{II}B}\\ Z_{\mathrm{I}B}; & Z''_{0B} \end{vmatrix}, \]

then the total resistance will be equal to

\[ Z=(Y_A+Y_B)^{-1} = \frac{ \begin{bmatrix} (Y''_{0A}+Y''_{0B}); & -(Y_{\mathrm{II}A}+Y_{\mathrm{II}B})\\ -(Y_{\mathrm{I}A}+Y_{\mathrm{I}B}); & (Y'_{0A}+Y'_{0B}) \end{bmatrix} }{ \begin{vmatrix} Y'_{0A}+Y'_{0B}; & Y_{\mathrm{II}A}+Y_{\mathrm{II}B}\\ Y_{\mathrm{I}A}+Y_{\mathrm{I}B}; & Y''_{0A}+Y''_{0B} \end{vmatrix} }. \]

Substituting the values for the conductances, we have

\[ Z= \frac{ \begin{bmatrix} \dfrac{Z'_{0A}}{D_A}+\dfrac{Z'_{0B}}{D_B}; & \dfrac{Z_{\mathrm{II}A}}{D_A}+\dfrac{Z_{\mathrm{II}B}}{D_B}\\[1.2ex] \dfrac{Z_{\mathrm{I}A}}{D_A}+\dfrac{Z_{\mathrm{I}A}}{D_A}; & \dfrac{Z''_{0A}}{D_A}+\dfrac{Z''_{0B}}{D_B} \end{bmatrix} }{ \begin{vmatrix} \dfrac{Z'_{0A}}{D_A}+\dfrac{Z'_{0B}}{D_B}; & \dfrac{Z_{\mathrm{II}A}}{D_A}+\dfrac{Z_{\mathrm{II}B}}{D_B}\\[1.2ex] \dfrac{Z_{\mathrm{I}A}}{D_A}+\dfrac{Z_{\mathrm{I}B}}{D_B}; & \dfrac{Z''_{0A}}{D_A}+\dfrac{Z''_{0B}}{D_A} \end{vmatrix} } = \frac{ D_A^{-1}\!\cdot \begin{bmatrix} Z'_{0A}; & Z_{\mathrm{II}A}\\ Z_{\mathrm{I}A}; & Z''_{0A} \end{bmatrix} + D_B^{-1}\!\cdot \begin{bmatrix} Z'_{0B}; & Z_{\mathrm{II}B}\\ Z_{\mathrm{I}B}; & Z''_{0B} \end{bmatrix} }{ D_A^{-1}\cdot D_B^{-1}\cdot \begin{vmatrix} Z'_{0A}+Z'_{0B}; & Z_{\mathrm{II}A}+Z_{\mathrm{II}B}\\ Z_{\mathrm{I}A}+Z_{\mathrm{I}B}; & Z''_{0A}+Z''_{0B} \end{vmatrix} }. \]

or

\[ Z=\frac{\mathrm{D}_{\mathrm{B}}\cdot Z_{\mathrm{A}}+\mathrm{D}_{\mathrm{A}}\cdot Z_{\mathrm{B}}}{\mathrm{D}_{\mathrm{A+B}}}, \tag{45} \]

which coincides completely with expression (31) for trinomials. Similar correspondences will also occur in other transformations.

If, from expression (45), one determines the value of the zero-sequence components

\[ Z'_0=\frac{\mathrm{D}_{\mathrm{B}}\cdot Z'_{0\mathrm{A}}+\mathrm{D}_{\mathrm{A}}\cdot Z'_{0\mathrm{B}}}{\mathrm{D}_{\mathrm{A+B}}}; \quad Z''_0=\frac{\mathrm{D}_{\mathrm{B}}\cdot Z''_{0\mathrm{A}}+\mathrm{D}_{\mathrm{A}}\cdot Z''_{0\mathrm{B}}}{\mathrm{D}_{\mathrm{A+B}}}, \]

then it is easy to see that they can also be obtained from the trinomial expression, finding the component

\[ Z_0=\frac{[Z]_{\mathrm{B}}\cdot Z_{0\mathrm{A}}+[Z]_{\mathrm{A}}\cdot Z_{0\mathrm{B}}}{[Z]_{\mathrm{A+B}}} \]

and taking, in one case, the values \(Z'_{0\mathrm{A}}\) and \(Z'_{0\mathrm{B}}\), and in the other case—\(Z''_{0\mathrm{A}}\) and \(Z''_{0\mathrm{B}}\), while assuming

\[ [Z]=\mathrm{D}=Z'_0\cdot Z''_0-Z_{\mathrm{I}}\cdot Z_{\mathrm{II}}. \tag{46} \]

Since all transformations for trinomials are simpler than for matrices, in the case of nonstatic circuits it is also more convenient to carry out all transformations and derivations in the form of trinomials, taking into account relations of type (46) and substituting various values of \(Z'_0\) and \(Z''_0\) only when finding the corresponding components. In this case expressions of the type \([Z]_{\mathrm{AB}}\) take the form

\[ [Z]_{\mathrm{AB}} = \frac{Z'_{0\mathrm{A}}\cdot Z'_{0\mathrm{B}}-Z''_{0\mathrm{A}}\cdot Z''_{0\mathrm{B}}}{2} - \frac{Z_{\mathrm{IA}}\cdot Z_{\mathrm{IIB}}+Z_{\mathrm{IIA}}\cdot Z_{\mathrm{IB}}}{2}, \tag{47} \]

while everything else remains unchanged.

The matrix form of the expressions can also be extended to circuits with zero components of currents, i.e., to cases where all three components of impedances and admittances have to be taken into account.

Multi-terminal networks

Four-terminal networks are widely known; they play a very important role in the investigation of single-phase circuits. These four-terminal networks possess a property that makes it possible to decompose any current and voltage into two parts, corresponding to the no-load condition and to the short-circuit condition of the given circuit, which permits simplification of the solution of various questions associated with this circuit.

In three-phase circuits, four-terminal networks are usually used only for one phase of a symmetrical circuit, although their properties can also be extended to any unsymmetrical polyphase circuit. Of course, the number of terminals then increases, and for a three-phase circuit without a neutral conductor reaches six, and with a neutral conductor seven or eight. In the absence of a neutral conductor, for using the properties of the resulting six-terminal networks, the application of trinomials is very convenient.

Let us suppose that there is a section of a three-phase circuit in which three terminals are input terminals, and three others are output terminals. The former are connected through some circuit to the generator, while to the latter a load or the subsequent part of the circuit is connected. We shall denote the voltage and current of the load end by \(U_2\) and \(I_2\), remembering that they are binomial expressions of type (39). Correspondingly, for the generator terminals we introduce the notation \(U_1\) and \(I_1\).

Let the given section of the circuit be reduced to the T-shaped form of the phase-substitution circuit shown in Fig. 5.

