Sequence networks of Y load, neutral current

For a balanced star load with neutral grounded through $Z_n$, the sequence networks are three separate circuits with impedances $Z_1 = Z_2 = Z_s$ and $Z_0 = Z_s + 3Z_n$, and the neutral carries $I_n = 3I_{a0}$.

Neutral current is three times zero sequence current

By the symmetrical component definitions:

$$\begin{aligned} I_a &= I_{a0} + I_{a1} + I_{a2} \\ I_b &= I_{a0} + a^2 I_{a1} + a I_{a2} \\ I_c &= I_{a0} + a I_{a1} + a^2 I_{a2} \end{aligned}$$

Adding and using $1 + a + a^2 = 0$:

$$\begin{aligned} I_n &= I_a + I_b + I_c \\ &= 3I_{a0} + (1+a^2+a)I_{a1} + (1+a+a^2)I_{a2} \\ &= 3I_{a0} \end{aligned}$$

So only zero sequence currents flow through the neutral, and the neutral current is $3I_{a0}$. Positive and negative sequence currents form balanced sets that cancel at the star point.

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

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

Sequence networks

 Positive         Negative         Zero
 a1 --Ia1-->      a2 --Ia2-->      a0 --Ia0-->
   |                |                |
  Zs               Zs               Zs
   |                |                |
   n1 (ref)         n2 (ref)        3Zn
                                     |
                                    n0 (ref)

• $V_{a1} = Z_s I_{a1}$, $V_{a2} = Z_s I_{a2}$, $V_{a0} = (Z_s + 3Z_n) I_{a0}$.
• The impedance $Z_n$ appears as $3Z_n$ in the zero sequence network only because the voltage drop across it is $Z_n \times 3I_{a0}$, while the network is drawn for one phase carrying $I_{a0}$.
• Solidly grounded neutral: $Z_n = 0$; ungrounded neutral: zero sequence network open between $Z_s$ and the reference.