**2. Star connected generator–star connected load without a neutral connection**

The equivalent load impedance shown in Figure 1 may be described as follows;

$$\mathbf{Z}\_a = \mathbf{Z}\_{\mathbf{L}\_a} \lor \mathbf{\prime} - \mathbf{j} \mathbf{X}\_{\mathbf{C}\_a} \tag{1}$$

<sup>a</sup> <sup>b</sup>

n

*<sup>V</sup> Van bn*

k

*Zc*

*Zb*

http://dx.doi.org/10.5772/53914

(5)

143

(6)

(7)

*VaK*

*Za*

*Vab*

*Ia*

*Ic*

0 021 0 1 102 1 2 210 2

Where the subscripts 0, 1, and 2 stands for zero, positive and negative sequence components, respectively. The symmetrical components of the load phase voltages may be found from the

2

On the other hand, the three phase voltages may be found in terms of their symmetrical

2

The transformation matrix shown in Eq. (6) can be used to find the symmetrical components of currents, namely, *Ia0, Ia1*, and *Ia2*. The symmetrical components of the three phase impedances

0

1

2

2

g g

2

é ù é ù é ù ê ú ê ú ê ú <sup>=</sup> ê ú ê ú ê ú ê ú ê ú ê ú ë û ë û ë û

g g

*ak ak bk ak ck ak*

1 1

11 1

*V V V V V V* g g

é ù é ù é ù ê ú ê ú ê ú <sup>=</sup> ê ú ê ú ê ú ê ú ê ú ê ú ë û ë û ë û

1 1

11 1

*ak ak ak bk ak ck*

*V V V V V V* g g

*ak a ak a ak a*

ë û ë ûë û

é ù é ùé ù ê ú ê úê ú <sup>=</sup>

*V ZZZ I V ZZZ I V ZZZ I*

**Figure 1.** Star connected generator - star connected load without neutral connection.

0

1

2

components by using the following transformation:

*Ib*

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

c

*Vcn*

three-phase values as follows:

where *γ*=1∠120

*Za, Zb,* and *Zc* are as follows;

$$\mathbf{Z}\_b = \mathbf{Z}\_{\mathbf{L}\_b} \parallel - \mathbf{j} \mathbf{X}\_{\mathbf{C}\_b} \tag{2}$$

$$\mathbf{Z}\_c = \mathbf{Z}\_{\mathbf{L}\_c} \lor \mathbf{\color{red}{-j $\mathbf{X}\_c$ }} \tag{3}$$

where,

Z*<sup>L</sup> <sup>a</sup>*,*b*,*<sup>c</sup>* = Load impedance at base frequency connected across phase *a*, *b* and *c*, respectively.

X*Ca*,*b*,*<sup>c</sup>* = Excitation capacitor reactance at base frequency connected across phase *a*, *b* and *c*, respectively.

Z*a*,*b*,*<sup>c</sup>* = Equivalent impedance of load and excitation capacitor at base frequency connected across phase *a*, *b* and *c* respectively.

At the load side, the phase voltages are:

$$
\begin{bmatrix} V\_{ak} \\ V\_{bk} \\ V\_{ck} \end{bmatrix} = \begin{bmatrix} Z\_a & 0 & 0 \\ 0 & Z\_b & 0 \\ 0 & 0 & Z\_c \end{bmatrix} \begin{bmatrix} I\_a \\ I\_b \\ I\_c \end{bmatrix} \tag{4}
$$

Since the load and/or the excitation capacitors are expected to be unbalanced, it is more appropriate to describe the different quantities involved in Eq. (4) in terms of their symmetrical components. Using the symmetrical components technique, the following is found:

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 143

**Figure 1.** Star connected generator - star connected load without neutral connection.

$$
\begin{bmatrix} V\_{\text{ak}0} \\ V\_{\text{ak}1} \\ V\_{\text{ak}2} \end{bmatrix} = \begin{bmatrix} Z\_0 & Z\_2 & Z\_1 \\ Z\_1 & Z\_0 & Z\_2 \\ Z\_2 & Z\_1 & Z\_0 \end{bmatrix} \begin{bmatrix} I\_{a0} \\ I\_{a1} \\ I\_{a2} \end{bmatrix} \tag{5}
$$

