First of all, I do not have any electrical / electronics background, I am just a self-taught, so there are very high chances that this question does not make any sense at all. Sorry in advance if that is the case.

I am trying to understand the concept of floating voltage.

Imagine you have two isolated power supplies and you connect their terminals in series while they are under load (like the light blue wire):

enter image description here

What happens to their voltage? Do they instantaneously match? Is there any current moving between them until their potential matches? What happens to electrons, do they move at all between the transformers?

In my simulation, it seems it takes some time until both potentials match. If that is true, can I assume that I could slow that down by adding a very large resistor like in the following picture:

enter image description here

  • 2
    \$\begingroup\$ reality is there is stray capacitance everywhere, and a small charge may flow to equalize it, but it's so low we choose to neglect it to simplify everything. \$\endgroup\$ Dec 5, 2022 at 19:22
  • 2
    \$\begingroup\$ Don't think that the circuit you have shown in your question will work practically at all. \$\endgroup\$
    – Andy aka
    Dec 5, 2022 at 19:26
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    \$\begingroup\$ The circuit with DC being fed to transformers does not really make sense. In real life it's much simpler. Just think of two batteries, 1.5V or 9V or 12V car batteries. They are completely floating and isolated from each other. So whatever static charges there would be would instantly equalize when connected and they would then be a single 3V or 18V or 24V battery. \$\endgroup\$
    – Justme
    Dec 5, 2022 at 20:34
  • \$\begingroup\$ expanding on @Justme comment ... connect a positive terminal of a 9 V battery to the negative terminal of a 1.5 V cell ... then use a voltmeter to investigate voltage levels \$\endgroup\$
    – jsotola
    Dec 6, 2022 at 0:57
  • \$\begingroup\$ Carlos, when you write "I am trying to understand the concept of floating voltage", is your question specifically about the more complex situation with two secondaries in two different transformers? Or are you just generally trying to understand even the simpler case of a single secondary where the secondary galvanically "floats" relative to the primary? I can't quite tell. \$\endgroup\$
    – jonk
    Dec 6, 2022 at 3:04

2 Answers 2


Yes. When you connect two isolated things, their voltages instantaneously match, in the ideal case.

The simulation is lying to you, twice. It is assuming a capacitor is connected between the two supplies. Which is sensible in some respect as there is always stray capacitance between any two conductive objects. It's not good didactically however, as the diagram should show all of the components that it's considering. It is also assuming the wire is resistive, possibly for much the same reasons, as the graph shows a typical RC time response, which will not occur with an ideal zero resistance wire.

When you make this connection, current will flow from that capacitor through the wire, for a short while. If you replace the wire with a resistor, less current will flow for longer, integrating to the same total charge that has to move to bring the two conductors to the same potential.


The behavior will be dominated by things that don't appear on the schematic. Interwinding capacitance, stray fields, leakage resistance. If you measure a floating power supply that's plugged into 120VAC 60Hz mains you might see 40 or 80 VAC relative to earth on the output wires. But if the transformer is healthy, almost no current can flow because it's all coupled by some capacitance.

Normal transformers don't work on DC so your power supplies as shown won't work at all- you would get a pulse when the DC was switched and then the transformer would burn out. If you replace the DC sources with AC sources you'll get more sensible results.

In reality, the stray capacitance will be of the order of pF to tens of pF, so a 1MΩ resistor would equalize the voltage in a few time constants of say 5pF * 10MΩ = 5 microseconds. So within some tens of microseconds there will be no measurable potential difference. Since a meter typically has an input resistance of 10MΩ, within some hundreds of microseconds it will measure nothing (assuming DC as TimW suggests, with AC there will be other effects, and especially with switching power supplies).

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    \$\begingroup\$ "Normal transformers ... If you replace" Or if you replace the transformers with batteries -- which is a better example of isolated DC circuits anyway. \$\endgroup\$
    – TimWescott
    Dec 5, 2022 at 19:27

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