The series impedances, respectively, are

\[ \begin{aligned} Z_1 &= Z_{01}+\alpha \cdot Z_{\mathrm{I}1}+\beta \cdot Z_{\mathrm{II}1};\\ Z_2 &= Z_{02}+\alpha \cdot Z_{\mathrm{I}2}+\beta \cdot Z_{\mathrm{II}2}, \end{aligned} \tag{48} \]

and the conductance to neutral is

\[ Y_n=Y_{0n}+\alpha \cdot Y_{\mathrm{I}n}+\beta \cdot Y_{\mathrm{II}n}. \tag{49} \]

Proceeding as for ordinary complexes in the consideration of four-poles, we find for the six-pole the voltage at point \(m\)

\[ U_m=U_2+I_2 \cdot Z_2, \]

the current in the branch

\[ I_n=U_m \cdot Y_n=U_2 \cdot Y_n+I_2 \cdot Z_2 \cdot Y_n, \]

the total current entering from the network

\[ I_1=I_2+I_n=U_2 \cdot Y_n+I_2 \cdot (Z_2 \cdot Y_n+1), \]

and the network voltage

\[ U_1=U_m+I_1 \cdot Z_1=U_2 \cdot (Z_1 \cdot Y_n+1)+I_2 \cdot (Z_1 \cdot Z_2 \cdot Y_n+Z_1+Z_2). \]

Fig. 5

The last two expressions show that in a six-pole, as in a four-pole, between the currents and voltages there are relations in the form of equations

\[ \begin{aligned} U_1&=A \cdot U_2+B \cdot I_2;\\ I_1&=C \cdot U_2+D \cdot I_2. \end{aligned} \tag{50} \]

For the T-shaped substitution circuit under consideration, the coefficients \(A, B, C\), and \(D\) (the constants of the six-pole) are equal to

\[ \begin{aligned} A&=(Z_1 \cdot Y_n+1); \qquad B=(Z_1 \cdot Z_2 \cdot Y_n+Z_1+Z_2);\\ C&=Y_n; \qquad\qquad\quad D=(Z_2 \cdot Y_n+1). \end{aligned} \tag{51} \]

For other circuits these coefficients will be correspondingly different [for example, expressions (86)].

Equations (50) are the fundamental relations for the six-pole, and the quantities entering into them may be expressed not only in trinomial form, but also in the form of matrices. In the latter case they can also be extended to circuits with a neutral conductor, i.e., to seven- and eight-poles.

In some cases equations (50) are conveniently combined into one matrix expression of the form

\[ \begin{bmatrix} U_1\\ I_1 \end{bmatrix} = \begin{bmatrix} A;B\\ C;D \end{bmatrix} \cdot \begin{bmatrix} U_2\\ I_2 \end{bmatrix}, \tag{52} \]

regardless of whether their terms are expressed in the form of trinomials or also in the form of matrices.

Since equations (50) and (52) do not differ in any way from the corresponding equations of a four-pole, all properties of the latter can be transferred in full to the coefficients of the six-pole. Thus, the relation must necessarily hold

\[ A \cdot D-B \cdot C=1 \qquad \text{or} \qquad \begin{vmatrix} A;B\\ C;D \end{vmatrix} =1. \tag{53} \]

If the six-terminal network is symmetrical with respect to the input and output terminals, then the coefficients \(A\) and \(D\) must be identical. This is clearly seen from expressions (51), and also from the inverse relation

\[ \begin{bmatrix} U_2\\ I_2 \end{bmatrix} = \begin{bmatrix} A;B\\ C;D \end{bmatrix}^{-1} \cdot \begin{bmatrix} U_1\\ I_1 \end{bmatrix} = \frac{ \begin{bmatrix} D;-B\\ -C;A \end{bmatrix} }{ \begin{vmatrix} A;B\\ C;D \end{vmatrix} } \cdot \begin{bmatrix} U_1\\ I_1 \end{bmatrix} = \begin{bmatrix} D;-B\\ -C;A \end{bmatrix} \cdot \begin{bmatrix} U_1\\ I_1 \end{bmatrix}, \]

which, for the reverse connection of the six-terminal network (i.e., for the interchange of \(U_1\) and \(U_2\)), gives

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

In order that, under such a reverse connection, the former relations be preserved, the condition \(A=D\) must be satisfied.

Equations (50) show that, for the no-load condition \((I_2=0)\), we obtain the relations

\[ U_{1\mathrm{xx}}=A\cdot U_2;\quad I_{1\mathrm{xx}}=C\cdot U_2, \tag{54} \]

and for the short-circuit condition \((U_2=0)\),

\[ U_{1\mathrm{kz}}=B\cdot I_2;\quad I_{1\mathrm{kz}}=D\cdot I_2 \tag{55} \]

or, under normal operation with load,

\[ U_1=U_{1\mathrm{xx}}+U_{1\mathrm{kz}};\quad I_1=I_{1\mathrm{xx}}+I_{1\mathrm{kz}}. \tag{56} \]

These relations make it possible to find the current distribution under load as the result of superposing the short-circuit condition on the no-load condition.

Fig. 6

Let us determine the change in the coefficients \(A\), \(B\), \(C\), and \(D\) when the phases of an unsymmetrical circuit are switched. Suppose that the values are given for some one connection of the terminals of the six-terminal network, and then replace them in cyclic order, as shown in Figs. 6a and 6b. For the first connection, the impedance trinomial will be

\[ Z_{(a)}= \frac{Z_m+Z_n+Z_k}{3} +\alpha\cdot \frac{Z_m+a\cdot Z_n+a^2\cdot Z_k}{3} + \beta\cdot \frac{Z_m+a^2\cdot Z_n+a\cdot Z_k}{3}, \]

and for the second connection

\[ Z_{(b)}= \frac{Z_m+Z_n+Z_k}{3} +\alpha\cdot \frac{a\cdot Z_m+a^2\cdot Z_n+Z_k}{3} + \beta \frac{a^2\cdot Z_m+a\cdot Z_n+Z_k}{3}. \]

Thus, if for the initial connection we had

\[ Z_{(a)}=Z_0+\alpha\cdot Z_I+\beta\cdot Z_{II}, \]

then after switching by one phase in the positive sequence order we shall have

\[ Z_{(b)}=Z_0+\alpha\cdot a\cdot Z_I+\beta\cdot a^2\cdot Z_{II}, \tag{57} \]

and, when switched by one more phase,

\[ Z_{(c)}=Z_0+\alpha\cdot a^2\cdot Z_{\mathrm{I}}+\beta\cdot a\cdot Z_{\mathrm{II}}. \]

The components for the admittances will change correspondingly. If we now turn to expressions (51), it is easy to see that the same regularity will also hold for the coefficients of the six-terminal network. If, for example, for some connection there was the value

\[ B_{(a)}=B_0+\alpha\cdot B_{\mathrm{I}}+\beta\cdot B_{\mathrm{II}}, \]

then, when the terminals are switched by one phase in the positive order, we shall have

\[ B_{(b)}=B_0+\alpha\cdot a\cdot B_{\mathrm{I}}+\beta\cdot a^2\cdot B_{\mathrm{II}}, \tag{58} \]

and, when switched by two phases,

\[ B_{(c)}=B_0+\alpha\cdot a^2\cdot B_{\mathrm{I}}+\beta\cdot a\cdot B_{\mathrm{II}}. \]

If in the given part of the circuit there are only series-connected impedances, i.e. there are no admittances between phases, then in the equations of the six-terminal network one should put \(Y=0\), which gives for the coefficients the values

\[ A=D=1;\quad B=Z_1+Z_2=Z;\quad C=0, \tag{59} \]

where

\[ U_1=U_2+B\cdot I_2;\quad I_1=I_2. \]

