|Serway says -q/R -IR = 0. Can anyone explain how that could be correct?
This is correct if i is defined as positive when it is CW (in your
previously drawn circuit), and q is the charge on the lower plate. Note
this also defines the relation i = dq/dt
| The only way I can get that is to draw the current through the resistor in
a
| direction that would result in charging the capacitor rather than
| discharging it.
You are not drawing the actual current, you are merely defining the meaning
of a positive result for i.
I agree that Serway should explain his conventions more specifically.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.winbeam.com/~trebor/
trebor@winbeam.com
----- Original Message -----
From: "Michael Edmiston" <edmiston@bluffton.edu>
To: "PHYS-L Maillist" <phys-l@carnot.physics.buffalo.edu>
Sent: Sunday, February 19, 2006 9:45 AM
Subject: Re: [Phys-l] RC Disharge Analysis
| My apologies for a typographical error in my late night typing. I
| should have known better than to fall asleep, wake up, and make a post
| before going back to bed.
|
| Of course when I wrote q/C - IV = 0 for the loop theorem, I had intended
| to write q/C - IR = 0. Worse, I typed the wrong thing more than once.
|
| I think John and I are talking past each other on drawing
| diagrams/circuits. I am talking about only one circuit; that is, only
| one way of organizing the resistor, capacitor, switch in a single-loop
| circuit. All ways of constructing the circuit yield either the
| identical circuit or the mirror image of the circuit, and the circuit
| and mirror-image circuits are analyzed in an identical procedure. John
| is talking about the annotated circuit diagram in which the analysis has
| already begun (the current direction is noted and +/- signs are placed
| on the diagram). I agree there are two conventional ways to draw the
| annotated diagram because it is arbitrary which capacitor plate is
| chosen as the positive plate.
|
| When you apply the loop theorem to a single loop, you have only two
| choices, CW or CCW. I hope we agree on that. And the reason I said you
| can apply the loop theorem after choosing the current direction is
| because if you apply the loop theorem before choosing the a current
| direction you don't know whether there is a voltage increase or decrease
| as you traverse the resistor. Assuming we are defining current as
| positive charge flow, the loop theorem either yields q/C - IR = 0 or it
| yields -q/C + IR depending on which way we navigate the loop. Serway
| says -q/R -IR = 0. Can anyone explain how that could be correct? The
| only way I can get that is to draw the current through the resistor in a
| direction that would result in charging the capacitor rather than
| discharging it.
|
| Bob Sciamanda says that properly written loop equations will be valid
| whatever the signs. Of course, because a valid loop can be traversed
| the opposite direction, or the algebraic equation can have both sides
| multiplied by -1.
|
| I still maintain that if you apply the loop theorem correctly, you must
| choose i = -dq/dt for this problem, and that bugs students.
|
| Michael D. Edmiston, Ph.D.
| Professor of Physics and Chemistry
| Bluffton University
| Bluffton, OH 45817
| (419)-358-3270
| edmiston@bluffton.edu
|
|
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