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Why does a phase-ground fault in an impedance grounded 3-phase system lead to variation of voltage in healthy phases but this variation doesn’t happen in solidly grounded systems?

I was thinking about using the idea of a neutral shift to explain it but turns out their both connected to ground so I see the neutral won’t float in either case so I can’t use that to explain it. Unless I’m missing something?

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Although your question is correct, I find it easier to look at it the other way around:

Suppose that originally L1-L2 = L1-L3 = L2-L3 = 100 kV and L1-N = L2-N = L3-N = 58 kV. Now you have a fault between L1 and ground.

L2 and L3 are not affected directly.

In a solidly grounded system, N = 0V by definition. L1 is now grounded, making L1 = N = 0. This leads to L2-N = L3-N = L2-L1 = L3-L1 = 58 kV, but L1-N = 0. The phase voltage of the healthy phases are unchanged relative to ground, but changed relative to the faulted phase.

In an isolated or impedance grounded system, N is not grounded, but equal to If*Z_ground (fault current times the ground impedance), relative to ground potention (0V). This time N=0 is not fixed, so instead of L1 = N = 0, you change it to N = L1 = 58 kV relative to distant ground potential. L2 and L3 are unchanged relative to distant ground, but the neutral voltage is now equal to L1. So, when you measure L2-N you will get 100 kV, instead of 58 kV.

This allows you to keep the system running normally, since the line-line voltages are unchanged. This is used in coil-compensated systems (Peterson-coils), where a faulted distribution system can operate for hours, while personell are fixing the fault. This is normal for 11 - 132 kV grids in Norway, but not many other countries that I know of.

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I think using this idealized example will help you visualize the neutral shift. Below is a three-phase ideal voltage source (e.g. 13kV system) in series with source impedances. There is a fault on A-phase as shown and no load on B&C to keep it simple.

enter image description here

The voltage from the neutral (star point) to ground is, $$V_N=-(375A\angle-5°)(20Ω)=7.5kV\angle175°$$

Knowing the B-phase source voltage is \$7.5kV\angle-120°\$ we can find \$V_B\$-to-ground (call it \$V_{BG}\$) with KVL around this loop (no B-ph load current so no drop across \$Z_B\$),

enter image description here

$$-V_N-V_B+V_{BG}=0$$

So,

$$V_{BG}= 7.5kV\angle175°+7.5kV\angle{-120}°$$

$$V_{BG}=12.65\angle{-152.5°}$$

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