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Residual-current device & isolation transformer

  1. Apr 11, 2010 #1
    Hi guys,

    I don't understand why when one touches both the live wire and neutral wire in the secondary of an isolation transformer, the current in the live wire and the neutral wire in the primary will be the same. This is described in my book as the reason why a residual-current device in the primary can't do its work in such situation. I understand that part. But why will the current in this situation be the same in both wires, while if its done without an isolation transformer the current will flow only to the earth, thus causing a current difference in the wires?
  2. jcsd
  3. Apr 13, 2010 #2
    Could someone please help me out? I'm stuck here. I really wish to be able to continue my studies of physics with the feeling I understand this part. It's not explained in my physics book and it's probably not so difficult to explain.
  4. Apr 13, 2010 #3
    The output of an isolation transformer is, well, isolated. Duh. Hence current will not flow to ground only between the two output wires. A normal mains circuit has one side grounded out at the street transformer. Hence if you touch the "live" side and are grounded a current will flow through YOU to ground and not back through the neutral wire to the mains ground. This can be detected and used to trip a safety breaker. OK?
  5. Apr 13, 2010 #4
    I understand how the Residual-current device works. I only asked about the part I've quoted. I am trying to understand how it's possible that the current going into the secondary of the isolation transformer is equal to the current coming out of it when you are touching both live wire and neutral wire. Could you elaborate on that please? Thank you very much for your help!
  6. Apr 13, 2010 #5
    An isolation transformer has two output terminals.

    If you grab onto both of them, what goes into one must come out of the other, as you complete a circuit.

    Hence the current is the same value in both as it is the same current - it is the current in the circuit.
  7. Apr 13, 2010 #6
    It has two output terminals? I don't understand. I'm trying to understand everything in terms of electrons flowing. Could you explain it like that please?
  8. Apr 13, 2010 #7
    Years ago I used an isolation transformer for safety, when working on "hot chassis" TV sets and monitors, I understand your question very well. I agree a "good" answer has not been given yet.

    I assume the electrons in the secondary winding are sloshing around this "wire" at 60Hz, if you grab one side of this winding, do your electrons simply "slosh" along and cause no problem?

    But if you grab a "hot" line, electrons 60Hz "slosh" through you to ground,(sometimes fatally). I am missing something here from an intuitive perspective. This has been a subject here before, apparently without a good "intuitive" explanation.

    Perhaps, "what are the electrons doing differently" is the question, with regard to an Isolation Transformer and Human Safety.
    Last edited: Apr 13, 2010
  9. Apr 13, 2010 #8
    Oh, I do hope someone comes with a clear explanation. Thanks for your concern :).
  10. Apr 13, 2010 #9
    I "googled" the question, the "clear explanation" seems elusive................
  11. Apr 13, 2010 #10
    What is to not understand?

    Two wires - two terminals - one for each?

    Forget about electrons, Do you know what an electric circuit is?
  12. Apr 13, 2010 #11
    NO there is NOT two wires, the secondary winding is ONE wire, with TWO ENDS.

    Please explain the electron flow differences, between the secondary 110VAC on the isolation transformer and the 110VAC "hot leg" line voltage. Both are induced, correct? your 110 leg is induced at the neighborhood transformer for your home, correct?

    Why is one 110VAC source lethal and the other NOT, I can go into more detail if you like

    Why WOULD the 110VAC secondary NOT be hot when referenced to ground......................

    What "IS" a circuit that produces 110VAC from a secondary, but no current flow to ground. I know "HOW" this circuit is constructed, but no theoretical reason for it not to be hot referenced to ground.
    Last edited: Apr 13, 2010
  13. Apr 13, 2010 #12
    Perhaps if you stopped shouting and started thinking you would see that the OP specifically said there are two wires and labelled them.

    I quoted this.

    And yes thank you I have a good knowledge of electric safety theory.

    Bjacoby already provided a pretty good answer to the OP, I was just trying to rephrase it in more everyday language.
  14. Apr 13, 2010 #13
    I know what an electric circuit is. I don't understand what you are trying to say. Also I think you don't understand what I'm trying to find out since you say that Bjacoby has provided a pretty good answer to me. Try to read my post carefully, maybe you will realize where I'm stuck and how to help me out. I'll try to rephrase my problem again though.

    It actually comes from a question posed in a book: Explain why when applying an isolation transformer, the residual-current device can't do its work.

    The answer given by the book is: If there is current leaking in the secondary of an isolation transformer, then there will only be an increase in demand of power in the secondary circuit. In the primary of the isolation transformer this power is delivered, there will be an equal current in the live wire and the neutral wire, thus not allowing the residual current device to respond.

