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20 ohms

Discussion in 'General Electronics Discussion' started by sector5, Jun 6, 2016.

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  1. sector5

    sector5

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    Jun 6, 2016
    Hi all,
    Have something that does not make sense to me.
    In laws had a breaker trip I put a meter on the cct and found 20ohms on all the GPOs, tracked it down to the door opener can't find any info on it
    But to me 240v 20ohms = 12 amp
    The service guy told them this is in spec
    Just wondering how this is correct


    Thanks

    Simon
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Welcome to EP.

    What kind of door opener?
    The small buzzing type that opens the latch or a motor that openes the door?
    For the buzzing type this sounds illogical. Are you sure the door opener is operated at 240V, not on 12V or a similar low voltage, that makes much more sense in terms of power consumption and safety. 12V/20Ω=0.6A, a reasonable short term value for acuating the door opener.
    For a motor this is possibly within specs. A quick Google search for motorized door openers turns up models that use from 3A to 15A, depending on the specific model.
     
  3. BobK

    BobK

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    Jan 5, 2010
    Also, for a motor, the resistance will not indicate to the current used when the motor is running.

    Bob
     
  4. sector5

    sector5

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    Jun 6, 2016
    Plugged in to the mains 240 that's where the 20ohms is, this is when unit is in standby. I am sure there would be a step down transformer I the unit but don't want to open the unit and void warranty. I can't see that you could have a device that puts 20ohms on all GPO's in he house.
     
  5. sector5

    sector5

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    Jun 6, 2016
    12A on stand by seems a bit high will put a clamp meter on it when I get a chance
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Definitely. On standby, the current should be near 0 A (depending on the controller for the motor).
    But beware of the inductive component. Assuming, as you do, there is a transfomer, you need to look at the AC current consumption, not the DC resistance.A transformer has a rather high inductive component (which will not show on your multimeter). The impedance (AC resistance) of this inductance is Xl=2*pi*f*L with f being mains frequency in this case and L being the inductance.
    See for example here.
     
  7. Minder

    Minder

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    Apr 24, 2015

    Measuring the resistance on an AC circuit does not indicate current if inductive devices are involved.
    Which on a door opener I would expect this.
    The Inductive reactance (ohms) will be the deciding factor.
    M.
     
    Last edited: Jun 9, 2016
  8. sector5

    sector5

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    Jun 6, 2016
    Ok I get it, but is it still 20ohms resistance and if I had 3 door openers on the same circuit would I not have a 6 ohms resistance on the circuit? Or do they not add on parallel ?
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    They do. But you should measure only the resistance of an active door opener (one that would start to operate if mains is present). When the door is closed or fully open, the motor of the door opener should be disconnected - no use to waste energy on an open or closed door.
     
  10. Minder

    Minder

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    Apr 24, 2015
    In a AC motor circuit/solenoid/contactor coil , the Only time the circuit 'see's' the device resistance is at the point of switch on for a fraction of a second, after this the current is limited by the Inductive reactance of the motor etc which is much higher than the measured resistance by ohm meter.
    M.
     
  11. sector5

    sector5

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    Jun 6, 2016
    So would you expect to see 20Ω when the unit is unplugged no power and 20Ω when on mains on all GPO
    on that cct ?
     
    Last edited: Jun 9, 2016
  12. Minder

    Minder

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    Apr 24, 2015
    Yes.
    It is normal to see very low resistance when measuring a non powered inductive device.
    A extreme prime example is a induction motor, not only is there the initial low resistance of the windings, but when powered the 'secondary', (rotor) appears as a shorted turn secondary to the stator windings, representing a shorted turn transformer, until it starts to rotate.
    M..
     
  13. sector5

    sector5

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    Jun 6, 2016
    sorry so a AC act won't see the DC reactance ?
    so the DC reactance should not be present when the unit is switched on?
    sorry if i ask the same thing in ten different ways, When i done my training I was told they use a very high winding so on the mains side so this would not happen
     
    Last edited: Jun 11, 2016
  14. Minder

    Minder

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    Apr 24, 2015
    On an AC inductive device at switch on, the current is dependent on whatever the DC resistance of the device is, this resultant high current occurs for a fraction of a second, less than 1 cycle, fuses and breakers have a certain time delay which does not respond to this, after this inrush, the inductive reactance is taken into account which has a frequency component, 50 or 60Hz.
    There are devices that extend the period until the final inductive reactance value occurs and this is Induction motors, solenoids and AC contactors, where in the case of the latter two, inrush is extended until the armature has shifted to its operated position.
    This is why it is generally preferable to use a DC counterpart for these devices.
    It is seen in practice where in the case of maintenance personnel servicing a malfunctioning piece of equipment equipped with AC solenoid valves.
    e.g. In an effort to trace a problem, the armature on an AC Hydraulic solenoid is pushed over manually to attempt to trace a problem, if the opposing coil is energized when the armature is physically pushed over, it results in a burnt coil. and/or a blown fuse.
    M.
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Ok, time for a basic lesson about inductors.

