Derivation of mechanism of charging capacitor

  • #1
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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:

Answers and Replies

  • #2
rl.bhat
Homework Helper
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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.
 
  • #3
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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.

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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