Conversely, if the six-terminal network consists only of admittances between phases, then, putting \(Z=0\), we obtain

\[ A=D=1;\quad B=0;\quad C=Y, \tag{60} \]

where

\[ U_1=U_2;\quad I_1=C\cdot U_2+I_2. \]

Fig. 7

Series connection of six-terminal networks

With the series connection of two six-terminal networks (Fig. 7), for each of them we have the relations

\[ \begin{bmatrix} U_m\\ I_m \end{bmatrix} = \begin{bmatrix} A_2;B_2\\ C_2;D_2 \end{bmatrix} \cdot \begin{bmatrix} U_2\\ I_2 \end{bmatrix} ;\quad \begin{bmatrix} U_1\\ I_1 \end{bmatrix} = \begin{bmatrix} A_1;B_1\\ C_1;D_1 \end{bmatrix} \cdot \begin{bmatrix} U_m\\ I_m \end{bmatrix}. \]

Substituting the first of these into the second, we obtain

\[ \begin{bmatrix} U_1\\ I_1 \end{bmatrix} = \begin{bmatrix} A_1;B_1\\ C_1;D_1 \end{bmatrix} \cdot \begin{bmatrix} A_2;B_2\\ C_2;D_2 \end{bmatrix} \cdot \begin{bmatrix} U_2\\ I_2 \end{bmatrix} = \begin{bmatrix} A;B\\ C;D \end{bmatrix} \cdot \begin{bmatrix} U_2\\ I_2 \end{bmatrix}, \]

whence

\[ \begin{bmatrix} A;B\\ C;D \end{bmatrix} = \begin{bmatrix} A_1;B_1\\ C_1;D_1 \end{bmatrix} \cdot \begin{bmatrix} A_2;B_2\\ C_2;D_2 \end{bmatrix}. \tag{61} \]

Carrying out the multiplication of the matrices, we find

\[ \begin{aligned} A&=A_1\cdot A_2+B_1\cdot C_2; &\quad B&=A_1\cdot B_2+B_1\cdot D_2;\\ C&=A_2\cdot C_1+D_1\cdot C_2; &\quad D&=C_1\cdot B_2+D_1\cdot D_2. \end{aligned} \tag{62} \]

If each of the six-terminal networks was symmetrical with respect to its terminals (i.e. \(A_1=D_1\) and \(A_2=D_2\)), but the six-terminal networks themselves were not identical, then, because of the inequality \(B_1\cdot C_2\ne B_2\cdot C_1\), the resulting common six-terminal network will no longer have symmetry (since in this case \(A\ne D\)).

With a larger number of six-terminal networks connected in series, we accordingly have

\[ \left[\begin{array}{c} A;B\\ C;D \end{array}\right] = \left[\begin{array}{c} A_1;B_1\\ C_1;D_1 \end{array}\right] \cdot \left[\begin{array}{c} A_2;B_2\\ C_2;D_2 \end{array}\right] \cdots \left[\begin{array}{c} A_n;B_n\\ C_n;D_n \end{array}\right], \tag{63} \]

moreover, if all the six-terminal networks are identical, we obtain

\[ \left[\begin{array}{c} A;B\\ C;D \end{array}\right] = \left[\begin{array}{c} A_1;B_1\\ C_1;D_1 \end{array}\right]^n \tag{64} \]

where \(n\) is their number. In this case it is convenient to pass to hyperbolic functions, as is done below in considering long lines.

In constructing transmission lines, in order to obtain phase symmetry, transposition of the conductors is used, calculated so that one third of the total length of unbalance is in one phase, the second third in another, and the third third in the last phase. Taking each section as a six-terminal network, in the simplest case we have three sections

\[ \left[\begin{array}{c} A_1;B_1\\ C_1;D_1 \end{array}\right]; \left[\begin{array}{c} A_2;B_2\\ C_2;D_2 \end{array}\right]; \left[\begin{array}{c} A_3;B_3\\ C_3;D_3 \end{array}\right], \]

and, owing to the interchange of phases, relations of the type (58) will hold between their coefficients. If the product of these three matrices is found, i.e., if the values of the coefficients for the common six-terminal network of the whole line

\[ \left[\begin{array}{c} A;B\\ C;D \end{array}\right], \]

are determined, then one can verify that these coefficients will nevertheless have terms with the factors \(\alpha\) and \(\beta\), i.e., the phases of the common six-terminal network will not have complete symmetry. To obtain complete symmetry, the middle six-terminal network must differ somewhat from the end ones; however, the remaining unbalance is usually very small and may be freely neglected.

In view of the complexity of the derivation in general form, it is not given here; it is easy to verify this in any particular numerical example.

In the absence of conductances between phases, the coefficients \(A\) and \(D\) become units (if these are not transformers), and \(C\) becomes zero [eq. (59)], and relation (63) gives

\[ \left[\begin{array}{c} A;B\\ C;D \end{array}\right] = \left[\begin{array}{cc} 1;B_1+B_2+B_3\ldots+B_n\\ 0&1 \end{array}\right], \]

that is,

\[ B = B_1+B_2+B_3+\ldots+B_n = \sum B_n \]

or

\[ Z = Z_1+Z_2+Z_3+\ldots+Z_n = \sum Z_n, \]

as in an ordinary series connection.

Parallel connection of six-terminal networks

In some cases it may be necessary to replace two parallel six-terminal networks by one, i.e., to find the common equivalent six-terminal network (for example, in the parallel operation of transformers). For such six-terminal networks the voltages \(U_1\) and \(U_2\) will be common, while the currents will be summed:

\[ I_1 = I_{1m}+I_{1n}; \qquad I_2 = I_{2m}+I_{2n}. \]

According to equations (50), for each of them

\[ \begin{aligned} U_1 &= A_m \cdot U_2 + B_m \cdot I_{2m}; & U_1 &= A_n \cdot U_2 + B_n \cdot I_{2n}. \\ I_{1m} &= C_m \cdot U_2 + D_m \cdot I_{2m}; & I_1 &= C_n \cdot U_2 + D_n \cdot I_{2n}. \end{aligned} \tag{65} \]

Eliminating from each pair of equations the currents \(I_{2m}\) and \(I_{2n}\), we find

\[ I_{1m}=\frac{D_m \cdot U_1-U_2}{B_m}; \qquad I_{1n}=\frac{D_n \cdot U_1-U_2}{B_n}. \tag{66} \]

The total current, equal to their sum, will be

\[ I_1=I_{1m}+I_{1n} =\frac{D_m \cdot U_1-U_2}{B_m} +\frac{D_n \cdot U_1-U_2}{B_n}, \]

or, after reduction to a common denominator,

\[ I_1= \frac{ \displaystyle \frac{D_m \cdot B_n + D_n \cdot B_m}{B_m+B_n}\,U_1-U_2 }{ \displaystyle \frac{B_m \cdot B_n}{B_m+B_n} }. \]

Comparing with expression (66), we see that for the common six-terminal network the relations must hold

\[ B=\frac{B_m \cdot B_n}{B_m+B_n}; \qquad D=\frac{D_m \cdot B_n+D_n \cdot B_m}{B_m+B_n}. \tag{67} \]

