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Homework Help: Derivation of mechanism of charging capacitor

  1. Mar 1, 2010 #1
    I am trying to derive the mechanism of a charging capacitor , V=Vo(1-e^(-t/CR))

    sorry , i couldn't upload the diagram here so i will briefly describe it , its a circuit with a battery (Vo) , connected to a capacitor (Vc) , and resistor (VR) and also a switch , all in series .

    i started with

    Q=CVc , then differentiate w r t to time t ,

    [tex]\frac{dQ}{dt}=\frac{d}{dt}(CV_c)=C\frac{dV_c}{dt}[/tex]

    using kirchoff law , [tex]V_o=V_R+V_c[/tex] (Refer to the diagram)

    [tex]=IR+V_c[/tex]

    [tex]=CR\frac{dV_c}{dt}+V_c[/tex]

    [tex]\frac{1}{RC}dt=\frac{dV_c}{V_o-V_c}[/tex]

    Integrate from time , 0 to t ,

    and also integrate from potential difference , 0 to V ,

    Here is my question , why integrate the pd from 0 to V ??

    I understand that the capacitor is initially at a constant potential difference , then when the switch is closed , the amount of charge increases , followed by the pd between the capacitors .

    is my thought process even correct ? Thanks in advance .
     
    Last edited: Mar 1, 2010
  2. jcsd
  3. Mar 1, 2010 #2

    rl.bhat

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    Homework Helper

    When the switch is closed , the amount of charge increases. PD across the capacitor also increases until its PD is equal to the applied PD.
     
  4. Mar 1, 2010 #3
    thank you , so when the capacitor is not charged , there is no pd , and when its charged , the pd becomes v (the applied pd) , so integrate within this range . Thanks !
     
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