step-by-step

Discussion in 'Homework Help' started by Voltboy, Jan 10, 2007.

  1. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    i was watching this and im confused about how this work.. what are the capacitors here for.. pllease someone explain me why you need each coponent in this circuit? :confused:


    [​IMG]
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    Hi,

    May I suggest you read Volume III of the e-book this site hosts?
     
  3. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    im reading it, thats how im learning all but for me using a capacitor there dont make sense because the battery is always producing current , so the capacaitors will always be charging and acting a a load always, so they when cahrged will never be discharged and work like a simple conductor/wire, so it dont make sense
     
  4. Distort10n

    Active Member

    Dec 25, 2006
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    The coupling capacitors or the tuning circuit?
     
  5. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    i dont know what is each of those :(
     
  6. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    can anyone tell me in what direction goes the current, i know where it ends but i cant find all the current path:eek:
     
  7. mikeonthegreen

    Member

    Jan 7, 2007
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    OK I am new to electronics, but as I understand it a capacitor blocks dc and allows ac to pass.
    Also I dont think caps act as a wire when charged do they?
    Dont they charge and discharge so there would not be a steady flow of current. I am speculating but I hope someone can correct me.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
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    Pretty much. The "current flow" in a capacitor is due to the changing voltage across the capacitor: i(t) = C(dV/dt)

    The voltage stops changing (i.e. dv/dt = 0), then the plates of the capacitor charge to a steady state condition and the "current" ceases to flow.

    Dave
     
  9. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    But, if the capacitor is charged and its still current flowing will the capacitor discharge or it will only discharge if the current stop flowing (if u disconnect the battery)
     
  10. thingmaker3

    Retired Moderator

    May 16, 2005
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  11. Dave

    Retired Moderator

    Nov 17, 2003
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    Thinking of a DC application, then when the capacitor is charged, the voltage across the capacitor is constant, i.e. dv/dt = 0 and therefore i = 0. The situation changes for AC application, because the opposing plates of the capacitor are constantly charging and discharging, i.e. dv/dt /= 0 and therefore i /= 0.

    The crucial point to note is, that the voltage across the capacitor must be changing for current flow to occur. Without the need to complicate things, but dependant on the dielectric properties of the dielectric between the plates you may experience some current flow from one plate to another, although this is negligible in capacitor components, it has a noticable impact on dielectric heating processes which exploit the same mechanism as charging and discharging of a capacitor.

    Dave
     
  12. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    oh thanks dave.. umm, last question, where is the negative end of the source in the circuit i showed in the post, the ground right?
     
  13. thingmaker3

    Retired Moderator

    May 16, 2005
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    Yup. Most common convention is to show any voltages at reference to ground, so if only one voltage is shown, the opposite power connection will be ground.
     
  14. Voltboy

    Thread Starter Active Member

    Jan 10, 2007
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    Thanks alot to everybody that helped me :D
     
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