Determining from the first equations (65) the values of \(I_{2m}\) and \(I_{2n}\),

\[ I_{2m}=\frac{U_1-A_m \cdot U_2}{B_m}; \qquad I_{2n}=\frac{U_1-A_n \cdot U_2}{B_n} \tag{68} \]

and summing them, we find

\[ I_2=I_{2m}+I_{2n} = \frac{ \displaystyle U_1-\frac{A_m \cdot B_n + A_n \cdot B_m}{B_m+B_n}\cdot U_2 }{ \displaystyle \frac{B_m \cdot B_n}{B_m+B_n} }, \]

whence we obtain for the common six-terminal network

\[ A=\frac{A_m \cdot B_n + A_n \cdot B_m}{B_m+B_n}. \tag{69} \]

Finally, for the coefficient \(C\) the value may be found from relation (53)

\[ C=\frac{A \cdot D-1}{B}=\frac{A \cdot D}{B}-\frac{1}{B}, \]

which, after substitution from (67) and (69), gives

\[ C= \frac{C_m \cdot B_n+C_n \cdot B_m}{B_m+B_n} + \frac{A_m \cdot D_n+A_n \cdot D_m}{B_m+B_n} - \frac{2}{B_m+B_n}. \tag{70} \]

All these expressions can be easily calculated with the aid of trinomials. Thus, for example, equation (67) gives for \(B\) the expression

\[ B=\frac{B_m \cdot B_n}{B_m+B_n} = \frac{[B]_m \cdot B_n + [B]_n \cdot B_m}{[B]_{m+n}}. \tag{71} \]

In the absence of admittances between phases, the coefficients \(A\) and \(D\) become unity (if these are not transformers), and \(C\) becomes zero, and equation (71) fully coincides with equation (31) for two parallel sections of a three-phase circuit.

Transformers

Every transformer for a three-phase circuit without neutral wires represents a six-terminal network; moreover, in most cases the latter will be symmetrical (for the referred windings) and with symmetrical phases.

Since for such a transformer there will be components of the impedances and admittances only of the zero sequence, \(Z'_{\mathrm{ot}} = Z''_{\mathrm{ot}}\) and \(Y'_{\mathrm{o}} = Y''_{\mathrm{o}}\), the equivalent circuit corresponding to Figure 5 will have

\[ Z_1 = Z_2 = \frac{1}{2} Z'_{\mathrm{ot}};\quad Y_{\mathrm{n}} = Y'_{\mathrm{o}}. \]

According to equations (51), the coefficients of such a six-terminal network are equal to

\[ A = D = \frac{1}{2}\, Z'_{\mathrm{ot}} \cdot Y'_{\mathrm{o}} + 1; \]

\[ B = \frac{1}{4}\,(Z'_{\mathrm{ot}})^2 \cdot Y'_{\mathrm{o}} + Z'_{\mathrm{ot}};\quad C = Y'_{\mathrm{o}}. \tag{72} \]

If, for simplification, instead of the T-shaped circuit one takes the \(\Gamma\)-shaped circuit, then, setting \(Z' = 0\) and \(Z_2 = Z'_{\mathrm{ot}}\), we find

\[ A = 1;\quad B = Z'_{\mathrm{ot}};\quad C = Y'_{\mathrm{o}};\quad D = Z'_{\mathrm{ot}} \cdot Y'_{\mathrm{o}} + 1 \simeq 1. \tag{73} \]

These values give simpler relations, while the error obtained in this case is very small and may freely be neglected. In most cases the magnetizing currents may in general be neglected, i.e., \(Y'_{\mathrm{o}}\) may be taken equal to zero, which gives the values \(A\) and \(D\) equal to unity, and \(C\) equal to zero.

In doing this, one must always take into account the possible rotation of the secondary vectors with respect to the primary ones, occurring in connection with the type of winding connection and the marking of the terminals. For connections of the first group (star—star or delta—delta) there will either be no rotation at all, or it will be \(180^\circ\). In the latter case all coefficients of the six-terminal network should be taken with opposite signs, i.e., multiplied by \(-1\).

For the second group of connections (star—delta or star—zigzag), with the marking of terminals by the opposing phase, a rotation of the vectors by \(90^\circ\) in one direction or the other is obtained, which makes it necessary to multiply all coefficients of the six-terminal network by \(+j\) or \(-j\). If the terminal marking is also changed in this case, then rotation factors of \(120^\circ\), i.e. the values \(a\) or \(a^2\), have to be introduced additionally.

For this group of transformers, all impedances of the secondary circuit (load) should be taken with the unbalance changed to the opposite one, i.e., instead of impedance or admittance trinomials, one should take the conjugate trinomials.

If the transformer has asymmetrical phases, which may occur in an open-delta circuit, nonuniformity of the magnetic circuits, incorrect connection of windings, etc., then the impedances \(Z_1\) and \(Z_2\) and the admittance \(Y_{\mathrm{n}}\) will have the form of trinomials. Thus, in an open-...

of a triangle with a common point in phase “a,” these trinomials for the T-shaped equivalent circuit will be

\[ Z' = Z_2 = \frac{1}{3}\cdot Z_{\mathrm{kz}} - (\alpha+\beta)\cdot \frac{1}{6} Z_{\mathrm{kz}};\quad Y_{\mathrm{p}} = 2Y_{\mathrm{xx}} + (\alpha+\beta)\cdot \frac{1}{2}\cdot Y_{\mathrm{xx}}, \tag{74} \]

where \(Z_{\mathrm{kz}}\) is the impedance under a short circuit of one phase of the transformer, and \(Y_{\mathrm{xx}}\) is the conductance of one phase at no load.

The coefficients of the six-pole network can also be found without reducing the transformer windings to one voltage, but with direct allowance for the transformation ratio. In this case the secondary voltages must be increased \(n=\dfrac{w_2}{w_1}\) times, and the secondary currents reduced by the same factor relative to the reduced values. Consequently, the new coefficients will have the values

\[ A' = A\cdot \frac{1}{n};\quad B' = B\cdot n;\quad C' = C\cdot \frac{1}{n};\quad D' = D\cdot n \tag{75} \]

with allowance for the connection of the windings, as indicated above.

When two transformers with different transformation ratios are connected in parallel, equalizing currents will flow between them; in the case of phase unbalance these currents will also be unbalanced. The determination of these currents is shown in the following example.

EXAMPLE 2. Two transformers are connected for parallel operation. One of them has a symmetrical star—star connection with a transformation ratio of 10 and a phase short-circuit impedance of 5 ohms, while the second is connected according to the open-delta scheme with a common vertex in phase “a,” a transformation ratio of 8, and a short-circuit impedance of 18 ohms. The transformers are step-up transformers; the primary line voltage is 220 volts. Find the currents in the windings and the secondary voltage at no load, and determine the coefficients of the six-pole network of the common connection. Neglect the magnetizing currents and the active resistances of the windings.

