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    Smoke in the Machine

    I've been a service commercial/industrial electrician for 21 years now. I've had a problem that has confused me since I was a 2nd year apprentice. As everyone knows, there are 5 types of electricians; those that don't know, those that don't care, those that don't know and don't care but only want a paycheck, those that don't know but can make-up logical fairy tales on the spot and finally, those that know but who we never seem to encounter when it comes to the NEC or electrical theory. Of all of these types...I've come across the entire lot, except of course, the ones who are truly educated. Ponder the following situation; you have a single phase 3 wire ckt, A and B phase are under load, the grounded conductor is carrying the un balanced current and someone before you didn't follow ART 300-13 "Device Removal"... you open a J-box, the solderless pressure connector "pops-off" (in search of freedom no doubt), the customers new 120v 16I machine gets force fed 208v, followed by the facility manager frantically asking, "What did you do??????" You answer, "Open a J- Box!!!!!!" The above actually happened to me in 1986 and ever since then, I've been wondering how one of the phases can double its voltage in an open grounded conductor scenario? I can recreate it "safely" with two keyless fixtures; one with a 60watt bulb and one with a 150 watt bulb, two separate ckts sharing one common grounded conductor, with a switch to open the grounded conductor. Can someone please explain to me, how does this elevation in voltage occur? I hope someone out there can explain this to me without telling me "in an open neutral...electron fairies get their wings!!!!!!" lol

    #2
    Draw the entire circuit out with the two loads and a neutral. Then calculate all four variables on each portion of each circuit... voltage (E), amperage (I), power (W) and resistance (R). Some will be a given, some you will need to calculate.

    Then erase the neutral and recalulate all the E, I, W and R varaibles again

    Ol' Georg will be proud of you when you figure it out.

    Comment


      #3
      Sparky has put you on the right track. Remember that when you lost the neutral, instead of having two parallel 120 volt loads you had two loads hooked in series on a 208-volt circuit.

      Comment


        #4
        The choice of the bulbs in your bench experiment could have a greater difference from each other.

        That is, try a 25 Watt and a 200 Watt (or even bigger) bulb.

        To get a high voltage swing on one of the two series loads, that load has to be a much higher total resistance than the other series load.

        Saying this another way, thinking of your experience back in the Eighties, the "brand new 16I machine" was probably in a low load state, that is, the machine, at its line terminals looked like a high resistance.

        And.

        Whatever was running on the other side of the multiwire branch circuit was at full load, that is, the line terminals saw a low resistance.

        Break the grounded conductor and the bulk (but not every last volt) of the voltage will appear on the line terminals of the 16I.
        Another Al in Minnesota

        Comment


          #5
          Once the grounded conductor is no longer in the circuit, what is left is the following circuit path: From the source, through load #1, through load #2, and back to the source. This is a series circuit, with the two loads (i.e., two resistors) being in series.

          The voltage provided by the source will be shared by the two loads. If the two loads are identical, then the half the voltage will be dropped across each load. If one load is much higher than the other (such as Al suggested with a 25 watt and 200 watt bulb), then the voltage will be divided between the two in accordance with their resistance values. The formula is that the voltage across each load is equal to its resistance divided by the total resistance, and that result multiplied by the available voltage. In this instance, the voltage across the 25 watt bulb would be 27 volts, and the voltage across the 200 watt bulb would be 213 volts.
          Charles E. Beck, P.E., Seattle
          Comments based on 2017 NEC unless otherwise noted.

          Comment


            #6
            The fun part about your bench experiment is the 25 Watt lamp will FLASH and blow. . .
            Another Al in Minnesota

            Comment


              #7
              Charlie,
              I think you have the drops across the lamps backwards. The 25 watt lamp will have the higher resistance and the greatest voltage across it.
              Don, Illinois
              (All code citations are 2017 unless otherwise noted)

              Comment


                #8
                Originally posted by charlie b
                In this instance, the voltage across the 25 watt bulb would be 27 volts, and the voltage across the 200 watt bulb would be 213 volts.
                Sorry, Charlie. The higher-wattage load has the lower resistance, and would have the lower voltage drop across it. The 25-watt bulb is the one that dies a premature death.

                Added: Don caught it first, but I didn't see his post until after I responded; he gets the cookie.
                Master Electrician
                Electrical Contractor
                Richmond, VA

                Comment


                  #9
                  Smoke in the Machine

                  I have read the posts for my question and am very grateful to all of you for offering your advice and trying to educate me. Thank you. I've put my code book on the shelf to cool down. It is so refresting to finally be amongst educated people in the field and I've got another question to pose. I currently have the biggest calculator I could buy and am now focused on Ohms Law.

