Sequence impedance of load with mutual coupling

For a balanced star load with self impedance $Z_s$ in each phase and equal mutual impedance $Z_m$ between phases, solidly grounded ($Z_n = 0$), the sequence impedances are $Z_0 = Z_s + 2Z_m$ and $Z_1 = Z_2 = Z_s - Z_m$.

Since the load is balanced: $Z_R = Z_Y = Z_B = Z_s$ and $Z_{RY} = Z_{YB} = Z_{RB} = Z_m$.

Phase equations

With $Z_n = 0$ the star point is at ground potential:

$$\begin{aligned} V_R &= Z_sI_R + Z_mI_Y + Z_mI_B \\ V_Y &= Z_mI_R + Z_sI_Y + Z_mI_B \\ V_B &= Z_mI_R + Z_mI_Y + Z_sI_B \end{aligned} \qquad Z_p = \begin{bmatrix} Z_s&Z_m&Z_m\\Z_m&Z_s&Z_m\\Z_m&Z_m&Z_s \end{bmatrix}$$

Transformation

$V_p = AV_s$ and $I_p = AI_s$ give $V_s = A^{-1}Z_pA\,I_s$.

Multiply $Z_p$ by each column of $A$, using $1 + a + a^2 = 0$:

• Column $[1\ 1\ 1]^T$: each row gives $Z_s + 2Z_m$
• Column $[1\ a^2\ a]^T$: row 1 gives $Z_s + Z_m(a^2+a) = Z_s - Z_m$, and the other rows give the same factor times $a^2$, $a$
• Column $[1\ a\ a^2]^T$: factor $Z_s - Z_m$

So

$$Z_{seq} = A^{-1}Z_pA = \begin{bmatrix} Z_s+2Z_m & 0 & 0 \\ 0 & Z_s-Z_m & 0 \\ 0 & 0 & Z_s-Z_m \end{bmatrix}$$

Result

$$Z_0 = Z_s + 2Z_m, \qquad Z_1 = Z_s - Z_m, \qquad Z_2 = Z_s - Z_m$$

• The matrix is diagonal, so the sequence networks are decoupled.
• If the neutral were grounded through $Z_n$, every element of $Z_p$ would gain $Z_n$ and only $Z_0$ would change: $Z_0 = Z_s + 2Z_m + 3Z_n$.