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I Capacitive reactance

  1. Sep 18, 2016 #1
    Hi Folks,

    I am reading some text about capacitive reactance that reads 'so we know that capacitors oppose changes in voltage'

    would this be the supply voltage or the voltage in the capacitor? I am struggling to see how the change is opposed

    surely the quoted statement would infer increasing impedance with increasing frequency
    any direction would be greatly appreciated.

    thanks
     
  2. jcsd
  3. Sep 18, 2016 #2

    David Lewis

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    With a capacitor (or inductor), reactance equals the voltage-to-current ratio. Now if applied voltage changes, there will be a corresponding change in capacitor voltage, but not right away. It takes time. Therefore current will be somewhat impeded over and beyond Ohmic resistance.
     
  4. Sep 18, 2016 #3

    sophiecentaur

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    I don't know where you read that but it is so vague that you cannot rely on it at all. The formulae for the reactance of a Capacitor (or Inductor) is what you need to refer to if you want a proper answer. Arm waving statements really don't help at all. "Oppose" means nothing in this context.
     
  5. Sep 18, 2016 #4

    CWatters

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    I agree it is vague but you can make some sense of it by looking at the questions relating charge and voltage on a capacitor...

    Q=CV
    Differentiate wrt time
    dQ/dt=CdV/dt
    Now dQ/dt=I so
    I=CdV/dt

    So if you apply a constant current to a capacitor the speed or rate at which the voltage on the capacitor changes is inversely proportional to the capacitance. The larger the capacitor the slower the voltage changes. It's possible to interpret this as the capacitor resisting changes in its voltage.
     
  6. Sep 18, 2016 #5
    It's a basic rule that you can't have two different voltages connected directly to each other in a circuit (Kirchoff's Voltage law). The current between them would be infinite if they were. In the real world if you connected batteries together with different voltages the current would not be infinite because there is a small resistance in the wire. The circuit you draw to model the connection of the batteries would have to include that wire resistance which separates the battery voltages.

    Like the batteries, you can't connect a source directly to a capacitor without some wire resistance.

    Let's say that you have a capacitor and touch the leads to a source. The capacitor will draw a high current which causes a voltage drop across the small resistance of the leads until the cap is fully charged. Now if you quickly short the leads the cap will output a high current across the resistance of the leads and maintain some voltage until it fully discharges. The bigger the cap is the longer this takes so you could say that caps oppose changes in voltage.

    You'll see caps used often in DC power supplies for this reason. Caps maintain a steady voltage by absorbing or emitting current when a supply changes in voltage.

    That's just one way to look at it. It's a bit unsophisticated and not very useful for real circuit design. In the case of DC power supplies it would be better to say that capacitors maintain a steady voltage by offering a low impedance path for high frequency noises. That's a whole topic by itself.
     
  7. Sep 18, 2016 #6

    sophiecentaur

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    The wonderful thing about Maths is that, if you can apply it to a Physical situation then it tells the story far better than any amount of arm waving and 'analogies'. Using a term like "opposes change" is so imprecise, compared with the definition of Reactance and Impedance, which give you the ability to predict the actual degree to which a Capacitor affects the Volts in a circuit. Without some actual figures you can't get close to understanding a subject like Electronic circuit theory.
    How big is the Capacitance value and what is the source resistance being discussed? The possible range of values of the two is vast and the effect could be virtually zero.
     
  8. Sep 18, 2016 #7

    David Lewis

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    When you first connect a 10V battery to an uncharged capacitor, cap voltage will start out at 0V. Then some time later, cap voltage will be 1V, and so on.
     
  9. Sep 19, 2016 #8

    CWatters

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    PS to post #4....

    An alternative way to view I= CdV/dt is to realise that the faster you want to change V the higher the current you need to force into the capacitor.

    By comparison if you make a step increase in the current through a resistor the voltage changes nearly instantly.
     
  10. Sep 19, 2016 #9

    sophiecentaur

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    There is always an exponential change in voltage, dictated by the fact that there will always be some finite resistance in the supply. Its is never instantaneous. The 'time constant' is RC so it is proportional to the Resistance times the Capacitance.
    People can Google "RC time constant" for more information about this.
     
  11. Sep 21, 2016 #10

    David Lewis

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    There will be some Ohmic (a.k.a. loss) resistance and inductance in the wires. There will also be radiation resistance due to charge acceleration.
     
  12. Sep 23, 2016 #11

    NascentOxygen

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    Perhaps a clearer way of expressing this use of a capacitor is to say capacitors can help smooth out fluctuations in voltage. They smooth out the voltage that you connect them to, and this is typically a parallel connection to the DC supply voltage which would otherwise look like a rectified sinewave.

    The introductory book where you read this probably went on to say that inductors, by contrast, can help smooth over fluctuations in current. You insert the inductor in series with the wire carrying the fluctuating current and it tends to smooth out that current.

    To understand this behaviour it really is necessary to dive into the mathematics describing how these electronics elements operate, as other respondants have indicated.
     
  13. Sep 23, 2016 #12

    sophiecentaur

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    That is a specific use, perhaps but the essential thing about a Capacitor in a circuit is to introduce a known exponential delay, when used in conjunction with a specific resistor or to provide a resonant frequency when used in conjunction with an Inductor. But, as with other electronic components, it's better to describe its function than mention just one application. Uninformed people can be very literal in their understanding of an abstract thing like electricity.
     
  14. Sep 23, 2016 #13

    NascentOxygen

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    Poster does appear to be asking about the rôle of smoothing capacitors, so I addressed only that use. We have no indication that OP will understand associated maths, incl. "exponential".
     
  15. Sep 23, 2016 #14
    Bringitondown are you still around?
    My reductionist version: The instant you apply a voltage to a capacitor the voltage "sees" a short. This short in effect looks like a small value resistor that changes into a larger and larger value of resistance as the capacitor is charged, finally becoming the leakage resistance value of that capacitor. From a functional view that is all that is happening. There is no opposition involved. The voltage is simply distributed across the circuit attached to the capacitor and only slowly appears across the capacitor as the charge (number of electrons on the capacitor plate) increases.
     
  16. Sep 23, 2016 #15

    sophiecentaur

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    I'd go along with that because you have included the idea that the Resistance value is key to the performance.
     
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