isolation transformer - electrocution

Question

I couldn't understand why can't you get electrocuted touching one wire of an isolation transformer, as showed in the image:

The man in the right touches one wire of the isolation transformer and presumably get electrocuted, however I couldn't figure why - He touches a live AC wire with 230V, the ground potential is 0, therefore for t such that 230sin(2pi50t) != 0, there is a potential difference between the wire and the ground, Therefore a current should flow from the wire to the man, and to the ground, no?

Answer 1

there is a potential difference between the wire and the ground,

no. Or rather, it's not fixed. The transformer doesn't connect primary and secondary side on a potential basis.

$$V(t)=230\sin(2\pi50t)$$

is wrong! You need to add a voltage offset,

$$V(t)=230\sin(2\pi50t)+V_0,$$

because the transformer's secondary is floating. That \$V_0\$ could really be anything. However, as soon as anything connects the circuit to ground, a very short (and probably not very high) current flows, bringing the secondary loop to be biased to ground.

So, the person touching the wire defines the ground point, and the rest of the circuit now swings around that. Since there's no voltage difference between hand of person and ground, no current flows.

Answer 2

You CAN get electrocuted, IF the capacitive coupling between primary and secondary provides 0.005 or 0.010 amperes.

But how large an area (overlap) between primary and secondary does this require, if the spacing is fishpaper insulation (old style transformers) of thickness 0.5 millimeter and relative dielectric constant of 5 (assumed)?

What is achievable in this coupling?

• At 230 VAC RMS, or 300 volts peak, given 10mA requires 100 ohms per volt, the impedance of the capacitive coupling at 60Hz (377 radians / second) is 30,000 ohms

• given 1uF at 1Hz is 160,000 ohms, and is 3,000 ohms at 60Hz, we'll assume we need 0.1uF capacitance between primary and secondary

• what area of overlap, at 0.5 mm spacing and Er = 5?

• Using C = Eo * Er * Area/Distance, we want the Area.

Thusly

• Area = C * 0.5 millimeter * 1meter/1000mm/ (9e-12 farad/meter * Er=5)

• Area = (0.1uF * 0.5 /1000) / 45e-12 farad/meter

and

• Area = [(0.1e-6 * 0.5/1000)/45] * 1e+12 == 1e-7 * 1e-3 * 1e-2 * 1e+12

and

• Area = 1 square meter

Thus a small transformer, operating at 230 VAC, will tingle you, perhaps badly, but should not provide the 0.01 amp (10mA) or the 0.005 amp needed to electrocute you.

Why? they may have 2" by 6" internal area of overlap of primary to secondary, but not the square meter of overlap.

=========================================

Notice the design of large power transformers will place the primary and secondary coils in separate regions of the iron core.

This separation makes the capacitive coupling much small, even if thru mineral oil.

And if the iron core is GROUNDED, the coupling is smaller yet.

Remember these large transformers have to handle lightning strikes.

Thus physical distance between the windings ---------- is a survival strategy.

Answer 3

The man in the right touches one wire of the isolation transformer and presumably get electrocuted, however I couldn't figure why

You mean this man: -

He doesn't get electrocuted because the output of the isolation transformer isn't naturally forming a current loop that involves ground/earth. In fact, one wire of the isolated secondary could be connected to ground/earth (ignoring the man) and no current would flow hence there is no electrocution risk.

However, once one wire is connected, if the man touched the other wire he would be electrocuted. But, this wouldn't happen without that initial connection because, the output is not galvanically connected to earth/ground i.e. it is floating.