Where the subscripts 0, 1, and 2 stands for zero, positive and negative sequence components, respectively. The symmetrical components of the load phase voltages may be found from the three-phase values as follows:

$$
\begin{bmatrix} V\_{ak0} \\ V\_{ak1} \\ V\_{ak2} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \gamma & \gamma^2 \\ 1 & \gamma^2 & \gamma \end{bmatrix} \begin{bmatrix} V\_{ak} \\ V\_{bk} \\ V\_{ck} \end{bmatrix} \tag{6}
$$

where *γ*=1∠120

model iteratively. An experimental setup has been built to verify the results obtained from the theoretical model. The model is generalized to cover more connection types of SEIG and/or load. It is clear that the theoretical results are in good agreement with those reported experi‐ mentally. The effect of the machine core losses is considered by representing the core resistance as a second order polynomial in terms of *Xm*. Furthermore, the magnetizing reactance has been included in the negative sequence equivalent circuit as a variable. The positive and negative sequence equivalent circuits are used to model the SEIG. The final characteristic equation is reached by equating both the positive-sequence and negative-sequence voltages across the

**2. Star connected generator–star connected load without a neutral**

The equivalent load impedance shown in Figure 1 may be described as follows;

*a a*

*b b*

*c c*

Z*<sup>L</sup> <sup>a</sup>*,*b*,*<sup>c</sup>* = Load impedance at base frequency connected across phase *a*, *b* and *c*, respectively.

X*Ca*,*b*,*<sup>c</sup>* = Excitation capacitor reactance at base frequency connected across phase *a*, *b* and *c*,

Z*a*,*b*,*<sup>c</sup>* = Equivalent impedance of load and excitation capacitor at base frequency connected

0 0 0 0 0 0

Since the load and/or the excitation capacitors are expected to be unbalanced, it is more appropriate to describe the different quantities involved in Eq. (4) in terms of their symmetrical

*ak a a bk b b ck c c*

*VZ I V ZI V ZI*

é ù é ùé ù ê ú ê úê ú <sup>=</sup>

ë û ë ûë û

components. Using the symmetrical components technique, the following is found:

*Z Z / / jX aL C* = - (1)

*Z Z / / jX bL C* = - (2)

*Z Z / / jX c L <sup>C</sup>* = - (3)

(4)

SEIG and the load.

142 New Developments in Renewable Energy

**connection**

where,

respectively.

across phase *a*, *b* and *c* respectively.

At the load side, the phase voltages are:

On the other hand, the three phase voltages may be found in terms of their symmetrical components by using the following transformation:

$$
\begin{bmatrix} V\_{ak} \\ V\_{bk} \\ V\_{ck} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & \mathcal{Y}^2 & \mathcal{Y} \\ 1 & \mathcal{Y} & \mathcal{Y}^2 \end{bmatrix} \begin{bmatrix} V\_{ak0} \\ V\_{ak1} \\ V\_{ak2} \end{bmatrix} \tag{7}
$$

The transformation matrix shown in Eq. (6) can be used to find the symmetrical components of currents, namely, *Ia0, Ia1*, and *Ia2*. The symmetrical components of the three phase impedances *Za, Zb,* and *Zc* are as follows;

$$
\begin{bmatrix} Z\_0 \\ Z\_1 \\ Z\_2 \end{bmatrix} = \frac{1}{3} \begin{vmatrix} 1 & 1 & 1 \\ 1 & \gamma & \gamma^2 \\ 1 & \gamma^2 & \gamma \end{vmatrix} \begin{bmatrix} Z\_a \\ Z\_b \\ Z\_c \end{bmatrix} \tag{8}
$$

Since in an isolated-neutral star-connected load, the zero sequence component of line current (phase current) equals to zero, substituting *Ia0*=0 in Eq. (5) and expanding yields:

$$V\_{a11} = Z\_0 I\_{a1} + Z\_2 I\_{a2} \tag{9}$$

$$\mathbf{V}\_{\rm ak2} = \mathbf{Z}\_1 \mathbf{I}\_{a1} + \mathbf{Z}\_0 \mathbf{I}\_{a2} \tag{10}$$

It can be shown using the symmetrical components technique that the relation between the positive and negative sequence components of both the line and the phase voltages are as follows:

$$V\_{LL1} = \left(1 \cdot \gamma^2\right) V\_{LN1} \tag{11}$$

$$V\_{LL2} = \left(1 - \gamma\right) V\_{LN1} \tag{12}$$

Furthermore, the air gap voltage may be approximated over the saturated region as a function

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

of *Xm* by the following sixth order polynomial. Figure 4 shows this variation.

