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Current Inrush

  1. Feb 15, 2009 #1
    When you have a current inrush from turning on a devise, does that mean there is actually less resistance for that brief amount of time?
     
  2. jcsd
  3. Feb 15, 2009 #2

    MATLABdude

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    In my experience (for lower powered electronics) it's usually capacitors charging up that cause the inrush current. (i = C * dv/dt) So you're actually seeing the resistance of the wires / traces leading up to the capacitors. I'm not sure if it's entirely correct to say that it's less resistance during this transient response, however (especially since the line voltage often sags too).
     
  4. Feb 15, 2009 #3
    Electric motors also have a large inrush.
     
  5. Feb 16, 2009 #4
    That's true. Even though the motor is made of inductive elements, the inertial of a load--including the armature, acts like a capacitance.
     
  6. Feb 16, 2009 #5
    In the case of a motor, it's easiest understood as a generator which generates a back emf (opposing the applied voltage) proportional to the speed. The actual resistance of the winding stays almost constant and the core losses (which can be seen as a parallel resistance) generally are less at start-up.
     
  7. Feb 16, 2009 #6
    That depends on the device. The most common example for most of us is the inrush current from turning on an incandescent lamp. The resistance really is much less when it's cold.
     
  8. Feb 16, 2009 #7
    I'm not sure that's a true statement. Capacitance implies a storage or electrical charge which in the case of a motor, you do not have.

    Significant "in rush" currents usually only occur with devices that have inductive loads or change in temperature rapidly. The motor for example has a large in rush current because when you start it, it is not moving and therefor has very low inductance. As mentioned before the majority of the loading is caused by the resistance of the windings alone which remains relatively constant.
     
  9. Feb 16, 2009 #8
    [tex]v(t)=L \frac{di}{dt}[/tex]

    Nope. No inrush current for inductive loads. Indctors dampen current changes. But conversely, interrupting current flow can cause large voltage changes, even arcing.

    [tex]i(t)=C \frac{dv}{dt}[/tex]

    Here it is. A finite change in voltage is accompanied by an infinite change in current for an ideal capacitor. For a real world capacitor, series resistance keeps the current in the realm of reality.

    For motors in general, and without going into the equivalent circuits, "voltage is rmp", "current is load".

    [tex]I = K_1 \tau[/tex]

    [tex]V = K_2 \omega[/tex]

    This last one is the so call back EMF. When the angular velocity of the shaft is zero, the back EMF riszs to the applied voltage, only limited by the series resistance. At motor turn-on, you are effectively appling the line voltage across the DC resistance of the windings.

    You would apply the rules for Thevenin or Norton equivalent circuits to see what something looks like electrically in simplifed parts to the AC line. You would need the motor equations for the rest.

    To first order, a motor appears as a capacitor in series with a resistor at start-up.
     
    Last edited: Feb 16, 2009
  10. Feb 16, 2009 #9

    stewartcs

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    Not sure what you are trying to say here...but there definitely is inrush current for inductive loads. Motors and transformers both are inductive loads and both have inrush currents.

    When power is first applied, a magnetic field must be established before the motor will operate. Until then, the only resistance is in the windings and power conductors which prevents a short circuit. Once the field is established the inductance (along with the resistance that was already there) acts to impede the current flow.

    In a transformer, once the core of the transformer saturates (which is what causes inrush in the first place), the inductive reactance effectively goes to zero. When this happens, the impedance of the transformer is equal to the resistance of the wire in the winding.

    CS
     
  11. Feb 16, 2009 #10

    dlgoff

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    High voltage transmission line transformers (and probably smaller voltage ones too), so I've been told, are designed to take the shock of inrush current by having movement in the windings.
     
  12. Feb 16, 2009 #11
    So......let me see if i got this....... In reality....for a motor, current inrush is the "true" current value respective to the resistance of the windings. however what we see when the current sags (after the inrush) is really the back emf generated from the motor actaully moving?

    and with the light buld the only reason current sags (after the inrush) is because the filament becomes hotter thus increasing the resistance?
     
  13. Feb 17, 2009 #12
    It's only simple electic circuits with some moving parts in the case of a motor.

    Any current establishes a magnetic field. The two are proportional in a simple inductor. There really isn't the idea of a delay involved. The primary of a transformer is an inductor. An induction motor is inductive as the armature completes the magnetic circuit.

    An examination of the integral for a inductor will tell you about all you need to know about the relation between voltage and current in normal applications.