Solution. In accordance with relations (73) and (75), for the first transformer we obtain

\[ A'_{\mathrm{m}}=0.1;\quad B'_{\mathrm{m}}=j\cdot 50;\quad C'_{\mathrm{m}}=0;\quad D'_{\mathrm{m}}=10, \]

and for the second, by the same relations and by (74),

\[ A'_{\mathrm{n}}=0.125;\quad B'_{\mathrm{n}}=j\cdot 48-(\alpha+\beta)j\cdot 3;\quad C'_{\mathrm{n}}=0;\quad D'_{\mathrm{n}}=8. \]

Equations (67) and (69) give three coefficients of the common six-pole network

\[ A' = 0.113 + (\alpha+\beta)\cdot 0.0004;\quad B' = j\cdot 24.42(\alpha+\beta)\cdot j\cdot 0.78; \]

\[ D' = 8.97 - (\alpha+\beta)\cdot 0.031. \]

We determine the fourth coefficient either from (70) or from relation (53). Taking the latter, we have

\[ C'=\frac{A'\cdot D'-1}{B'} = -j\cdot 0.000511-(\alpha+\beta)\cdot j\cdot 0.0000157, \]

i.e., a value different from zero, despite the fact that \(C'_{\mathrm{m}}\) and \(C'_{\mathrm{n}}\) are equal to zero. From equations (54) we find, for no load, the secondary voltage

\[ U_2=\frac{U_1}{A'}=\alpha\cdot 1127-\beta\cdot 3.9 \]

and the total no-load current

\[ I_{1\mathrm{xx}}=C\cdot U_2=-\alpha\cdot j\cdot 0.575-\beta\cdot j\cdot 0.0157. \]

Thus, at no load the secondary phase voltage will have a positive-sequence component of 1127 volts, which corresponds to the overall transformation ratio

\[ n=\frac{1127}{127}=8.85. \]

The currents separately for each of the transformers can be found from equations (50). For the first of them we have

\[ \mathbf I_{2m}=\frac{\mathbf U_1-\mathbf A'_m\cdot \mathbf U_2}{\mathbf B'_m} =-\alpha\cdot j\cdot 0.286-\beta\cdot j\cdot 0.0078. \]

Since there is no load,

\[ \mathbf I_{2n}=-\mathbf I_{2m} =\alpha\cdot j\cdot 0.286+\beta\cdot j\cdot 0.0078. \]

For the primary circuit, from equations (50), we obtain

\[ \mathbf I_{1m}=-\alpha\cdot j\cdot 2.860-\beta\cdot j\cdot 0.0780 \]

\[ \mathbf I_{1n}=\alpha\cdot j\cdot 2.296+\beta\cdot j\cdot 0.0624, \]

which corresponds to the power values

\[ P_m=-j\cdot 630\,V A;\qquad P_n=j\cdot 505\,V A. \]

Thus, the first transformer takes reactive power from the network, while the second returns part of it back to the network.

Circuits with Uniformly Distributed Parameters

Let us determine the values of the hexapole constants for a long line with asymmetrical phases. Typical representatives of such a line are a two-wires–earth system or a system with a grounded phase.

Having determined the three-term expression of the resistance per unit length of the line from equations (10), and the conductance between phases, also in the form of a three-term expression, we may write expressions for the change of voltage and current along the line, taking distances from the load end:

\[ d\mathbf U=\mathbf I\cdot \mathbf Z\cdot dl;\qquad d\mathbf I=\mathbf U\cdot \mathbf Y dl. \]

From these relations we find

\[ \frac{d\mathbf U}{dl}=\mathbf I\cdot \mathbf Z;\quad \frac{d\mathbf I}{dl}=\mathbf U\cdot \mathbf Y, \tag{76} \]

or, differentiating and eliminating the current or voltage,

\[ \frac{d^2\mathbf U}{dl^2}=\mathbf U\cdot \mathbf Z\cdot \mathbf Y;\quad \frac{d^2\mathbf I}{dl^2}=\mathbf I\cdot \mathbf Z\cdot \mathbf Y. \tag{77} \]

Since \(\mathbf Z\) and \(\mathbf Y\) refer to different elements of the circuit, their product is, of course, not equal to unity.

It is known that the solution of equations of this type will have the form

\[ \mathbf U=\mathbf M\cdot \operatorname{Ch}\gamma l+\mathbf P\cdot \operatorname{Sh}\gamma l. \]

Differentiating twice, we obtain

\[ \frac{d^2\mathbf U}{dl^2}=\mathbf M\cdot \gamma^2\cdot \operatorname{Ch}\gamma l+\mathbf P\cdot \gamma^2\cdot \operatorname{Sh}\gamma l. \]

Substituting these expressions into equation (77), we obtain

\[ (M \cdot \operatorname{Ch}\gamma l + P \cdot \operatorname{Sh}\gamma l)\cdot \gamma^2 = (M \cdot \operatorname{Ch}\gamma l + P \cdot \operatorname{Sh}\gamma l)\cdot Z\cdot Y \]

or

\[ (\gamma^2 - Z\cdot Y)=0. \tag{78} \]

Since the value of \(\gamma\) must not include the sequence coefficients, whereas the product \(Z\cdot Y\) does include them, the determination of the values can be carried out by introducing the identity matrix

\[ E=\begin{bmatrix} 1;0\\ 0;1 \end{bmatrix}. \]

If the values \(Z\) and \(Y\) are also taken in matrix form, then the determinant of the matrix

\[ \left|\gamma^2\cdot E - Z\cdot Y\right|=0 \]

will be the characteristic equation. Substituting the values, we have

\[ \left| \begin{matrix} (\gamma^2 - Z_0\cdot Y_0 - Z_{\mathrm{II}}\cdot Y_{\mathrm{I}});(-Z_0Y_{\mathrm{II}}+Z_{\mathrm{II}}\cdot Y_0)\\ (-Z_{\mathrm{I}}\cdot Y_0 - Z_0\cdot Y_{\mathrm{I}});(\gamma^2 - Z_0\cdot Y_0 + Z_{\mathrm{I}}\cdot Y_{\mathrm{II}}) \end{matrix} \right|=0, \]

whence

\[ \gamma^4 - 2\cdot \gamma^2\cdot [ZY] + [Z]\cdot [Y]=0. \]

In this expression \([Z]\) and \([Y]\) are the ordinary products of conjugate trinomials, while \([ZY]\) represents a similar expression,

\[ [ZY]=Z_0\cdot Y_0+\frac{1}{2}\cdot (Z_{\mathrm{I}}\cdot Y_{\mathrm{II}}+Z_{\mathrm{II}}\cdot Y_{\mathrm{I}}). \tag{79} \]

Solving the resulting biquadratic equation with respect to \(\gamma^2\), we find

\[ \gamma^2=[ZY]\pm \sqrt{[ZY]^2-[Z]\cdot [Y]}. \tag{80} \]

Thus, the equation will have four roots, differing in pairs only by their signs,

\[ \gamma_1=\pm \sqrt{[ZY]+\sqrt{[ZY]^2-[Z]\cdot [Y]}}, \]

\[ \gamma_2=\pm \sqrt{[ZY]-\sqrt{[ZY]^2-[Z]\cdot [Y]}}. \tag{81} \]

The complete solution of differential equation (77) for the voltage takes the form

\[ U_1=M\cdot \operatorname{Ch}\gamma_1 l+N\cdot \operatorname{Ch}\gamma_2 l+P\cdot \operatorname{Sh}\gamma_1 l+Q\cdot \operatorname{Sh}\gamma_2 l. \tag{82} \]

For the current we obtain a similar equation, using relation (76),

\[ I=M\cdot \frac{\gamma_1}{Z}\cdot \operatorname{Sh}\gamma_1 l +N\cdot \frac{\gamma_2}{Z}\operatorname{Sh}\gamma_2 l +P\cdot \frac{\gamma_1}{Z}\operatorname{Ch}\gamma_1 l +Q\cdot \frac{\gamma_2}{Z}\cdot \operatorname{Ch}\gamma_2 l. \tag{83} \]

To determine the constants we have the boundary condition for the voltage and current of the load. Setting \(l=0\), we obtain

\[ U_2=M+N;\qquad I_2=\frac{1}{Z}(P\cdot \gamma_1+Q\cdot \gamma_2). \tag{84} \]

Thus, only the sums of the constants in pairs are determined, and not each of them separately.