654321 0

*V / F b X b X b X b X b X b X b p.u. g m mmmmm*

=+++++ +

65 4 3 2 1 0

= =- = =- = =- =

*bb b b b b b*

0.95, 8.28, 28.91, 51.78, 50.18, 25.07, 6.21

0.8 1.2 1.6 2 2.4 *Xm* - p.u.

Experimental readings Fitted Curve

(16)

http://dx.doi.org/10.5772/53914

145

65432

**Figure 2.** Induction generator equivalent circuits: (a) +ve seq. (b) –ve seq.

,

0

**Figure 3.** Variation of core resistance as a function of magnetizing reactance.

20

*Rc* - p.u.

40

60

*where*

Hence,

$$V\_{ab1} = \left(\mathbf{1} \cdot \boldsymbol{\gamma}^2\right) V\_{ak1} \tag{13}$$

$$\left(V\_{ab2}\right) = \left(1 - \chi\right)V\_{ak2}\tag{14}$$

Now looking at the generator side, the following positive and negative sequence circuits of Figure 2 can be used to model the generator. As can be seen in Figure 2, the core loss resistance is taken into consideration in the positive-sequence equivalent circuit of the SEIG.

As the core loss is variable according to saturation, the core loss resistance is expressed as a function of the magnetizing reactance (*Xm*) as shown in Eq. (15). Figure 3 shows the variation of the core resistance as a function of magnetizing reactance. Although this will increase the complexity of the model but the model will be closer to the actual case.

$$\begin{aligned} R\_{\mathcal{C}}(X\_{\mathcal{m}}) &= a\_2 X\_{\mathcal{m}}^2 + a\_1 X\_{\mathcal{m}} + a\_0 \quad \text{p.u.}\\ \text{where,} \\ a\_2 &= -17.159, a\_1 = 33.372, a\_0 = 35.699 \end{aligned} \tag{15}$$

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 145

**Figure 2.** Induction generator equivalent circuits: (a) +ve seq. (b) –ve seq.

0

1

2

follows:

144 New Developments in Renewable Energy

Hence,

2

*a b c*

*V ZI ZI ak a a* 1 01 22 = + (9)

*V ZI Z I ak a a* 2 11 0 2 = + (10)

(11)

(12)

(13)

(14)

(8)

(15)

2

Since in an isolated-neutral star-connected load, the zero sequence component of line current

It can be shown using the symmetrical components technique that the relation between the positive and negative sequence components of both the line and the phase voltages are as

> ( ) <sup>2</sup> 1 1 1 - *V V LL LN* = g

( ) 2 1 1 *V V LL LN* = g

( ) <sup>2</sup> 1 1 1 - *V V ab ak* = g

( ) 2 2 1 *V V ab ak* = g

is taken into consideration in the positive-sequence equivalent circuit of the SEIG.

complexity of the model but the model will be closer to the actual case.

,

*where*

2

*R (X ) a X a X a p.u. c m <sup>m</sup> <sup>m</sup>*

=+ +

2 10

=- = =

*a aa*

21 0

17.159, 33.372, 35.699

Now looking at the generator side, the following positive and negative sequence circuits of Figure 2 can be used to model the generator. As can be seen in Figure 2, the core loss resistance

As the core loss is variable according to saturation, the core loss resistance is expressed as a function of the magnetizing reactance (*Xm*) as shown in Eq. (15). Figure 3 shows the variation of the core resistance as a function of magnetizing reactance. Although this will increase the

g g

é ù é ù é ù ê ú ê ú ê ú <sup>=</sup> ê ú ê ú ê ú ê ú ê ú ê ú ë û ë û ë û

11 1

*Z Z Z Z Z Z* g g

(phase current) equals to zero, substituting *Ia0*=0 in Eq. (5) and expanding yields:

Furthermore, the air gap voltage may be approximated over the saturated region as a function of *Xm* by the following sixth order polynomial. Figure 4 shows this variation.

