I might have added that it follows from the choice 1) that there is a
potential DROP of V = iR across the resistor in the direction chosen for
positive i. Likewise, it follows from the choice 2) that there is a
potential DROP of V = q/C from the capacitor plate chosen in 2) to carry q,
to the other capacitor plate. These signed potential differences must be
heeded in writing the loop equation.
I also add that in making the choices 1) and 2) you are NOT forecasting the
direction of i, or the sign of q; you are simply DEFINING the meaning
(including sign) of these quantities. Properly written loop equations will
be valid whatever circumstances dictate to be the actual signs of these
quantities.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.winbeam.com/~trebor/
trebor@winbeam.com
----- Original Message -----
From: "Bob Sciamanda" <trebor@winbeam.com>
To: "PHYS-L Maillist" <phys-l@carnot.physics.buffalo.edu>
Sent: Sunday, February 19, 2006 9:04 AM
Subject: Re: [Phys-l] RC Disharge Analysis
| One must always DEFINE the positive sense of all signed quantities and
stick
| with those definitions in writing equation models and in interpreting
| results of calculations. Just as this applies to coordinates (eg: x,y,z)
in
| kinematics, it applies no less in this capacitor discharge situation:
|
| 1) One must specify the positive direction of the current i. You may
choose
| either direction as positive;
| 2) One must specify which capacitor plate carries the signed quantity q.
| You may choose either plate for this purpose; the other plate then carries
| the charge (-q).
| If you are faithful to your specifications, and interpret your results
| accordingly, you may then traverse the circuit in either direction and
your
| "loop theorem" statement will be valid.
|
| Note if one chooses positive i in 1) to be the direction INTO the
capacitor
| plate chosen to carry q in 2), then it follows that i = dq/dt. If instead
| one chooses positive i in 1) to be the direction OUT OF the capacitor
plate
| chosen to carry q in 2), then it follows that i = - dq/dt.
|
| Note here the unspoken convention that the positive direction of i refers
to
| the flow of POSITIVE charge carriers.
|
| Bob Sciamanda
| Physics, Edinboro Univ of PA (Em)
| http://www.winbeam.com/~trebor/
| trebor@winbeam.com