    Normally a transformer is operated below saturation. Saturation is an overloaded condition for a line transformer and will cause it to self distruct. There are some few applications for saturating cores. A Helmholz oscillator is one. Saturation is not a sharp change normally, where the inductor will suddenly look like the resistance of the conducting wire; it has a knee. A notable exception is a so called square ferrite that was used in computer core memory, that also an accompanying large hysteresis.
     
    Last edited: Feb 17, 2009
  14. Feb 17, 2009 #13
    An electrical appliance can usually be considered a 'two terminal device'. This just means that it has two wires going into it. How the current and voltage behave with respect to one another for any given device can often be described by a few simple parts (resistors, capacitors and inductors) rather than the entire internal circuity.

    In the case of a motor, we're assuming it's not initially turning and the currents and voltages inside the motor are all zero to begin with. We also get to simplify things so we don't have to get an exact equivalent circuit. In such a case, the motor appears like three parts: a resistor in series with a capacitor and another resistor in parallel. To this we can add an inductor across the capacitor for the motor inductance.

    The so-called inrush current is the relatively large current experienced through the two terminals (the two wires) when a voltage is placed across the terminals.

    Edit: I see the model has to change some for an AC motor, as the effective capacitance doesn't charge and discharge each cycle. The idea here was to consider the period of time during the inrush.
     
    Last edited: Feb 17, 2009
  15. Feb 17, 2009 #14
    Series resistance. When the velocity is zero at start-up what happens?
     
  16. Feb 17, 2009 #15

    stewartcs

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    What?

    There absolutely is a delay involved. Inrush current is a transient state. It only lasts for a few milliseconds and requires a "build up" time. Magnetic fields do not magically appear, they must be created, and that takes a little bit of time. Hence, they are transient phenomenon. After the transient, it goes to a steady state, and, in the case of a motor, the average current is the locked rotor current (which is less than the inrush but more than the full load amps). Once the motor is up and running at rated speed, the current is at the full load current (presuming it is operated at full load). So for a motor you'll have a transient inrush and a "steady state inrush" so to speak. Although I do not like using the term steady state inrush, it is often used mistakenly by EE's and in some literature. Inrush should be confined to transient state discussions to avoid confusion.

    Transformers are designed to be operated just below the saturation point for optimum efficiency. There comes a point when the core cannot sustain a proportional increase in flux density even with more MMF. That point is the operating point of the transformer and begins the non-linear region of the core.

    When an AC voltage is applied across the primary winding of a transformer, the transformer must create an opposing voltage drop to balance against the source voltage by rapidly increasing the magnetic flux. This results in a rapid increase in the winding current.

    Now, during normal operation this isn't a problem since the voltage and current are out of phase by 90 degrees. However, when the transformer is cold started and the voltage is at a zero crossing (the magnetic flux and winding current start at zero now), the magnetic flux must increase in response to the rising voltage. But since it is starting from zero it will be about double what it is normally. In an ideal transformer, the magnetizing current would rise to approximately twice its normal peak value as well, generating the necessary MMF to create required flux. Since transformers are not ideal they lack the ability to avoid saturation. Since the core would now be saturated, disproportionate amounts of MMF are needed to generate magnetic flux. Hence, the winding current (which creates the MMF to cause flux in the core) will rise to a value many times that of its normal peak.

    So, as I said earlier, this causes real transformers to have a transient inrush due to the non-linearities in the core. That is, it enters the saturation region. Although it is indeed saturated, it is only for a very very short period (typically a half-wave cycle). Hence, the transformer does not destroy itself. Only nuisance tripping of breakers or fuses occurs. Thus one uses time delays or other appropriate inrush mitigation.

    If the transformer was being operated in the saturation region under normal operating conditions then there will be problems. However, we're talking about an abnormal transient start-up condition and not normal operation.

    CS
     
  17. Feb 17, 2009 #16
    [tex]\Phi = NI[/tex]
    No time delay here.

    [tex]H=B/ \mu[/tex]
    None here.

    Do you want to argue that the time required to flip magnetic domains is the source of inrush current? Or maybe the time it takes for a flux change to propagate through space from one side of an inductor to another?

    ----------------------

    Look, before this goes on any further, I'm really T.O.ed at myself that I didn't know this before.

    When the AC line is applyed to an inductor when the voltage is at zero, an entire cycle is applied to build the flux up from zero. Normally a quarter cycle takes the flux to full. So the double flux saturates the core. It's that simple and I didn't know about it.