$$\begin{array}{l} V \;/\; F = b\_6 X\_m^6 + b\_5 X\_m^5 + b\_4 X\_m^4 + b\_3 X\_m^3 + b\_2 X\_m^2 + b\_1 X\_m + b\_0 \; p.u. \\\\ where \; \\ b\_6 = 0.95, b\_5 = -8.28, b\_4 = 28.91, b\_3 = -51.78, b\_2 = 50.18, b\_1 = -25.07, b\_0 = 6.21 \end{array} \tag{16}$$

**Figure 3.** Variation of core resistance as a function of magnetizing reactance.

**Figure 4.** Variation of air gap voltage as a function of magnetizing reactance.

The terminal voltage of the positive and negative sequence equivalent circuits is given by;

$$\mathbf{V}\_{\text{an1}} = \begin{array}{c} -\mathbf{I}\_{\text{G1}} \; \mathbf{Z}\_{\text{G1}} \end{array} \tag{17}$$

*a G* 1 1 *I I* = (23)

http://dx.doi.org/10.5772/53914

147

*a G* 2 2 *I I* = (24)

1 0 1 22 0 *G aa (Z Z ) I Z I* + += (25)

11 0 2 2 0 *a Ga Z I (Z - Z )I* + = (26)

0 1 0 2 12 0 *G G ( Z Z ) ( Z - Z ) - Z Z* + = (27)

*V Z I Z I Z I ak*0 00 21 12 *a aa* = ++ (28)

*V Z I Z I Z I ak*1 10 01 22 *aaa* = ++ (29)

Substituting Eq. (23) and Eq. (24) into Eq. (21) and into Eq. (22) and rearranging, yields the

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

This is the characteristic equation of an isolated-neutral star connected induction generator. It consists of two parts, namely, the real part and the imaginary part. *Ia1* and *Ia2* does not equal zero because self-excitation is assumed to occur, hence, the real and imaginary parts of Eq. (27) must equal to zero. By substituting the machine parameters, speed, excitation capacitor values, a nonlinear equation with constant coefficients in *F* and *Xm* can be found. Solving iteratively to find the real roots of the equation that satisfies the constraints, the values of *F* and *Xm* are found; hence, the performance of the generator under these conditions can be determined. This procedure is carried out using MATHCAD® software. The flow chart

**3. Star connected generator–star connected load with a neutral connection**

The connection for this case is shown in Figure 6. In this type of connection the zero sequence component of line currents is present (i.e. *Ia0 ≠ 0*) while the zero sequence component of phase

following:

Solving these two equations simultaneously yields,

describing the performance evaluation is shown in Figure 5.

voltages (*Vak0 = 0*) equals zero.

Expanding Eq. (5) yields:

$$\mathbf{V\_{ar2}}\_{\rm ar2} = \mathbf{I\_{G2}} \; \mathbf{Z\_{G2}} \tag{18}$$

Eq. (11), Eq. (12), Eq. (17), and Eq. (18) yield the following,

$$V\_{ab1} = -\left(\mathbf{1} \cdot \boldsymbol{\gamma}^2\right) I\_{G1} Z\_{G1} \tag{19}$$

$$V\_{ab2} = \left(1 - \gamma\right) I\_{G2} Z\_{G2} \tag{20}$$

where

*ZG*1, *ZG*2Input impedance of positive and negative sequence equivalent circuits, respectively. Equating symmetrical components of line-to-line voltages yields:

$$-I\_{G1}Z\_{G1} = Z\_0I\_{a1} + Z\_2I\_{a2} \tag{21}$$

$$\mathbf{I}\_{\rm G2} \, Z\_{\rm G2} = \mathbf{Z}\_1 \mathbf{I}\_{a1} + \mathbf{Z}\_0 \mathbf{I}\_{a2} \tag{22}$$

Since the phase current in a star connected generator is the same as the line current, hence,

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 147

$$I\_{a1} = I\_{G1} \tag{23}$$

$$I\_{a2} = I\_{G2} \tag{24}$$

Substituting Eq. (23) and Eq. (24) into Eq. (21) and into Eq. (22) and rearranging, yields the following:

$$(\left(\mathbf{Z}\_{G1} + \mathbf{Z}\_0\right)\mathbf{I}\_{a1} + \mathbf{Z}\_2\mathbf{I}\_{a2} = \mathbf{0} \tag{25}$$

$$(\mathbf{Z\_1 I\_{a1}} + (\mathbf{Z\_0} \cdot \mathbf{Z\_{G2}})I\_{a2} = \mathbf{0}) \tag{26}$$

Solving these two equations simultaneously yields,

0.8 1.2 1.6 2 2.4 *Xm* - p.u.