Taking from equations (82) and (83) the second derivatives and substituting them into equation (77), we then obtain

\[ U=-\frac{1}{Z\cdot Y}\cdot M\cdot \gamma_1^{2}\operatorname{Ch}\gamma_1 l +N\cdot \gamma_2^{2}\operatorname{Ch}\gamma_2 l +P\cdot \gamma_1^{2}\operatorname{Sh}\gamma_1 l +Q\cdot \gamma_2^{2}\operatorname{Sh}\gamma_2 l); \]

\[ I=-\frac{1}{Z\cdot Y}\cdot \left( M\cdot \frac{\gamma_1^{3}}{Z}\operatorname{Sh}\gamma_1 l +N\cdot \frac{\gamma_2^{3}}{Z}\operatorname{Sh}\gamma_2 l +P\cdot \frac{\gamma_1^{3}}{Z}\operatorname{Ch}\gamma_1 l +Q\cdot \frac{\gamma_2^{2}}{Z}\operatorname{Ch}\gamma_2 l \right). \]

If the boundary conditions are now again taken, then

\[ U_2=\frac{M\cdot \gamma_1^{2}+N\cdot \gamma_2^{2}}{Z\cdot Y}; \qquad I_2=\frac{1}{Z}\cdot \frac{P\cdot \gamma_1^{3}+Q\cdot \gamma_2^{3}}{Z\cdot Y}. \tag{84a} \]

The simultaneous solution of equations (84) and (84a) gives the values of all four constants

\[ M=\frac{U_2}{2} \left( 1+\alpha\cdot \frac{S}{R}+\beta\cdot \frac{T}{R} \right); \]

\[ N=\frac{U_2}{2} \left( 1-\alpha\frac{S}{R}-\beta\cdot \frac{T}{R} \right); \]

\[ P=\frac{I_2}{2}\cdot \frac{Z}{\gamma_1}\cdot \left( 1+\alpha\cdot \frac{S}{R}+\beta\cdot \frac{T}{R} \right); \]

\[ Q=\frac{I_2}{2}\cdot \frac{Z}{\gamma_2}\cdot \left( 1-\alpha\cdot \frac{S}{R}-\beta\cdot \frac{T}{R} \right), \tag{85} \]

where

\[ R=\sqrt{[ZY]^2-[Z]\cdot [Y]},\qquad S=Z_0\cdot Y_{\mathrm{I}}+Z_{\mathrm{I}}\cdot Y_0;\qquad T=Z_0\cdot Y_{\mathrm{II}}+Z_{\mathrm{II}}\cdot Y_0. \]

Thus, the coefficients of the six-terminal network will be equal to

\[ A=D=\frac{1}{2}\cdot \left[ (\operatorname{Ch}\gamma_1 l+\operatorname{Ch}\gamma_2 l) +\alpha\frac{S}{R}(\operatorname{Ch}\gamma_1 l-\operatorname{Ch}\gamma_2 l) + \beta\frac{T}{R}(\operatorname{Ch}\gamma_1 l-\operatorname{Ch}\gamma_2 l) \right]; \]

\[ B=\frac{Z}{2}\cdot \left[ \left( \frac{1}{\gamma_1}\operatorname{Sh}\gamma_1 l + \frac{1}{\gamma_2}\operatorname{Sh}\gamma_2 l \right) +\alpha\frac{S}{R} \left( \frac{1}{\gamma_1}\operatorname{Sh}\gamma_1 l - \frac{1}{\gamma_1}\operatorname{Sh}\gamma_2 l \right) + \beta\frac{T}{R} \left( \frac{1}{\gamma}\operatorname{Sh}\gamma_1 l - \frac{1}{\gamma_2}\operatorname{Sh}\gamma_2 l \right) \right]; \tag{86} \]

\[ C=\frac{1}{2Z}\cdot \left[ (\gamma_1\operatorname{Sh}\gamma_1 l+\gamma_2\operatorname{Sh}\gamma_2 l) +\alpha\frac{S}{R}(\gamma_1\operatorname{Sh}\gamma l-\gamma_2\operatorname{Sh}\gamma_2 l) + \beta\frac{1}{R}(\gamma_1\operatorname{Sh}\gamma_1 l-\gamma_2\operatorname{Sh}\gamma l) \right]. \]

These expressions are the exact solution of the problem of a six-terminal network replacing a line with asymmetrical phases. In the case of phase symmetry, i.e. when \(Z_{\mathrm{I}}=Z_{\mathrm{II}}=0\) and \(Y_{\mathrm{I}}=Y_{\mathrm{II}}=0\), all expressions are simplified and assume the usual form

\[ \gamma_1=\gamma_2=\sqrt{Z_0\cdot Y_0}=\gamma;\qquad A=D=\operatorname{Ch}\gamma l;\quad B=Z_c\operatorname{Sh}\gamma l;\quad C=\frac{1}{Z_c}\operatorname{Ch}\gamma l, \tag{87} \]

where

$$ Z_c=\frac{Z}{\gamma}=\sqrt{\frac{\bar Z_0}{\bar Y_0}} $$

represents the characteristic impedance of the line.

Whereas in a symmetrical line there is one system of incident and reflected waves, having propagation coefficient \(\gamma\) and characteristic impedance \(Z_c\), in a line with asymmetrical phases there will be two systems of waves with propagation coefficients \(\gamma_1\) and \(\gamma_2\) and characteristic impedances \(Z_{c1}=\dfrac{Z}{\gamma_1}\) and \(Z_{c2}=\dfrac{Z}{\gamma_2}\), the latter moreover having the form of trinomials, i.e., contributing to the appearance of asymmetry in the currents and voltages.

Circle diagrams for a six-terminal network of an asymmetrical three-phase circuit

Circle diagrams are an excellent graphical means for investigating the variable operating conditions of an electric circuit and are widely known for single-phase and symmetrical three-phase circuits. With a very simple method of construction and sufficient accuracy of the result, they make it possible easily and visually to determine the magnitude and phase shifts of any currents and voltages of the circuit under consideration, the values of powers, and a number of other quantities.

The application of these diagrams to the study of three-phase circuits with asymmetrical phases, although known, occurs very rarely and only as separate particular cases. The introduction of trinomials makes it possible to give a general method for constructing such diagrams, which greatly broadens the possibility of using them. Below is presented a general theory of circle diagrams for circuits with asymmetrical phases and a method for constructing them.

The finding on the diagram of segments corresponding to the various currents of the circuit, voltages, powers, etc., and the determination of their scales is considered in another work by the author and is not given here¹.