                  I see how a parallel ckt losing the grounded conductor now becomes a series 208v ckt. So, am I to understand it that the elevated voltage is on the lowest resistive load? What I don't get is, if the two ckts that have lost the common grounded conductor are on A and B phase and A phase has the elevated voltage, how does the voltage of B transfer to A? And does that mean that all the circuits on A phase in the supply panel also experience the elevated voltage? Or is it only limited to the circuit(s) with the 16I machine now in series? If the latter is true, this makes no sense to me, then again...this is why I'm here! :grin:

                  Comment


                    #10
                    Originally posted by Karl H
                    So, am I to understand it that the elevated voltage is on the lowest resistive load?
                    No, the elevated voltage is on the largest resistive load.
                    Originally posted by Karl H
                    What I don't get is, if the two ckts that have lost the common grounded conductor are on A and B phase and A phase has the elevated voltage, how does the voltage of B transfer to A?
                    When the common grounded conductor is disconnected from the two loads, while the two loads remain connected to each other and the A and B phase, the two load's resistance become a simple "voltage divider".

                    That is, there is a single voltage between A and B. The two resistive loads add together, as far as the voltage between A and B is concerned. You have 208 Volts between A and B. If the two resistive loads total to, say, 104 Ohms, then every 1.0 Ohm will have 2.0 Volts across it.
                    Originally posted by Karl H
                    And does that mean that all the circuits on A phase in the supply panel also experience the elevated voltage?
                    No, the voltage swing on the two resistive loads occurs only on the other side of the break in the common grounded conductor.
                    Originally posted by Karl H
                    Or is it only limited to the circuit(s) with the 16I machine now in series?
                    No, it is not limited to the 16I machine. Any loads will do.
                    Another Al in Minnesota

                    Comment


                      #11
                      I cant believe it

                      Originally posted by don_resqcapt19
                      Charlie,
                      I think you have the drops across the lamps backwards. The 25 watt lamp will have the higher resistance and the greatest voltage across it.
                      I agree twice in the same day. It must be a full moon . Itis too late for me to explain tonite But this is one of my favorites I will be back.
                      The tail does NOT wag the dog.

                      Comment


                        #12
                        I will be back tomorrow

                        Originally posted by Karl H
                        I've been a service commercial/industrial electrician for 21 years now. I've had a problem that has confused me since I was a 2nd year apprentice. As everyone knows, there are 5 types of electricians; those that don't know, those that don't care, those that don't know and don't care but only want a paycheck, those that don't know but can make-up logical fairy tales on the spot and finally, those that know but who we never seem to encounter when it comes to the NEC or electrical theory. Of all of these types...I've come across the entire lot, except of course, the ones who are truly educated. Ponder the following situation; you have a single phase 3 wire ckt, A and B phase are under load, the grounded conductor is carrying the un balanced current and someone before you didn't follow ART 300-13 "Device Removal"... you open a J-box, the solderless pressure connector "pops-off" (in search of freedom no doubt), the customers new 120v 16I machine gets force fed 208v, followed by the facility manager frantically asking, "What did you do??????" You answer, "Open a J- Box!!!!!!" The above actually happened to me in 1986 and ever since then, I've been wondering how one of the phases can double its voltage in an open grounded conductor scenario? I can recreate it "safely" with two keyless fixtures; one with a 60watt bulb and one with a 150 watt bulb, two separate ckts sharing one common grounded conductor, with a switch to open the grounded conductor. Can someone please explain to me, how does this elevation in voltage occur? I hope someone out there can explain this to me without telling me "in an open neutral...electron fairies get their wings!!!!!!" lol
                        But recreate this with one 100w lamp on 1 phase and 5 100w lamps on the other phase .This will also make this more realistic of reallife situations. I also made the math so simple you are gonna puke . I will be back tommorow. I made this exact presentation in 1984 for a jatc convention no electron fairies here. 208 was a lie at least until the equipment on the other phase melted into a direct short then the see saw goes the other way
                        Last edited by quogueelectric; 08-12-07, 01:21 AM.
                        The tail does NOT wag the dog.

                        Comment


                          #13
                          For those with PowerPoint, I created a basic demonstration of this.

                          You can view it from this link:

                          http://www.code-elec.com/content/00/...st Neutral.ppt

                          Left-click to advance the slides when you are ready.
                          Hopefully, you can understand what I did with it, and that it makes sense.
                          I may get ambitious and make this into an animated .gif, but this will have to do for now.
                          I did this late at night, so I may have made some errors. Please correct me if it is wrong.
                          Last edited by 480sparky; 08-12-07, 01:37 AM.

                          Comment


                            #14
                            480, I like that. Very simple, and makes it easy to explain.
                            John from Baltimore "One Day at a Time"

                            Comment


                              #15
                              "Electricity is actually made up of extremely tiny particles called electrons, that you cannot see with the naked eye unless you have been drinking."

                              Comment

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