    I'm trying to understand how the isolation transformer causes this to happen. And why this problem supposedly isn't there without the isolation transformer. And I think the only way I will really understand this is if someone could tell me the physics in both situations in terms of electrons.
  15. Apr 14, 2010 #14
    So you have two input terminals I1 and I2, the wires from there both go through a rcd and afterwards I1 is grounded. This way touching I1 will not do any harm. When touching only I2 the loop is closed through the ground tap, so the current goes in through the rcd and back outside. Thus the rcd activates.

    The two outgoing terminals O1 and O2 are being fed by the secondary coil, which gets its energy through the magnetic field. In principle this potential difference can be applied to anything. It is like a battery and you can connect it to a point of any potential and add that voltage. The voltage difference is only between these two terminals. If it is ungrounded you can touch either one of O1 and O2 and nothing happens. If O1 is grounded and you touch it nothing happens, if you touch O2 current will flow through ground to O1 not to I1!
    There it closes the loop of the secondary coil, sucking energy from the magnetic field, making the current flow in the primary, which will go through terminals I1 and I2 and thus through the rcd and not around it. Thus the rcd doesn't react.
  16. Apr 14, 2010 #15
    Sure there's a reason for it not to be referenced to ground. Since you know how it's constructed think about it. An isolation transformer has a single piece of wire run through it around and round the core. This wire is INSULATED from everything! All the voltage appears BETWEEN the two ends of the wire. And like any insulated piece of metal if you touch it, you ground it through you but since it is just insulated metal (and not charged...I'm NOT going into leakage in an isolation transformer!) there is no current flow. However if you grab each end of the secondary wire with a hand you CAN get killed as the output of the wire that is the secondary goes through you. But a safety device will pickup no difference and won't trip. Look. Ever see a clamp-on ammeter? Why don't you get a shock using it? It's an isolation transformer! The primary to the transformer has one side (neutral) connected to ground. Hence there are TWO paths back to the transformer in the street. One is back the neutral wire and the other is through YOU and back through ground. Notice that when you are being electrocuted the current in the neutral wire is not equal to that in the "hot" wire because part of what normally flows in it is now going back through the ground.
  17. Apr 14, 2010 #16
    I dropped out of College my Senior Year,(Solid State Physics), I actually completed requirements for P.E. "Electrical" in Colorado. Most of my 20+ year career has been systems integration for Automation, I have years of application experience, and almost four years of Electrical Engineering and Solid State Physics, yes I have a Fluke Amp Clamp and am aware of it's operation from a Practical and Theoretical perspective.

    It is interesting, that only two people,(oops, now three), yourself and two others have attempted an answer to this question, there are a lot of smart people here, correct? I believe one other person attempting an answer still doesnt understand the "concepts", Perhaps I didnt understand the question and made my own "isolation transformer" question from this post,LOL.

    I spent a fair amount of time last night trying to figure out why no Voltage difference to ground, from the ends of the "Insulated" Secondary Winding,(wire).

    It "seemed" to me from an "intuitive" perspective, the "unloaded" Secondary has electrons sloshing back and forth at 60Hz, causing a 60Hz difference in potential at the "each" end of the wire, a "Standing Wave". I now believe this to be INCORRECT. I believe NO voltage difference exists "BETWEEN the two ends of the wire",nothing is induced,no wave, UNTIL a load is applied.

    Perhaps, better said, this "wire" is "balanced",(like a faraday cage), from the inducing primary, until a load is applied.

    I believe a person can become an extention of this "open" secondary, even when touching ground, but the extention means nothing,(from a Potential viewpoint), UNTIL the secondary is closed. This is the answer I was looking for yesterday. Is this a more "intuitive" confirmation of your above statements? Any comments are welcome.

    Now Ill start chewing on the "original" leakage question..................................

    Funny more people have not chimed in, hopefully they will. Regards, John
    Last edited: Apr 14, 2010
  18. Apr 14, 2010 #17
    Sooooo, back to the "actual" question.............

    Explain why when applying an isolation transformer, the residual-current device can't do its work.

    The answer given by the book is:

    If there is current leaking in the secondary of an isolation transformer, then there will only be an increase in demand of power in the secondary circuit. In the primary of the isolation transformer this power is delivered, there will be an equal current in the live wire and the neutral wire, thus not allowing the residual current device to respond.

    Wiki seems to have a decent RCD page:


    This quote is from the Wiki page:

    "RCDs operate by measuring the current balance between two conductors using a differential current transformer. This measures the difference between the current flowing out the live conductor and that returning through the neutral conductor. If these do not sum to zero, there is a leakage of current to somewhere else (to earth/ground, or to another circuit), and the device will open its contacts."

    bjacoby made reference to the difference in current flow..........

