- #1
thereddevils
- 438
- 0
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 kirchhoff 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 .
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 kirchhoff 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: