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Capacitor Charge/Discharge boundary conditions for time

  • #1
Member advised to use the homework template for posts in the homework sections of PF.

Homework Statement




Homework Equations

3. The Attempt at a Solution [/B]
Capture.PNG

I know dV=1/C∫idt and that we integrate the voltage from V to V0. What I don't get are the boundary conditions for t - How do we get what we get in the parenthesis? My closest assumption is that the t/T values refer to the areas described by the voltage graph - triangles and rectangles. I have noticed that the we section the period every time there's a change in current, and therefore, in voltage (0<t<2T, 2T<t<3T, etc.)

Also what use am I supposed to make of Q=-IT? From this equation and the first one we get i=dQ/dt, but so what?
 

Answers and Replies

  • #2
Grinkle
Gold Member
656
180
Not sure I understand your question.

I see a piece-wise definition of v(t) on the graphic in the lower right. Is that what you are asking about?
 
  • #3
Yes. I don't get why we have what we have in the parenthesis - I thought for 0<t<2T we'd have I(2T-0)/C but we have I(t-T)/C instead. Same goes for the rest.
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,071
At each change of the current, the capacitor voltage equation changes to follow a new curve---here it's a new straight line. A straight line is a linear function of t, but if the graph is rising then t will have a positive coefficient, and if falling t will have a negative coefficient in that section. The steeper the rise or fall, the greater will be the magnitude of the coefficient of t in the describing equation. In each section of the graph, the capacitor voltage usually doesn't start from 0V, it starts with an initial voltage determined by where it ended in the previous section. All of these three considerations must be taken into account when writing each straight-line equation for all the pieces that together describe VC.
 

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