RC circuit - Capacitor discharge

  • Thread starter jpas
  • Start date
  • #1
45
0
When a capacitor discharges, the potential difference between its plates and the resistor's terminals is the same. Hence,

[tex] \frac{Q}{C}=R \frac{dq}{dt}[/tex]
Solving this equation we get

[tex] q(t)= Q_{0} exp \frac{t}{RC} [/tex]

Obviously, this isn't the solution. It is actually [tex] q(t)= Q_0 exp \frac{-t}{RC} [/tex]. So, I'm missing a minus sign on the original equation.

But why should it be there?
 

Answers and Replies

  • #2
116
1
According to Kirchoff's Second Law, the sum of the voltages is equal to zero. R(dQ/dt) + (Q/C) = 0, therefore you need -R(dQ/dt) = (Q/C). I apologize for the lack of proper script.
 
  • #3
45
0
Thank you.
Didn't know such laws.
 
  • #4
2,055
319
If the capacitor discharges, then dq/dt is negative, so according toyour originial equation, Q/C wich is positive equals something negative
 

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