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Capacitor initial condition

  1. Apr 27, 2015 #1

    cnh1995

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    Suppose we have an RC series dc circuit with two capacitors C1 and C2 and resistance R. If the switch is closed at t=0, all the voltage appears across R initially. Fine.. But how does it reach across R through two insulation barriers (breaks) in the circuit? I can understand the mechanism for single capacitor but for multiple capacitors, how does the voltage signal travel through the wire and insulation (technically 'dielectric') of capacitors? I can do the differential-integral math and can solve RC networks but this initial condition baffles me...There is a mechanism called surface charge feedback but I can apply it only for single capacitor. It's well explained in "Matter & Interactions" by Chabey-Sherwood but I don't have that book..If someone knows anything please help...Thanks in advance:smile:
     
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  3. Apr 27, 2015 #2

    davenn

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    draw your circuit and post it

    then we all can be sure we know what you are referring to
     
  4. Apr 27, 2015 #3

    cnh1995

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    photo0257.jpg
    Circuit just after the switch is closed...How does the voltage V appear across the resistor?? There are two breaks in the circuit, dielectric of C1 and dielectric of C2. I know the differential equation of RC circuit but I don't understand this initial condition..The resistor looks isolated from the source due to the dielectrics..
     
  5. Apr 27, 2015 #4

    davenn

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    Think about the flow of charges on and off the capacitor plates
    before they reach equilibrium

    Write out a description of what is occurring as the capacitors charge up and you will see what is happening with the resistor

    Dave
     
  6. Apr 27, 2015 #5

    cnh1995

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    Okay..I understand the behavior of the circuit...Caps charge exponentially,decreasing the current in the same manner..But in a purely resistive circuit, the voltage reaches the resistor through surface charge feedback mechanism, totally through the conducting medium. Here there are insulators in the circuit so how does the feedback occur there? No motion of electrons is allowed inside the dielectric...So how does the voltage reach across the resistor in the first place,getting past the dielectric barrier?? I'm interested in the physics behind this...
     
  7. Apr 27, 2015 #6

    davenn

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    OK

    I favour describing these things as the movement of electrons carrying negative charge, OK
    lets just take a single capacitor and switch across a battery
    At T=0, the switch is off, the capacitor is not energised
    when the switch closes, electrons flow from the negative battery terminal and onto one plate of the capacitor and an electric field develops between the plates and across the dielectric.
    That field forces/repels electrons off the other plate and towards the positive terminal of the battery. That plate now becomes more positively charged, hence we now have an increasing voltage potential across the capacitor and when it reaches equilibrium, it equals the battery voltage.

    Now start again with 2 caps in series ... what is now happening to the movement of the charges on the plates of the 2 capacitors ?

    now put a resistor between the 2 caps, can you now see how there is a voltage potential difference developed across the resistor ?
    the cap plate on the left side of your resistor is more negative than the cap plate on the right side of the resistor in your diagram

    there has been a flow of charge ( current) between the plates via the resistor ..... that cannot happen unless there is a potential difference across the resistor


    Dave
     
    Last edited: Apr 27, 2015
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