|
Home > Archive > Electrical Engineering > July 2007 > Calculus and a capacitor question
You are viewing an archived Text-only version of the thread.
To view this thread in it's original format and/or if you want to reply to
this thread please [click here]
| Author |
Calculus and a capacitor question
|
|
|
| This is about the equation: C = I* dT/dE.
I've taken Calculus and currently taking Diff EQ. At the moment I
understand about applying Diff EQ techniques to turn that equation into a
solution for all values of t (time). What does the equation mean in the
form it's currently in?
I understand the d means "instantaneous", however, I'm use to seeing
equations such as: x d/dx and that tells me to differentiate 'x' with
respect to x and the answer would be '1' (X^n = n * times X^n-1).
Does the capacitor equation state: C = (I * T) d/dE ? if that's the case,
the derivative would be 0 and the equation would C = 0.
Theoretically it can be said that dT/dE means: (t1-t2) / (E1 - E2).
Typically you can say that, but theoretically Calculus should be applied
and I'm a bit confused about that equation. The inductor equation is
another one that confuses me, however, I'm sure it would make sense once I
understand the above.
| |
| G \Guglielmo\ Evans G4SDW 2007-07-29, 3:25 am |
|
"Peter" <private@private.com> wrote in message
news:GbSdnUyGpNDQnDHbnZ2dnUVZ_hKdnZ2d@comcast.com...
> This is about the equation: C = I* dT/dE.
>
> I've taken Calculus and currently taking Diff EQ. At the moment I
> understand about applying Diff EQ techniques to turn that equation into a
> solution for all values of t (time). What does the equation mean in the
> form it's currently in?
>
Perhaps you're difficulty arises from having a transposed equation?
In its original form, it reads I = C * dV/dT, where dV/dT is no longer
a quotient that can be split into two separate terms, but a function
that results from the limiting process of reducing a time difference
to an infinitesimal amount.
In its original form, it means that the instantaneous current, I, flowing
into
a capacitor, C, is proportional to the instantaneous rate-of-change of
the voltage across the capacitor.
In the transposed form that you presented, C = I * dT/dV, it means that
you can calculate the capacitance if you can simultaneously measure the
current and rate-of-change of voltage. Your wish to express the time
difference as
(t1 - t2) and the change of voltage as (v1 - v2) would only be valid in
the extremely simple case of a constant current and a linear rise of
voltage.
(It would certainly be a true use of the facts, but the exercise I think
you're
undertaking is trying to understand the application of calculus, and whereas
the
use of a straight line is part of the derivation of calculus, it is only the
derivation and once the limiting process of infinitesimal time has been
applied
the intention is that you should then work in a continuously changing world
and not a world of straight lines that you are trying to go back to.
By all means revert back to the straight line derivation of calculus to
remind
you what's going on, but don't stay trapped in there when trying to use
calculus.
|
|
|
|
|