Electronic – Transformer Insulation – High Voltage can Jump Through Ferrite Core

is it normal to get false results with those probes (or voltage dividers) on high-frequency voltages?

Yes, without a proper frequency compensation. It happens because resistors have small parasitic capacitance, which can be modelled as a capacitor in parallel with a resistor. These parasitic capacitors form a capacitive voltage divider for high-frequency signals. If the ratio of the parasitic divider differs from the ratio at DC, you will get wrong measurements, since the overall ratio becomes frequency-dependent.

Usually this is not a problem at kHz range. But not in the case of high voltage, which implies high-value resistors. The capacitance of a typical resistor is approximately 1.5 pF, which gives 3.3 MΩ at 32 kHz for a pure sine wave. Because you are using high-value resistors, the parasitic capacitance becomes the dominant factor even at kHz-range frequencies. If a signal is not a pure sine wave, i.e. it contains high-frequency harmonics, the parasitic capacitance dominates even more.

Do deal with the problem, add a compensating capacitor (typically, it is a variable capacitor). To get a frequency compensation the following condition must be met $$\frac{R_2}{R_1 + R_2} = \frac{C_1}{C_1 + C_2}$$

This can derived from the ratio for a capacitive divider $$\frac{\frac{1}{j\omega C_2}}{{\frac{1}{j\omega C_1} + \frac{1}{j\omega C_2}}}$$

The easiest way to test a divider is to look at a divided square wave signal via an oscilloscope. With the right compensation, the square wave looks like the scaled square wave. Without the right compensation, your will see a signal with a strange shape. That's because the ratio of uncompensated divider depends on a harmonic number, and after the division the harmonics do not sum up to the square wave.

I'm not sure that the frequency compensation is the only problem; probably there are other issues related to a noise in the measurement circuit.

Also, typical 1/8W resistors are not suitable for 1.2 kV RMS. The maximum allowed voltage for such resistors does not exceed 100 V RMS, if I remember correctly. Consult the datasheet for the exact value.

edit

One way to get proper division is to use 10 nF capacitor as a part of the divider

^{simulate this circuit – Schematic created using CircuitLab}

Note that $$\frac{100\,\text{kΩ}}{10000\,\text{kΩ} + 100\,\text{kΩ}} = \frac{10\,\text{nF}}{10\,\text{nF} + 1000\,\text{nF}}$$

So you have a pretty standard SWER setup. Turns out because SWER is considered "low-cost", local electricians tend to build it "even lower cost" and skip the secondary grounding. Like this:

^{simulate this circuit – Schematic created using CircuitLab}

From what you described, my bet is the shared ground rod is simply too small or too rotten or the earth humidity is too low. The reason why the phases on the secondary behave differently is because the secondary current is in phase or in counterphase with the primary ground current.

The problem with this shared ground rod is, the grounding needs are different for the primary and secondary.

The utility company electricians (who I expect they know what they are doing) installed a primary ground rod for a constant current of a few amperes, and a loss of a few dozen volts from 10kV doesn't matter much. So, primary grounding doesn't give you headaches. A small rod suffices, even in dry earth.

On the secondary, you don't have a constant ground current of a few amperes, but instead, you want to have a ground fault current of a few thousand amperes which should trip the breaker immediately. Three problems with the above setup arise:

• The primary grounding is built much too weak for these ground currents. You will have a lift of the neutral to the 10kV level in case of ground fault. This can break even more insulations on the secondary. Random appliances in the house may be broken after a ground fault.

• The ground fault current may damage the primary ground rod, which isn't made for currents of thousands of amperes, not even for a second.

• The breaker may not trip because of the above two problems.