Tension on a capacitor from Laplace domain to time domain.

Click For Summary

Discussion Overview

The discussion revolves around the analysis of a capacitor's voltage in the context of Laplace transforms, specifically after a switch has been opened. Participants explore the relationship between current and voltage in the s-domain and the implications of different equations used to derive voltage from current.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • The original poster (OP) describes their approach to finding the voltage across a capacitor using the equation I(s) = C(sV(s) - v(0)), expressing confusion over a different equation presented in the book: V(s) = (1/sC) I(s) - v(0)/s.
  • Some participants question the dimensional correctness of the OP's equation, suggesting that subtracting a voltage from a derivative of voltage is problematic.
  • Another participant argues that the OP's equation is not dimensionally wrong, explaining that V(s) has units of volt-seconds and that the terms involved are consistent in units.
  • There is a noted frustration regarding the lack of complete problem disclosure from the OP, limiting the ability to provide further assistance.

Areas of Agreement / Disagreement

Participants express differing views on the dimensional correctness of the OP's equation. The discussion remains unresolved regarding the best approach to derive the voltage from the current, as the OP ultimately states they have figured it out without further elaboration.

Contextual Notes

The discussion highlights potential limitations in understanding due to incomplete problem details provided by the OP, which affects the ability to fully analyze the equations in question.

maCrobo
Messages
51
Reaction score
1
The problem ask me to find the tension on a capacitor after a switch has been opened.

I have everything in terms of equations in s-domain and I'm sure they aren't wrong because I checked on the book. My unique problem is to understand a certain passage necessary to find the voltage knowing the current.

I found I(s) and I know v(0)/s of the capacitor. Then I just used this relation to find V(s) and compute its value in time: I(s)=C (sV(s)-v(0)). Here I solved for V(s) and I got the wrong answer. In fact, the book uses a different equation, actually just I sign changes, the following: V(s)= (1/sC) I(s) - v(0)/s. The first thing I thought was it was a Typing error, but checking other exercises showed it wasn't.

So, my question is: why do we use it?
 
Physics news on Phys.org
If you could provide the entire problem and not just some of it?
 
maCrobo said:
The problem ask me to find the tension on a capacitor after a switch has been opened.

I have everything in terms of equations in s-domain and I'm sure they aren't wrong because I checked on the book. My unique problem is to understand a certain passage necessary to find the voltage knowing the current.

I found I(s) and I know v(0)/s of the capacitor. Then I just used this relation to find V(s) and compute its value in time: I(s)=C (sV(s)-v(0)). Here I solved for V(s) and I got the wrong answer. In fact, the book uses a different equation, actually just I sign changes, the following: V(s)= (1/sC) I(s) - v(0)/s. The first thing I thought was it was a Typing error, but checking other exercises showed it wasn't.

So, my question is: why do we use it?

Your equation is dimensionally wrong!
You wrote sV(s)-v(0).
You have a voltage (v(0)) subtracted from the derivative of a voltage (sV(s)).
 
CEL said:
Your equation is dimensionally wrong!
You wrote sV(s)-v(0).
You have a voltage (v(0)) subtracted from the derivative of a voltage (sV(s)).

His equation is not dimensionally wrong.

V(s) has units of volt-sec since V = ∫v*exp(-st)*dt, s has units of sec-1, so sV(s) has units of volts, just like v(0).

However, since the OP won't disclose the problem there is little else we can do for the chap or lass.
 
Thank you, guys for the answer, but I have already figured it out.
Bye.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
23
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
13K
  • · Replies 5 ·
Replies
5
Views
3K