Sequence impedances of star load with Zn

Sequence impedance is the impedance offered by a circuit to currents of one sequence only: $Z_0 = V_{a0}/I_{a0}$, $Z_1 = V_{a1}/I_{a1}$, $Z_2 = V_{a2}/I_{a2}$. We start from the phase-domain voltage equations of the load and transform them with the symmetrical component matrix $A$.

Circuit and phase equations

   Ia -->  Zs
 a o-----/\/\---+
   Ib -->  Zs   |
 b o-----/\/\---+ n
   Ic -->  Zs   |
 c o-----/\/\---+
                |  In = Ia+Ib+Ic
               Zn
                |
              ground

Voltage of each phase to ground = drop in $Z_s$ + drop in $Z_n$:

$$\begin{aligned} V_a &= Z_s I_a + Z_n (I_a + I_b + I_c) \\ V_b &= Z_s I_b + Z_n (I_a + I_b + I_c) \\ V_c &= Z_s I_c + Z_n (I_a + I_b + I_c) \end{aligned}$$

In matrix form $V_p = Z_p I_p$ with

$$Z_p = \begin{bmatrix} Z_s+Z_n & Z_n & Z_n \\ Z_n & Z_s+Z_n & Z_n \\ Z_n & Z_n & Z_s+Z_n \end{bmatrix}$$

Transformation to sequence quantities

Put $V_p = A V_s$ and $I_p = A I_s$:

$$A V_s = Z_p A I_s \;\Rightarrow\; V_s = A^{-1} Z_p A\, I_s = Z_s^{seq} I_s$$

with $A = \begin{bmatrix} 1&1&1\\1&a^2&a\\1&a&a^2 \end{bmatrix}$, $A^{-1} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}$.

Write $Z_p = Z_s U + Z_n J$, where $U$ is the unit matrix and $J$ is the matrix of all ones. Then $A^{-1}(Z_sU)A = Z_sU$. For $J$: every row of $J A$ is $[3\ \ 0\ \ 0]$ (since column sums of $A$ are $3, 1+a^2+a=0, 1+a+a^2=0$), and $A^{-1}$ times that gives

$$A^{-1} J A = \begin{bmatrix} 3&0&0\\0&0&0\\0&0&0 \end{bmatrix}$$

Hence

$$Z_{seq} = A^{-1} Z_p A = \begin{bmatrix} Z_s+3Z_n & 0 & 0 \\ 0 & Z_s & 0 \\ 0 & 0 & Z_s \end{bmatrix}$$

Result

Writing the three rows separately:

$$\begin{aligned} V_{a0} &= (Z_s + 3Z_n) I_{a0} &\Rightarrow\; Z_0 &= Z_s + 3Z_n \\ V_{a1} &= Z_s I_{a1} &\Rightarrow\; Z_1 &= Z_s \\ V_{a2} &= Z_s I_{a2} &\Rightarrow\; Z_2 &= Z_s \end{aligned}$$

Physical meaning

• Positive and negative sequence currents are balanced sets; their sum is zero, so no current flows in $Z_n$ and the impedance is just $Z_s$.
• Zero sequence currents are equal and in phase in all three phases, so the neutral carries $3I_{a0}$. The drop across $Z_n$ is $3I_{a0}Z_n$, which appears as an extra $3Z_n$ in the zero sequence circuit of each phase.
• If the neutral is solidly grounded ($Z_n = 0$), $Z_0 = Z_s$; if it is isolated ($Z_n \to \infty$), $Z_0 = \infty$ and no zero sequence current can flow.
• Since the sequence impedance matrix is diagonal, the three sequence networks are independent (decoupled).