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How do Capacitors and Inductors Add/Remove VARS in AC?

  1. Feb 26, 2016 #1
    To my understanding, capacitors cause the current to lead the voltage which adds VARS to the circuit and inductors cause the current to lag behind the voltage which removes VARS (in AC circuits). Also, it is my understanding that VARS increase voltage. I work for one of the largest utilities in the nation as a system operator and literally no one I ask can give me a halfway decent answer on this.

    My questions are the following,
    How does the current leading the voltage (adding VARS) raise voltage?
    How does the current lagging the voltage (removing VARS) lower voltage?
    Why is Reactive Power (VARS) isolated to a specific area of a circuit and is unable to travel freely through the circuit like Real Power (Watts)?
    What is physically happening in the capacitor that causes current to lead voltage and add VARS?
    What is physically happening in the inductor that causes current to lag voltage and add VARS?

    If I'm wrong about any of my assumptions or understanding please call me out on it. Also, if you only know the answer to some of my questions or have any info, please share what you know. I'm very interested in getting any information on any of these topics.
     
  2. jcsd
  3. Feb 26, 2016 #2

    BvU

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    Hello rugscrub, :welcome:

    Can you tell us what you mean with VARS ? I'm pretty sure that I am not the only one who has no idea what you mean with that ...
     
  4. Feb 26, 2016 #3
    Sure, VAR is Volt Ampere Reactance, which is a measurement of reactive power.
     
  5. Feb 26, 2016 #4

    BvU

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    Haha, I googled VARS and ended up with a skiing resort. Should have googled electrical vars and that would have given your link.

    OK, here goes: For a halfway decent answer you need something that you can understand, but then you'll need to indicate a bit better what that could be. A school book on AC circuits ? Perhaps hyperphysics can help a bit ? There is plenty of material on the net, so it's more a matter of helping you to find that than to explain things to you that are being explained much better elesewhere.
     
  6. Feb 26, 2016 #5

    Hesch

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    A "VAr" is the unit for reactive power, which means that the direction of the power is switched for every half period ( example: 120 times per second by 60 Hz ac ).
    So reactive power will not be accumulated over time to e.g. energy that may heat up a kettle of water.

    You can charge a battery and then discharge it again: After that the battery will have no energy left.

    A capacitor acts like a battery: Connecting it to ac-voltage, it will be charged and discharged in turns.

    http://www.regentsprep.org/regents/math/algtrig/att7/phases17.gif

    In the above figure, the red curve is voltage and the blue curve is current. Whenever the voltage is constant ( red curve at top/bottom ) the current will be zero because the capacitor is not charged/discharged. Now, because the current is ahead of the voltage, we say that the capacitor produces reactive power ( just a convention ).

    Connecting an inductor to ac-voltage, the voltage will be ahead of the current, and we say that an inductor consumes reactive power.

    Reactive currents in e.g. power lines create active power losses ( P = I2 * R ). So we want to get rid of reactive power/current, mainly created by transformers and electric motors, containing coils/inductors that consumes reactive power. So if we connect a capacitor in parallel to a transformer/motor, the capacitor will locally produce reactive power consumed by the transformer/motor.

    Hence no reactive power/current will flow in the power lines, creating active power losses. That's clever.

    Adding/removing VAr's does not automatically increase/decrease voltage. You may say that minimizing reactive power/current will increase voltage ( decrease voltage drop in power lines ).

    Reactive may travel freely in a circuit.
     
    Last edited: Feb 26, 2016
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