# Understanding Differential Mode Voltage of a Floating Circuit?

I am currently trying to work on studying for the ECE NCEE FE exam, but I came across the problem and don't quite understand the answer.

Essentially we have an ideal transformer with an arbitrary load on the secondary side, along with the knowledge that the secondary side is floating, and there is a 120V, 60Hz source on the primary side. Now the question is asked "at which of the following points has the highest risk of electrical shock if a person were to touch the circuit?" (Assuming the person is standing on the ground)

A: Primary side before transformer

B: Primary side GND (after transformer)

C: Secondary side after transformer and before load

D: Secondary side after transformer and load

Now the solution is point A - as point A has an established ground and voltage source (and a shared ground between the circuit and person), so touching point A basically creates a short in the primary side of the transformer, which causes the full source to flow through the person touching it.

What I don't understand is the elaboration of the solution: "...The transformer is ideal, and the load is floating. A fault condition does not exist. Therefore, there is no differential voltage to ground at point C or D."

What I don't understand is how is this possible? Doesn't the transformer provide power to the secondary side, thus creating a flow of current and potential across the load? Wouldn't grabbing point C also be dangerous as well, and possibly D since that node isn't common with earth/person ground - as the potential difference again is basically creating a short when the person touches point C to where they stand?

Basically my confusion is to how the floating side is somehow harmless when you have energy being input from the primary side and creating a potentially dangerous, unknown voltage across the load? I've always been taught that floating circuits are dangerous, not harmless.

• what do you mean "before" and "after" the transformer? Nov 3, 2020 at 3:15
• @Hearth Tracing the primary and secondary loops' circuit paths, the way 4 comes "after" 3 on an analog clock, and 1 is "before" 2 on an analog clock. Since circuits are conventionally traced from the positive side, assume A, "before", is on the positive side. Refer to the attached image as well. Nov 3, 2020 at 3:35

The secondary circuit is coupled to the primary supply by the magnetic flux across the transformer. They do not otherwise share a common reference to ground.

If you as a grounded person touch either C or alternatively D you will effectively become the ground reference which anchors the relative potential of the secondary circuit. That is the secondary circuit will no longer be floating.

Touching both C and D will give you a shock as there will clearly be a voltage between them irrespective of whether you are grounded or not.

Clearly the winner in this case is A as the primary circuit has a ground reference (is not floating) and neither are you (as you're grounded) so a shock is guaranteed.

• My question is to focus on your second paragraph - the case where a person touches point C OR D: does this not still shock the person, since they are grounding the floating circuit through their body? Nov 3, 2020 at 3:33
• If the circuit were perfect with no interwinding capacitance and no capacitance to ground then yes. But this perfect circuit does not exist in the real world. See my answer to explain why this is a bad FE question. A is the answer they are looking for btw. Nov 3, 2020 at 3:45
• This is one of those "Best answer in the given circumstances..." type questions where other answers are potentially true but less than the one they're after. Irrespective of the other possibilities of risk for a shock the guaranteed answer is A. Nov 3, 2020 at 4:05
• The question as asked seems to imply that there is no potential between C/D and B. While it's true that (in the theoretical ideal) you wouldn't measure a voltage between any of the primary or secondary terminals, it's also true that when a connection is established, the connected node in particular has no voltage relative to B, but now that a reference has been established, other nodes on the secondary side will have a voltage relative to B. (The same is true for A, but B is used in this question because it's considered a ground reference.) Nov 3, 2020 at 16:58

I just have to say, this is a really bad question for the FE. It is bad, because it is dangerous. Every point shown is a risk of shock in the real world. The answer A is correct from the standpoint of highest ground fault current available. But, the whole question can imply that it is "safer" to touch B, C, or D. I have seen more than one fatality from contact with ungrounded or supposedly de-energized equipment.

If they had simply framed the question as one of "how much ground fault current is available at this point" etc. it would have been much better.

Lastly, there is no such thing as an ungrounded circuit. There is always parasitic capacitance. See my answer to this question for details.

p.s. I know this is “just” 120V and less likely to kill someone. But it gives a false impression that someone could apply elsewhere.

The risk factors for an electrical shock would be

1. One of the high voltage supply terminals being earthed.

2. Person standing barefoot on the ground.

3. Person coming in contact with a live point with respect to earth.

Hence the point of maximum risk would be A, being the only live point with respect to earth.

A 'floating secondary' implies no direct contact with the primary and with earth. A leakage through the capacitance between the primary and the secondary would be taken care of by an earthed shield between them.

Any point on the secondary having contact with earth, would be a necessary condition for it to cause an electric shock. In other words, a closed circuit,conductive path through the body and earth would be a must.

The risk at any point on the secondary would be minimum since it would be isolated from the primary and no point on it would be earthed.

Of course, one should not rule out the risk of coming in contact with both the terminals of the secondary. Added to that would be the risk of a ground fault on the secondary or insulation failure between the primary and the secondary, with a simultaneous failure of fault tripping mechanisms.