Line currents with isolated and grounded neutral

For a symmetrical star load, $Z_1 = Z_2 = Z$ and $Z_0 = Z + 3Z_n$. With an isolated neutral $Z_n = \infty$, so no zero sequence current flows; with a solidly grounded neutral each phase current is simply $V/Z$.

Assumption: the applied voltages are phase-to-ground (supply neutral grounded) voltages, $Z = 10\ \Omega$ resistive.

Sequence components of the applied voltages

$$\begin{aligned} V_a &= 220\angle0^\circ = 220 + j0 \\ V_b &= 200\angle{-110^\circ} = -68.40 - j187.94 \\ V_c &= 180\angle110^\circ = -61.56 + j169.14 \end{aligned}$$ $$\begin{aligned} V_0 &= \tfrac13(V_a+V_b+V_c) = 30.01 - j6.26 = 30.66\angle{-11.79^\circ}\ \text{V} \\ V_1 &= \tfrac13(V_a+aV_b+a^2V_c) = 198.08 + j1.16 = 198.08\angle0.33^\circ\ \text{V} \\ V_2 &= \tfrac13(V_a+a^2V_b+aV_c) = -8.09 + j5.11 = 9.56\angle147.73^\circ\ \text{V} \end{aligned}$$

(a) Neutral isolated

Zero sequence current cannot flow ($I_0 = 0$); the zero sequence voltage appears as the neutral shift $V_{nN} = V_0 = 30.66\angle{-11.79^\circ}$ V.

$$\begin{aligned} I_1 &= \frac{V_1}{Z} = 19.81 + j0.12\ \text{A}, \qquad I_2 = \frac{V_2}{Z} = -0.81 + j0.51\ \text{A} \\ I_a &= I_1 + I_2 = 19.00 + j0.63 = 19.01\angle1.89^\circ\ \text{A} \\ I_b &= a^2I_1 + aI_2 = -9.84 - j18.17 = 20.66\angle{-118.44^\circ}\ \text{A} \\ I_c &= aI_1 + a^2I_2 = -9.16 + j17.54 = 19.79\angle117.57^\circ\ \text{A} \end{aligned}$$

Check: $I_a + I_b + I_c = 0$ ✓

(b) Neutral solidly grounded

$Z_0 = Z_1 = Z_2 = 10\ \Omega$, so each phase sees its own voltage:

$$\begin{aligned} I_a &= \frac{220\angle0^\circ}{10} = 22.0\angle0^\circ\ \text{A} \\ I_b &= \frac{200\angle{-110^\circ}}{10} = 20.0\angle{-110^\circ}\ \text{A} \\ I_c &= \frac{180\angle110^\circ}{10} = 18.0\angle110^\circ\ \text{A} \end{aligned}$$

Neutral current: $I_n = I_a + I_b + I_c = 3V_0/Z = 9.00 - j1.88 = 9.20\angle{-11.79^\circ}$ A.