For the derivation of the basic relation we shall use the method usually employed for four-terminal networks, but we shall carry it out, in accordance with the six-terminal network, in the form of trinomials. From the basic relations of the six-terminal network (50) we eliminate the voltage \(U_2\) by substituting the load impedance

$$ Z_n=\frac{U_2}{I_2}, \tag{88} $$

which gives two new equations

$$ U_1=(A\cdot Z_n+B)\cdot I_2;\qquad I_1=(C\cdot Z_n+D)\cdot I_2. \tag{89} $$

Dividing them by one another, we eliminate also the current \(I_2\)

$$ \frac{U_1}{I_1}=\frac{A\cdot Z_n+B}{C\cdot Z_n+D}, $$

which gives for the current \(I_1\) the expression

$$ I_1=U_1\cdot\frac{C\cdot Z_n+D}{A\cdot Z_n+B}, \tag{90} $$

which is the basis for constructing the diagram. With the line voltage constant, the current \(I_1\) is a function of one load resistance \(Z_n\). Under certain conditions, the end of the current vector \(I_1\) will, for each of the phases, describe a circle, which will be the desired diagram.

At no load, for \(Z_n=\infty\), the current will be equal to

\[ I_{1xx}=U_1\cdot\frac{C}{A}, \tag{91} \]

and in a short circuit, for \(Z_n=0\),

\[ I_{1kz}=U_1\cdot\frac{D}{B}. \]

Transforming equation (90), we obtain

\[ I_1=U_1\cdot\frac{C}{A}+U\cdot \frac{1}{A\cdot B\left(1+\frac{A}{B}\right)\cdot Z_n}. \]

Since in this case

\[ I_{1kz}-I_{1xx} = U_1\cdot\frac{D}{B} - U_1\frac{C}{A} = U_1\cdot\frac{A\cdot D-B\cdot C}{A\cdot B} = U_1\cdot\frac{1}{A\cdot B}, \]

this equation can be written in the form

\[ I_1=I_{1xx}+ \frac{I_{1kz}-I_{1xx}}{1+\frac{A}{B}\cdot Z_n}, \tag{92} \]

well known for hexapole diagrams.

To obtain the circle diagram of a hexapole, it is necessary to introduce the condition that the load resistance \(Z_n\), as it varies, preserves the character of its asymmetry and the constancy of the phase-shift angle. This can be expressed as

\[ Z_n=z\cdot e^{j\varphi_n}\cdot\left(k_0+\alpha k_I+\beta k_{II}\right) = z\cdot e^{j\varphi_n}\cdot k, \tag{93} \]

where \(z\) is the varying magnitude of the resistance (a real number having the dimension of resistance), \(\varphi_n\) is the angle of the phase shift produced, and \(k_0\), \(k_I\), and \(k_{II}\) are constant complex coefficients determining the asymmetry of the load by phases. In this case equation (92) takes the form

\[ I_1=I_{1xx}+ \frac{I_{1kz}-I_{1xx}}{1+\frac{A}{B}\cdot k e^{j\varphi_n}\cdot z} = I_{1xx}+\left(I_{1kz}-I_{1xx}\right)\cdot F. \tag{94} \]

As is known from the theory of circle diagrams for four-poles, a circle is obtained only when the variable \(z\) enters the denominator only to the first power. Since equation (94) includes in the denominator the sequence coefficients \(\alpha\) and \(\beta\), in order to determine the actual degree of \(z\) it is necessary to get rid of them and reduce the equation to a form with separated sequences.

Transform the expression of the multiplier \(F\):

\[ F= \frac{1}{1+\frac{A}{B}\cdot k\cdot e^{j\varphi_n}\cdot z} = \frac{\frac{B}{A}}{\frac{B}{A}+k\cdot e^{j\varphi_n}\cdot z} = \frac{Z_{okz}}{Z_z+k\cdot e^{j\varphi_n}\cdot z}, \]

where \(Z_{\mathrm{okz}}=-\dfrac{B}{A}\) will correspond to the impedance of the reverse short circuit. Transforming further, we have

\[ F=\frac{[Z]_{\mathrm{okz}}+Z_{\mathrm{okz}}\cdot \hat{k}\cdot e^{j\varphi_{\mathrm{n}}}\cdot Z} {[Z]_{\mathrm{okz}}+[k]\cdot e^{j\varphi_{\mathrm{n}}} z^2+2[kZ_{\mathrm{okz}}]\cdot e^{j\varphi_{\mathrm{n}}}z}. \tag{95} \]

From this expression it is easy to see that, in order that the resistance \(z\) enter the denominator only to the first degree, it is necessary to satisfy the single condition for load unbalance:

\[ [k]=k_0^2-k_{\mathrm{I}}\cdot k_{\mathrm{II}}=0. \tag{96} \]

This condition determines the character of the load, and only three cases can occur:

\[ \begin{aligned} \text{a)}\quad & k_0=1;\quad k_{\mathrm{I}}=1;\quad k_{\mathrm{II}}=1,\\ \text{b)}\quad & k_0=1;\quad k_{\mathrm{I}}=a;\quad k_{\mathrm{II}}=a^2,\\ \text{c)}\quad & k_0=1;\quad k_{\mathrm{I}}=a^2;\quad k_{\mathrm{II}}=a. \end{aligned} \]

The first of these corresponds to a permanent connection of phases “b” and “c” and the insertion of a variable resistance in phase “a.” The second case corresponds to a similar insertion of a resistance in phase “b,” and the third to insertion in phase “c,” with a permanent connection of the other two phases. For other relations between the factors \(k_0\), \(k_{\mathrm{I}}\), and \(k_{\mathrm{II}}\), conditions are obtained that are not practicable.

If the load is expressed through admittance

\[ Y_{\mathrm{n}}=y\cdot e^{-j\varphi_{\mathrm{n}}}\cdot(Q_0+a\cdot Q_{\mathrm{I}}+\beta\cdot Q_{\mathrm{II}}) = y\cdot e^{-j\varphi_{\mathrm{n}}}\cdot Q, \]

then equation (94) takes the form

\[ i=I_{1\mathrm{xx}}+ \frac{(I_{1\mathrm{kz}}-I_{1\mathrm{xx}})\cdot y\cdot e^{-j\varphi_{\mathrm{n}}}\cdot Q} {y\cdot e^{-j\varphi_{\mathrm{n}}}\cdot Q+\dfrac{A}{B}} =I_{1\mathrm{xx}}+(I_{1\mathrm{kz}}-I_{1\mathrm{xx}})\cdot \varphi, \tag{97} \]

where

\[ \varphi= \frac{y\cdot e^{-j\varphi_{\mathrm{n}}}Q} {y\cdot e^{-j\varphi_{\mathrm{n}}}Q+\dfrac{A}{B}} = \frac{[Q]\cdot y^2\cdot e^{-j\varphi_{\mathrm{n}}}+y e^{-j\varphi_{\mathrm{n}}}\cdot Q\cdot \hat{Y}_{\mathrm{okz}}} {[Q]\cdot y^2\cdot e^{-j\varphi_{\mathrm{n}}}+2[QY_{\mathrm{okz}}]\cdot y\cdot e^{-j\varphi_{\mathrm{n}}}+[Y]_{\mathrm{okz}}}, \tag{98} \]

where \(Y_{\mathrm{okz}}=\dfrac{A}{B}\) represents the admittance for the reverse short circuit. This expression shows that, in order to obtain a circular diagram, the load-unbalance condition must be satisfied:

\[ [Q]=Q_0^2-Q_{\mathrm{I}}\cdot Q_{\mathrm{II}}=0. \]

As above, the existence of only three cases is possible:

\[ \begin{aligned} \text{a)}\quad & Q_0=1;\quad Q_{\mathrm{I}}=-1;\quad Q_{\mathrm{II}}=-1;\\ \text{b)}\quad & Q_0=1;\quad Q_{\mathrm{I}}=-a;\quad Q_{\mathrm{II}}=-a^2;\\ \text{c)}\quad & Q_0=1;\quad Q_{\mathrm{I}}=-a^2;\quad Q_{\mathrm{II}}=-a. \end{aligned} \]

The first of them corresponds to the insertion of a variable resistance between phases “b” and “c,” with phase “a” left open; the second—to the same insertion of a resistance between phases a and c; and the third—between phases “a” and “b,” with the third phase open.

