basic - Why does isolation transformer protect from shock?

Perhaps I can best explain the question with two examples:

(1) there is PC powered directly from 230V mains (unprotected by RCD or similar), and there is fault in PC that causes line voltage to connect to PC case. When user touches the case, he will be shocked (perhaps fatally)

(2) the same PC with same fault is connected to 230V mains via isolation transformer. Now user touches the case, but (s)he is safe from electrical shock (or at least much safer).

Now I understand why (at least I think I do), from the electricians point of view:

In first case, star of the transformer that powers the mains is connected to the ground, so electrical path is closed by human touching the case - current can go from power transformer line through case through human to the ground and back to the star of the power transformer - as explained in this question

In second case, there is no connection from secondary windings of isolation transformer to the earth, so current will have no incentive to flow from case to the the ground via human touching the case - electrical path will thus be open, so no shock.

Correct so far? I guess that measuring voltage between ground (on which uses is standing, which is same as protective grounding in electrical outlet, right?) and either end of secondary winding will show 0 volts (otherwise user standing on ground would get shocked by touching either end of secondary winding of isolation trafo).

What I am having problem is physics explanation WHY is that so? (I realize this might also belong on physics.SE, but thought this might be better place to ask than to explain isolation transformers and star topologies in p.se).

As I understand it, voltage is simple difference between two electrical potentials, and every object (be it any of secondary winding of isolation transformer or the ground) must be at SOME potential. But I guess that is probably flawed, because if one end secondary winding was at potential A, and other end of secondary winding was at potential B, and ground was at potential G, than it should stand:

• difference between potential of A and potential of B is 230V (or we couldn't power the PC from isolation transformer)


• difference between potential of A and potential of G is 0V (eg. they are at the same potential) -- or the user standing on G and touching A would be shocked.


• difference between potential of B and potential of G is 0V (eg. they are at the same potential) -- or the user standing on G and touching B would be shocked.

So, if A=G, and B=G, it would follow that A-B=0, which is obviously not true (as first point above says that A-B=230). So I know I'm wrong somewhere in my assumptions.

Could someone explain in basic physics (like movement of electrons and stuff) where I went wrong - what happens in first example, and why it doesn't happen in second example?

Answer

It's not really meaningful to talk about the voltage between two points that are completely isolated from each other. Any attempt to measure the voltage changes it.

Any real-world voltmeter requires a small current to pass in order to measure the voltage, which means it acts like a resistor. For a modern digital meter, this might be about 10MΩ. So by attempting to measure the voltage between two isolated points, you're changing the circuit by connecting them together.

If you actually tried to measure the voltage between A and G, it probably would be about 0V, but only because you've connected them together through the voltmeter. The same goes for B and G. If you connected two voltmeters at once (A to G and B to G), they would probably both read about 115V.