The terminal voltage of the positive and negative sequence equivalent circuits is given by;

( ) <sup>2</sup> <sup>1</sup> 1 1 - 1- *V IZ ab G G* = g

( ) <sup>2</sup> 2 2 1 *V IZ ab G G* = g

*ZG*1, *ZG*2Input impedance of positive and negative sequence equivalent circuits, respectively.

Since the phase current in a star connected generator is the same as the line current, hence,

Experimental readings Fitted curve

V I Z an1 G1 G1 = - (17)

V I Z an2 G2 G2 = (18)

(19)

(20)

*GG a a* 1 1 01 22 *- I Z Z I Z I* = + (21)

*GG a a* 2 2 11 02 *I Z Z I Z I* = + (22)

0.4

**Figure 4.** Variation of air gap voltage as a function of magnetizing reactance.

Eq. (11), Eq. (12), Eq. (17), and Eq. (18) yield the following,

Equating symmetrical components of line-to-line voltages yields:

where

0.6

0.8

*Vg/F* - p.u.

1

1.2

146 New Developments in Renewable Energy

$$(\mathbf{Z}\_0 + \mathbf{Z}\_{\rm G1})(\mathbf{Z}\_0 \cdot \mathbf{Z}\_{\rm G2}) \cdot \mathbf{Z}\_1 \mathbf{Z}\_2 = \mathbf{0} \tag{27}$$

This is the characteristic equation of an isolated-neutral star connected induction generator. It consists of two parts, namely, the real part and the imaginary part. *Ia1* and *Ia2* does not equal zero because self-excitation is assumed to occur, hence, the real and imaginary parts of Eq. (27) must equal to zero. By substituting the machine parameters, speed, excitation capacitor values, a nonlinear equation with constant coefficients in *F* and *Xm* can be found. Solving iteratively to find the real roots of the equation that satisfies the constraints, the values of *F* and *Xm* are found; hence, the performance of the generator under these conditions can be determined. This procedure is carried out using MATHCAD® software. The flow chart describing the performance evaluation is shown in Figure 5.

#### **3. Star connected generator–star connected load with a neutral connection**

The connection for this case is shown in Figure 6. In this type of connection the zero sequence component of line currents is present (i.e. *Ia0 ≠ 0*) while the zero sequence component of phase voltages (*Vak0 = 0*) equals zero.

Expanding Eq. (5) yields:

$$V\_{ak0} = Z\_0 \ I\_{a0} + Z\_2 \ I\_{a1} + Z\_1 \ I\_{a2} \tag{28}$$

$$V\_{a\&1} = Z\_1 I\_{a0} + Z\_0 I\_{a1} + Z\_2 I\_{a2} \tag{29}$$

**Figure 5.** Flow chart describing performance evaluation.

$$V\_{\rm ak2} = Z\_2 \ I\_{a0} + Z\_1 \ I\_{a1} + Z\_0 \ I\_{a2} \tag{30}$$

2

**Figure 6.** Star connected generator - star connected load with a neutral connection.

Since the phase voltage of both the generator and the load are equal, hence,

*Ia1* and *IG2 = Ia2*, yield,

*Vcn*

b

c

n *Vbn*

*Van*

a

system equals to zero;

2 1 2 2 1 10 2 0 0 *ak a a <sup>Z</sup> Z Z V Z IZ I Z Z* æ ö æ ö =ç - ÷ + - ç ÷ ç ÷ è ø è ø

*Ia*

*Ib*

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions

*Ic*

Substituting Eqs. (34) and (35) into Eqs. (32) and (33), and taking into consideration that *IG1 =*

1 2 1

*Z Z*

2 1 2

0 0

0 *G G*

*Z Z* æ ö æ ö =ç - ÷ + - ç ÷ ç ÷ è ø è ø

Since sequence currents does not equal to zero, hence, the characteristics equation of this