    You clearly understand what a resistor does. It (in non mathematical terms) "resists" the flow of current based on the instantaneous voltage. (yeah, "flow of current" is like "ATM machine"). You're also aware that this results in heat being generated and power being lost.

    So a resistor with 12A flowing through it and 240V across it will both get very hot AND make your power meter spin like a top.

    Inductors are different. They're so different they're even mathematically analogous to walking along reality and taking a sharp left into the twilight zone.

    inductors have the property of resisting any change to the flow of current. You might consider them like a flywheel. In DC terms, all you see is the losses in the bearings (the turning resistance). But this is (or should be) a very small figure that represents a deviation from perfect behaviour (as opposed to brakes where you actually want them to impart a turning resistance)

    So what determines the "effect" of a flywheel? The answer is inertia. We can derive a measure of this which will have something to do with mass and distance.

    Another thing about this flywheel is that it doesn't get hot while operating. Why not? Because although it requires energy to change the speed of rotation, it actually gives back energy when you try to slow it.

    so back to inductors. They resist a change in current. The amount they do this is measured in different units than a flywheel, but are analogous.

    lets imagine a perfect inductor connected to a perfect dc source and a perfect flywheel "torque source". (A torque source would be a source of twisting force). The perfect flywheel (presumably initially not rotating) will start to spin, spinning faster and faster and faster and faster... It is a perfect flywheel and a perfect source of torque, so the flywheel continues to go faster without end. The same thing with a perfect voltage source and a perfect inductor. The current rises from zero and continues to rise and rise, and rise... without end.

    How fast does the flywheel change its speed? This has to do with the magic measure we had previously. We can predict this. What about the inductor and the rate of change of current? Well, I said it had an equivalent measure, thus this is also something we can predict -- i.e. it will be orderly, it's not a short circuit.

    Now imagine that after some time we flip a switch and our voltage source is swapped end over end and the torque source is reversed. What will happen? The flywheel will do what flywheels do. The torque is now in the opposite direction, so the flywheel will slow down, BUT it does this while the direction of rotation is the opposite to the torque -- so energy is flowing from the flywheel to the torque "source". The flywheel will be giving up the energy it stored to what is now a load. The same thing happens with the inductor. The voltage gets reversed, but the current continues to flow in the same direction -- the current is flowing in a way that appears backwards. Power is coming FROM the inductor and being passed back to the voltage source. At the same time, this current is reducing.

    Over time, in the above cases, the flywheel will slow to a stop and begin rotating in the reverse direction. Likewise the current in the inductor will fall to zero and then begin to flow in the opposite direction.

    If we start looking when the flywheel is stopped (and the current through the inductor is zero) then apply the sources for a while, then reverse them. We stop looking when the flywheel stops and the current through the inductor falls to zero. If we measure the amount of energy put into the flywheel, we find the total is zero. Sure we put some in, but it gave it all back. We find the same with the inductor. From this we can conclude that both had a net zero load on our source.

    Let's consider a more typical flywheel and inductor. The flywheel has imperfect bearings (they have resistance) and the inductor has electrical resistance. If we repeat the above experiment we find that there is some net effect, and that is due to resistance. If we push ourselves even further into reality we find that the flywheel explodes if we spin it too fast and the inductor melts if we allow it to carry too much current.

    So we know that both of these are energy storage devices, and that a measure of "resistance" is one measure of their imperfection.

    In the real world you have one of these imperfect inductors and it is places across an alternating voltage source (which we can consider perfect). The current rises through the inductor to some value, and then falls through zero to a negative value, and then rises through zero to some positive value, and so on. We know that this action has a net zero effect on the voltage source (and your power meter will confim this). The losses come from the resistance.

    How much do these losses contribute? To answer this we need to know more about the magic and unspecified units that we measure inductance with and how this measure can be used to determine the current through the inductor.

    I'm not going to get into the math, but if you measure the inductance then, knowing the frequency of your AC source and it's voltage, we can determine an AC equivalent to resistance which will allow us to talk about current. This equivalent is reactance (which has been mentioned before) and the important thing about it is that it involves energy storage and release, not dissipation.

    Through various tricks that allow us to compare an AC waveform to an equivalent DC voltage, we can determine the losses through the DC resistance that you measure.

    SO, if you can measure the inductance of the device, knowing the voltage and frequency of your mains, you can plug in these values (along with the resistance) and determine the total power dissipated. I can guarantee the max current will not be 18A, and the losses will not be 240*240/20 W

    Incidentally, you may have thought you understood the flywheel but wonder how the inductor stores energy. The answer is more complex than you think. The inductor is simple, it creates a magnetic field (which requires energy) and pulls this energy back (collapsing the field). The flywheel is far more complex and probably involves the Higgs field. Whilst we have the mathematics to model both easily, we do not have the physics to explain the flywheel as adequately!
     
    Harald Kapp likes this.
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