Thus, the circle diagram in the case of phase asymmetry of a six-terminal network is possible only when there is just one varying resistance. As is known, in the case of phase symmetry one more condition is possible: the simultaneous variation of three equal resistances inserted symmetrically in all three phases (i.e., preservation of complete symmetry of the circuit).

Construction of Circle Diagrams of Six-Terminal Networks

For constructing a circle diagram one may use the usual method employed for four-terminal networks. For the quantity for which the circle diagram is to be constructed, one takes the values of the no-load and short-circuit vectors and constructs an arc of a circle on their difference, having first determined the angle between the tangent and the chord.

Fig. 8

This latter angle \(\delta\) (Fig. 8) is determined as the argument of the second term of the denominator of equation (95), and will be the same for all components of any currents and voltages of the given circuit. After eliminating \(z^2\), we obtain

\[ F = \frac{[Z]_{\mathrm{okz}} + Z_{\mathrm{okz}}\cdot \hat{k}\cdot e^{j\varphi_{\mathrm{н}}}\cdot z} {[Z]_{\mathrm{okz}} + 2\cdot [kZ_{\mathrm{okz}}]\cdot e^{j\varphi_{\mathrm{н}}}\cdot z} = \frac{ 1+\dfrac{Z_{\mathrm{okz}}\cdot \hat{k}}{[Z]_{\mathrm{okz}}}\cdot e^{j\varphi_{\mathrm{н}}}\cdot z }{ 1+2\dfrac{[kZ_{\mathrm{okz}}]}{[Z]_{\mathrm{okz}}}\cdot e^{j\varphi_{\mathrm{н}}}\cdot z }. \]

If we take

\[ \psi = \arg \frac{[kZ_{\mathrm{okz}}]}{[Z]_{\mathrm{okz}}} = \arg \frac{ k_{0}\cdot Z_{0\mathrm{okz}}-\dfrac{1}{2}\left(k_{\mathrm{I}}\cdot Z_{\mathrm{IIokz}}+k_{\mathrm{II}}Z_{\mathrm{Iokz}}\right) }{ Z_{0\mathrm{okz}}^{2}-Z_{\mathrm{Iokz}}\cdot Z_{\mathrm{IIokz}} }, \tag{100} \]

then the required angle will be equal to

\[ \delta = \psi + \varphi_{\mathrm{н}}. \tag{101} \]

For a positive value, this angle will be laid off at the no-load point clockwise, and at the short-circuit point counterclockwise.

In the expression obtained for \(F\), the numerator has the additional term

\[ \frac{\hat{k}\cdot Z_{\mathrm{okz}}}{[Z_{\mathrm{okz}}]}\cdot e^{j\varphi_{\mathrm{н}}}\cdot z, \]

which, at no load, i.e. for \(z=\infty\), gives a value of \(F\) different from zero,

\[ F_{xx}= \frac{\hat{k}\cdot Z_{\mathrm{okz}}}{2\cdot [kZ_{\mathrm{okz}}]}, \]

and, consequently, also changes the magnitude of the no-load current in equation (94):

\[ I'_{xx} = I_{xx} + \left(I_{\mathrm{kz}}-I_{xx}\right)\cdot F_{xx} = I_{xx} + \left(I_{\mathrm{kz}}-I_{xx}\right)\cdot \frac{\hat{k}Z_{\mathrm{okz}}}{2\cdot [kZ_{\mathrm{okz}}]}. \]

This difference is obtained because, for the values \(I_{1xx}\), the ideal no-load condition is taken, when all phases of the load are open; the value \(z=\infty\), however, corresponds to the opening of one phase when the other two are short-circuited. In constructing the circular diagram, one must take the vector of the actual no-load condition, and not the vector for the ideal case, i.e., represent equation (94) in the form

\[ I_1 = I'_{1xx} + (I_{1kz} - I'_{1xx}) \cdot \frac{1}{1 + 2 \frac{[kZ_{0kz}]}{[Z]_{0kz}} \cdot e^{j\varphi_n} \cdot z}. \]

The short-circuit current in this case does not differ from the current under an ideal short circuit of all phases.

If we turn to a load expressed in conductances according to equations (97) and (98), then all the relations remain the same, but instead of the term \(Z_{0kz}\) there will enter \(Y_{0kz}\), and instead of \(k\) there will enter \(Q\). In addition, in this case a difference between the actual and the ideal case will occur under short circuit, since part of the phases remains open. The equations take the form

\[ I_1 = I_{1xx} + (I'_{1kz} - I_{1xx}) \cdot \frac{1}{1 + 2 \frac{[Q \cdot Y_{0kz}]}{[Y_{0kz}]} \cdot e^{-j\varphi_n} \cdot y}, \]

where the required angle is determined as

\[ \delta = \psi_1 + \varphi_n, \]

where

\[ \psi_1 = -\arg \frac{[QY_{0kz}]}{[Y]_{0kz}}. \]

As indicated above, the quantity \(Z_{0kz}\) corresponds to the circuit resistance under an ideal reverse short circuit, i.e., when the voltage is applied from the load end, and the generator terminals are completely short-circuited. If the circuit is symmetrical, i.e., if for the six-terminal network the equality \(A = D\) holds, then the values will be identical to those for a direct short circuit of all three phases. The same, of course, also applies to the conductance \(Y_{0kz}\) of the reverse short circuit.

Conclusion

The method considered above for using trinomial expressions makes it possible to considerably simplify all derivations of relations and calculations in three-phase circuits that do not have paths for the components of zero-sequence currents. With their aid it has been possible to obtain simple ways of using the properties of six-terminal networks and their circular diagrams, to reduce to the general case the questions of asymmetrical loading of three-phase transformers, and to give relations for the exact calculation of transmission lines with asymmetrical phases.

All the results obtained can be fully extended also to the general case of three-phase circuits having a neutral conductor or connected in a ring, i.e., capable of conducting zero-sequence currents as well. In this case it is necessary to pass to matrix notation, although for most circuits trinomial expressions can also be partly used by applying techniques analogous to the Thévenin–Helmholtz method (no-load and short-circuit conditions). For lack of space, these questions cannot be considered in the present article and will be published separately later.

1. R. A. Voronov, “Circle Diagrams,” Proceedings of TII, Vol. 59, 1941, Tomsk. ↩↩