0 0

11 0 12 2

*a G a a Z Z <sup>Z</sup> IZ Z I Z I*

22 1 1 0 2

1 2 1 2 2 1

0 0 00

0 1 0 21 2

*Z Z Z Z Z Z Z ZZ Z Z Z Z Z ZZ* <sup>æ</sup> ö æ <sup>ö</sup> æ öæ ö <sup>ç</sup> - + - - -ç - ÷ ç - ÷ = ÷ ç <sup>÷</sup> <sup>ç</sup> ÷ ç <sup>÷</sup> ç ÷ç ÷ <sup>è</sup> ø è <sup>ø</sup> è øè ø

*a G a a <sup>Z</sup> Z Z IZ Z I Z I*

æ ö æ ö - = - +ç - ÷ ç ÷ ç ÷ è ø è ø

2

*V V - I Z ak an G G* 1 1 11 = = (34)

*V V I Z ak an G G* 2 2 22 = = (35)

2 2

2

(33)

149

*Zc*

k

*Vak*

*Zb*

http://dx.doi.org/10.5772/53914

*Za*

*Vab*

(36)

(37)

(38)

Substituting *Vak0* = *0* in Eq. (28) and rearranging, yield:

$$\mathbf{I}\_{a0} = - (\mathbf{Z}\_2 \mathbf{I}\_{a1} + \mathbf{Z}\_1 \mathbf{I}\_{a2}) / \mathbf{Z}\_0 \tag{31}$$

Substituting this result in Eqs. (29) and (30), yields:

$$V\_{ak1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right) I\_{a1} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right) I\_{a2} \tag{32}$$

Steady State Modeling of Three Phase Self–Excited Induction Generator Under Unbalanced/Balanced Conditions http://dx.doi.org/10.5772/53914 149

**Figure 6.** Star connected generator - star connected load with a neutral connection.

*V Z I Z I Z I ak a a a* 2 20 11 02 = ++ (30)

*C = C + ΔC*

Phase currents, sequence components of load phase voltages, and load phase voltages

*C ≤ Cmax*

YES

Stop

NO

*a aa* 0 21 12 0 *I - (Z I Z I ) / Z* = + (31)

(32)

2

1 2 1 10 12 2 0 0

*Z Z* æ ö æ ö = - +ç - ÷ ç ÷ ç ÷ è ø è ø

*ak a a Z Z <sup>Z</sup> V Z IZ I*

Substituting *Vak0* = *0* in Eq. (28) and rearranging, yield:

Start

148 New Developments in Renewable Energy

machine parameters and guess values

 load parameters and excitation capacitors

 Symmetrical components of the three phase load, *ZG1* and *ZG2*

*F* and *Xm* under the constraints: *0 < Xm < Xu 0 < F ≤ υ F* and *Xm* are real values

*Vg , Im , Ir , IG1* and *Vp* 

Load impedances, SEIG impedances at *F* and *Xm*

 *IG2* and *Vn*

**Figure 5.** Flow chart describing performance evaluation.

Substituting this result in Eqs. (29) and (30), yields:

$$V\_{ak2} = \left(Z\_1 - \frac{Z\_2}{Z\_0}\right)I\_{a1} + \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)I\_{a2} \tag{33}$$

Since the phase voltage of both the generator and the load are equal, hence,

$$V\_{ak1} = V\_{am1} = -I\_{G1}Z\_{G1} \tag{34}$$

$$V\_{ak2} = V\_{an2} \ = I\_{G2} \ Z\_{G2} \tag{35}$$

Substituting Eqs. (34) and (35) into Eqs. (32) and (33), and taking into consideration that *IG1 = Ia1* and *IG2 = Ia2*, yield,

$$-I\_{a1}Z\_{G1} = \left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0}\right)I\_{a1} + \left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right)I\_{a2} \tag{36}$$

$$I\_{a2} Z\_{G2} = \left( Z\_1 - \frac{Z\_2}{Z\_0} \right) I\_{a1} + \left( Z\_0 - \frac{Z\_1 Z\_2}{Z\_0} \right) I\_{a2} \tag{37}$$

Since sequence currents does not equal to zero, hence, the characteristics equation of this system equals to zero;

$$\left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0} + Z\_{G1}\right)\left(Z\_0 - \frac{Z\_1 Z\_2}{Z\_0} - Z\_{G2}\right) - \left(Z\_1 - \frac{Z\_2}{Z\_0}\right)\left(Z\_2 - \frac{Z\_1^2}{Z\_0}\right) = 0\tag{38}$$
