Why does isolation transformer protect from shock?

Question

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 1

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.

Answer 2

The keys to understanding this are

1. what causes shocks is current through the body.
2. the introduction of any extra components into a system can change the voltages. The human body is no exception to this.

Lets consider an experiment. Suppose I have two terminals, one is at ground potential and one is connected to a 230V (relative to ground) supply. Now suppose I have a 10 megohm resistor, lets call the two ends of that resistor A and B. Firstly I connect A to the 230V supply terminal and leave B disconnected for now. Both ends of that resistor are at 230V.

Now suppose I connect my multimeter between end B and ground. My multimeter has a 10 megohm input resistance (this seems to be a de-facto standard on virtually every digital multimeter) so the voltage on B will drop to 115V.

Now suppose (DO NOT DO THIS) I remove the multimeter and touch B and ground. Assuming the resistor is not faulty I do not get a shock because the current will be less than 23uA (the exact value will depend on the resistance of the body which is somewhat variable).

Now getting back to the isolation transformer case. Lets ignore the PC, it just complicates matters.

Suppose there is nothing connected to the output of the transformer. We know the voltage between the two output terminals. We do not know the voltage between the output terminals and ground, with no low impedance path to tie it down it could be virtually anywhere.

Now suppose you touch one end of the transformer and ground. The voltages now shift around so that the end of the transformer you just touched is at damn near ground potential. If you touch the other end then the same thing happens. It's only if you touch both ends at once or you touch one end and the other end inadvertently ends up connected to ground that a significant voltage can be maintained into the load that is your body and hence deliver a current that will give you a shock..

Answer 3

i think that the point is that the secondary side is floating , that's all.

Floating reference

We say that the isolated part of the circuit is "floating" with respect to the power source. Since there is no direct path for current to flow back to the power source, there is no way to keep it aligned to any reference relative to the power source. It's similar to a balloon that has nothing tethering it.

Instead it weakly finds its own potential according to whatever tiny leakage currents are available - if you connect it to something, it will immediately align itself to the potential of whatever you connected it to. In itself, it won't put up any resistance to adopting that reference; it is not connected directly to anything that could provide resistance.

A familiar case of galvanic isolation is a battery-powered instrument like a multimeter, since it has nothing fixing its reference potential it will float to whatever reference it is connected to; for example in itself a battery powered multimeter can be directly connected to the mains for measurements. It still works because it is floating at the mains potential, so its "0V" is at the mains potential, and its battery is "mains potential + 9V". Local to the multimeter nothing changed.

source

Answer 4

What you suppose about that A and B terminals of the insulation transformer secondary, are at 0 volts with respect to earth, it is incorrect.
It is very difficult to determine a priori the potential difference between any of the terminals of the transformer secondary and ground.

The earthing system you describe with the transformer is called IT system. This scheme assumes that there is a high impedance between any point in the secondary and ground. The protection system suitable for this type of installation, includes a line isolation monitor.
This connection system is the only one that allows operation occur after the first failure, unlike other systems, which in the first failure, the power supply (circuit breaker) is disconnected.

In the case of indirect contact, although a ground fault occurs that can be monitored, the user is not at risk because the isolated circuit is not closed by ground.