    "If there is current leaking in the secondary of an isolation transformer, then there will only be an increase in demand of power in the secondary circuit."This is obvious, correct?

    "In the primary of the isolation transformer this power is delivered, there will be an equal current in the live wire and the neutral wire,"This is obvious also, more demand on secondary, more demand on primary, correct?

    "there will be an equal current in the live wire and the neutral wire, thus not allowing the residual current device to respond"This is the "REAL" question. Perhaps they should specify RCD on the "primary". "IF" RCD was on the "primary" the RCD would be not "see" the "leakage" on the "secondary". We "assume" no leakage on the primary, so no matter what the secondary does, the PRIMARY line and neutral have EQUAL current flow, hence RCD will NOT trip.

    Would'nt RCD on the primary defeat the purpose? Perhaps this is the "point" of the question..................

    Sound Correct?
    Last edited: Apr 14, 2010
  19. Apr 14, 2010 #18
    Ok let us settle this with the aid of some sketches.

    It is almost always a good idea to draw a diagram - I was rather hoping someone else would do one.

    Sketch 1
    Shows an isolation transformer with two terminals labelled L & N

    Sketch 2
    Shows what happens when a load is placed across the terminals. This load may occur if a human grabs one terminal in each hand and stands on an insulating mat.

    It also shows the equivalent circuit. Since there is one loop or circuit a single current, I, flows in the entire loop through the load or human.

    Sketch 3
    I have inserted a residual current breaker in the circuit. There is still one single loop so a single current flows.

    In the equivalent circuit I have shown how the breaker operates.
    The current passes through two solenoidal coils which operate a common plunger switch.
    However as you can see the current from the L terminal flows
    in the opposite direction
    to the return current into the N terminal.

    These two currents produce opposing forces on the plunger.

    So long as the currents remain equal the forces are in balance and the switch is held closed by a spring.

    If for some reason the currents are not equal then the forces will not be equal and the spring will release the switch.

    Note I have not needed to mention electrons or ground in this explanation.
    Once the principle of operation is understood the effect of ground can be discussed.

    Attached Files:

  20. Apr 14, 2010 #19
    Is the "answer" to the "question" anywhere in there?

    Explain why when applying an isolation transformer, the residual-current device can't do its work.?
  21. Apr 14, 2010 #20
    The OP actually wrote:-

    Still too busy shouting to read other posts properly?

    The OP clearly understood the device can't do its work because there is no residual current since the current is the same in both wires.
    What he didn't understand was why the current is the same.
    Last edited: Apr 14, 2010
  22. Apr 14, 2010 #21
    Did your "settle this" post, in anyway address this or provide resolution?
  23. Apr 14, 2010 #22
    Yes the answer to the question
    "Why is the current the same in bothe wires?"

    remains that they are the same because they are part of the same loop or circuit and there is only one such loop.
  24. Apr 14, 2010 #23
    I believe the previously "missing" answer is that the device was on the "primary", hence current being the "same".....................
  25. Apr 14, 2010 #24
    Credentials are irrelevant here. The important thing is an understanding of what is going on.

    Since everyone is stuck on "electrons" let's go there. What happens in the transformer secondary (an isolated, insulated piece of wire) is that an electric field is induced in the wire by the transformer and it's primary currents. This electric field is primarily along the wire. The electric field is a function of time following a sin function at 60 Hz. There are no "waves". (although electric fields do travel at the speed of light) This electric field is ALWAYS there so long as the transformer is "on".

    So if the secondary wire is not connected (open) This electric field causes electrons to move in the wire in the opposite direction to the field and they "slosh" toward the end. But they can't go anywhere and build up there. That in turn creates an electric field in the opposite direction balancing the first field. Hence there is a POTENTIAL between the ends of the wires. The potential is a function of time varying at 60Hz. So yes, there is a voltage there even if the ends are not connected. But there is no current except for the very minor "sloshing" needed to build the canceling field.

    Now if a load is connected between the ends of the wire the electrons being sent to the wire end have a place to go. They go through the load and back in the other end of the wire. In other words the electric field in the wire is like a propulsion unit sending electrons round and round the "circuit" consisting of the secondary wire and it's load. The flow of electrons is a current and hence a current flows.

    Note that this current is AROUND THE WIRE AND LOAD. The whole business is isolated from ground by "induction" and grounding it does nothing.

    Does that help?
  26. Apr 14, 2010 #25
    I "think" I like it! Thanks
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