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Real power not the only useful form of electrical power?

  1. Apr 17, 2008 #1
    "Real power" not the only useful form of electrical power?

    Reactive power has been associated with "imaginary current" and "magnetizing power". From my understanding, this power would tend to move at right angles to the flow electrical charges. In a system that is powered by a battery, continued operation depends on the voltage difference that remaining just after the "dam has been broken" by the closing of the electrical circuit. The growth of the magnetic field depends on the coil properties and the flow of charges in that coil. If current and voltage were 90 degrees out of phase of each other, no power would be transmitted through the wire, but there may still be signficant reactive power. There would still be a flow of charges, so there would still be a changing magnetic field, though it would have to flip signs. Does the reactive power consume true power? If so, you would expect reactive power to increase in great proportion to the true power. But there are obviously many other variables. I believe there are systems are meant to run off a changing magnetic field. The changing magnetic field is due to a variable reactive power, right? Does a varying "reactive power" mathematically require "real power"? In other words, can we have a changing magnetic field when power factor is 0? Is it the case that battery powered devices deplete according only to the "real power" and that how they are depleted has nothing to do with the "reactive power" that may exist within the same circuit?
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  3. Apr 17, 2008 #2


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    Batteries are DC. They don't have reactive power. Reactive power is what you get when the amperage leads/lags the voltage.

    Anyway, in real life situations, I think reactive power and real power happen together.
  4. Apr 17, 2008 #3
    But if I add an inductor to a circuit, then I will have reactive power. This will have an effect of slowing the current fluctuations, in exchange for reactive power. The thing is that the real power is what the drains the batteries, right? So what happens if the reactive power is more than the real power?

    I agree.
  5. Apr 18, 2008 #4


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    No. Reactive power, not true power, is consumed and then returned to the source by the expanding then collapsing field (in either the magnetic or electrostatic field).

    In your example if the voltage and current were out of phase by 90 electrical degrees, then no true power would be consumed by the load.

  6. Apr 18, 2008 #5
    But this magnetic field is changing. This can produce a magnetomotive force which can cause a magnet to turn. Also, some of the energy can be lost in the form of eddy currents, which also produces a counter emf. Whatever the case, the counter emf will be sent back to the source right? But wouldn't there be traces of this power left? For example, the magnet (as well as the medium containing the eddy currents) may be in a different position (favorable or unfavorable). Yet do you mean that ALL of the power would be returned? How many times could this repeated without depleting the source? It seems that if we want to draw minimal current and maximize reactive power, we would use a long wire since reactive power can be integrated (summed) over the length of the wire while the true power is not integrated (summed) over the length of the wire, no?

    Would it be the case then that if I had batteries in this circuit that their energy content overall would not be affected by the utilization of reactive power?
  7. Apr 18, 2008 #6


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    The term reactive power has a specific meaning in industry as mentioned previously. It is normally only used in AC systems. The reactive power is temporarily stored in the magnetic field then returned to the source so no true power is lost. Of course in real life, some joule heating occurs in the conductors and a negligible amount of power is lost. But the definition of reactive power doesn't include that little fact (or least I don't think it does).

    Here is a little more reading on the subject.


    Hope that helps.

  8. Apr 18, 2008 #7


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    Once it reaches steady state (a few miliseconds), there are no current fluctuations: it is DC. So....
    There is no reactive power drawn from batteries.
  9. Apr 18, 2008 #8
    Yes. What I have though is a simple permanent magnet motor. When rotor rotates to a certain angle, the circuit is cut, creating a back spike that is over 200 volts (reduced L/R time constant). Since the time constant goes down, the resistance increases as the current increases (requiring voltages to increase), thus the huge AC voltages and currents. This produces much joule heating (although spread throughout a very large coil).
  10. Apr 22, 2008 #9
    Is there direct relationship between the reactive power and the energy of the magnetic field? Say, if you double the reactive power, you double rate change of the magnetic field energy?

    I have another question to ask. Knowing that when power factor is less than 0.5, the reactive power is greater than the true power, how much heat can be produced by an inductive DC circuit (whose charge source is a single AA battery) where 50% or more of the power is dissipated by backspikes caused by sudden breaks in the circuit?
    Last edited: Apr 22, 2008
  11. Apr 23, 2008 #10


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    The energy stored in the inductor is equal to the amount of work done by the source to establish the current in the inductor.

    Mathematically stated as

    [tex] W = L \int_0^I i \cdot di = \frac{LI^2}{2}[/tex]

    Hope this helps.

    EDIT: Here is a link with more information on the energy in the magnetic field: http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/engfie.html#c1

    Last edited: Apr 23, 2008
  12. Apr 23, 2008 #11
    Oh good idea!

    Now, I know that the reactive power is [itex]Q=I^2 X[/itex]. I would divided by sides by reactance [itex]X[/itex] so [itex]\frac{Q}{X}=I^2[/itex]. Then I plug this in to get:

    [tex]W = L \int_0^I i \cdot di = \frac{LQ}{2X}[/tex]

    Inductive reactance [itex]X[/itex] is [itex]L[/itex] times [itex]2 pi f[/itex], so:

    [tex]W = L \int_0^I i \cdot di = \frac{Q}{4 pi f}[/tex]

    Therefore it appears that is it not just the reactive power that determine the energy stored in the magnetic field but also the length of each period. During each period (assuming it is not complex), the reactive power would be directly proportional to the rate change in the magnetic field energy.
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