Fault currents of isolated-neutral 380 V generator

Data: $V_L = 380$ V, so phase emf at no load $E = 380/\sqrt3 = 219.39$ V. $Z_1 = j2\ \Omega$, $Z_2 = j0.5\ \Omega$, $Z_0 = j0.25\ \Omega$. Neutral is isolated, so the neutral impedance $Z_n = \infty$ and the zero-sequence network is open: no zero-sequence current can flow.

(i) L-G fault (phase a to ground)

$$I_f = 3I_{a0} = \frac{3E}{Z_1 + Z_2 + Z_0 + 3Z_n}$$

With $Z_n \to \infty$:

$$I_f = 0\ \text{A}$$

There is no return path for the current, so no fault current flows. Only the potential of the neutral shifts: the faulted phase goes to ground potential and the healthy phase voltages to ground rise to line voltage (380 V).

(ii) L-L-G fault (phases b, c to ground)

$I_{a0} = 0$ because the zero-sequence network is open. The fault behaves like an L-L fault:

$$\begin{aligned} I_{a1} &= \frac{E}{Z_1 + Z_2} = \frac{219.39}{j2.5} = -j87.76\ \text{A} \\ |I_b| = |I_c| &= \sqrt3\,|I_{a1}| = \frac{\sqrt3 \times 219.39}{2.5} = 152.0\ \text{A} \end{aligned}$$

Current to ground $= 3I_{a0} = 0$. The line currents are 152.0 A each (flowing from b to c).

(iii) L-L-L-G fault

Balanced fault, only the positive sequence network acts:

$$I_f = \frac{E}{|Z_1|} = \frac{219.39}{2} = 109.70\ \text{A}$$

Results

Comments

• With an isolated neutral, faults involving ground give no zero-sequence current; an L-G fault gives no fault current and an L-L-G fault reduces to an L-L fault.
• The L-L (and L-L-G) current, 152 A, is larger than the three-phase current, 109.7 A, because $Z_2$ (0.5 Ω) is much smaller than $Z_1$ (2 Ω): $I_{LL}/I_{3\phi} = \sqrt3 Z_1/(Z_1+Z_2) = 1.386$.
• If the neutral were solidly grounded, the L-G current would be $3E/(Z_1+Z_2+Z_0) = 3\times219.39/2.75 = 239.3$ A, the highest of all. Isolating the neutral therefore removes the severe earth fault current but lets healthy phases rise to line voltage, which stresses insulation.