    If the AC voltage is applied to the inductor at peak voltage, everything starts out in the right phase relationship.

    This doesn't change the nature of mechanical inertial that translates as electrical current.
     
    Last edited: Feb 17, 2009
  18. Feb 17, 2009 #17
    sorry but I don't think this is the answer my question.
     
  19. Feb 18, 2009 #18

    stewartcs

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    I presume phi is meant to represent the MMF and not the flux density??

    At any rate, that equation gives the MMF at any point for a given number of turns and current. However, in an inductor, the current requires a certain amount of time before reaching its maximum value. Typically around 5 time constants. Take a look at some texts on transient circuits or inductors and you'll see that the max current in the inductor is not reached instantaneously.

    Therefore, it should seem intuitive that since the value of "I" is increasing exponentially (and thus requiring a build up time) and that the MMF is a function of "I", that the magnetic field resulting from the current must also require a build up time.

    Anyway, it seems as if the OP has his answer so this is probably a moot point to him.

    CS
     
  20. Feb 19, 2009 #19
    Sorry about that. I cut a few corners.

    I'm sure you've seen this
    [tex]V = V_o \left( 1 -exp(-t/RC\right)[/tex]

    The complementary L and R circuit would be something like an inductor and resistor in series fed by a constant current source, maybe. I've never required an LR circuit of this kind, so your guess is as good as mine. The time constant is -L/R. L/R has units of seconds. So where L is considered a constant it's not a run-away exponential increase; it's an exponential decay to some finite value. But we care about when L isn't a constant, but decreases as the core begins to saturate.

    In any case your implied question is a good one, "What is the magnetic flux as a function of current for an inductor or an unloaded transformer?"

    There are dozens of transformer/inductor equations I haven't used in a long time, so it took some time recall and assemble the right ones. So instead of cutting corners as I did above...

    [tex]M = N \cdot i[/tex]

    [tex]M = H \cdot l_e[/tex]

    [tex]B = H \mu[/tex]

    [tex]\Phi = B \cdot A_e[/tex]

    [itex]M[/itex] - magnetomotive force.
    [itex]\Phi[/itex] - magnetic flux
    [itex]l_e[/itex] - effective magnetic path length
    [itex]A_e[/itex] - effective area of the core.
    [itex]\mu[/itex] - the permeability of the core -- not a constant, but a function of H

    This should be good for a typical inductor/transformer without a gapped core, where the initial permeability is greater than about 200 or so, and we can ignore hysteresis.

    [tex] \Phi = i \left(\frac{A_e N \mu}{l_e}\right)[/tex]

    This tells us that the flux is proportional to the current at initial permeability. Because we might know that the permeability decreases with increasing flux, we know the current will increase at an ever increasing rate with respect to flux.

    It's a nice start but it doesn't really get us what we want. We want to apply DC voltage or a sinusoidal voltage across an inductor and see how the current runs away exponentially as mu decrease, or spikes-up, respectively.
     
    Last edited: Feb 19, 2009
  21. Nov 23, 2011 #20
    I cast: "Revive thread":

    I am working on a device that could be considered an electric generator, but it can be simply viewed as a transformer. In investigating the magnetic field I have been informed that "an inductor is not an inductor all the time".

    Apart from knowing that a real world inductor can be approximated by resistive capacitive and inductive components I have always understood that the inductance value of an object exhibiting inductance was constant however you model it (Extra resistances, capacitances and inductances).

    This thread and some other sources suggest that, for whatever reason, the inductance value itself has transient properties!?1?!!?1!?1?1!?11/1/1!?!?!!?!?!

    I made a simulation in SCILAB that produces similar results to: "http://www.opamp-electronics.com/tutorials/inrush_current_2_09_12.htm" and understand that the startup condition of the inductor can be in a different state than when it is in 'steady state', as indicated in the link.

    If the thread I have replied to implied this then Im happy as cake and Ill continue with my research, however I have an inkling that the thread was hinting at magnetization, for which I have:

    B=μ(H+M), the M representing the magnetisation

    Now depending on the material M depends on H, but nonlinearly in ferromagnetic materials due to hysteresis. Magnetisation deals with moments.

    My question is: Has there been any breakthough ideas regarding this topic?

    I am especially interested in how this applies to a single low frequency loop in vacuum and a transformer with vacuum for a core – these two situations are just for my understanding.

    -CP
     
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