Understanding Charge Formulation in Discharging RC Circuits: A Quick Question

Click For Summary

Homework Help Overview

The discussion revolves around the formulation of charge on a capacitor plate as a function of time in a discharging RC circuit. The original poster expresses confusion regarding the signs in the loop rule equation for a discharging capacitor, specifically questioning the negative signs in the expression -q/C - IR = 0.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the reasoning behind the negative signs in the loop rule equation and questions whether the current's direction affects the signs in the equation. Other participants suggest considering standard conventions for voltage increases and drops in loop equations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the loop rule and the implications of current direction on the signs in the equation. Some guidance has been offered regarding standard practices in writing loop equations, but consensus has not been reached.

Contextual Notes

The original poster mentions a specific setup involving a capacitor, resistor, and switch without a battery, indicating a focus on the discharging process. There is a lack of a diagram, which some participants note may hinder clarity in the discussion.

Ja4Coltrane
Messages
224
Reaction score
0
So, I understand the charging of an RC circuit perfectly, but I am a tiny bit bothered by one part of the formulation of charge on a capacitor plate as a function of time for a discharging capacitor in a simple RC circuit (charged with one battery, one resistor and one capacitor all in series and discharged without the battery).
What bothers me is that the original loop rule equation seems to have a negative sign that doesn't make sense. The book gives the expression
-q/C-IR=0

I don't understand why both of these should be negative, in looking at the direction that the charge would flow, it seems that the equation should be

q/C-IR=0 or IR-q/C=0

The books explanation is that the formula is the same as the one used for charging the capacitor just without the battery, but this explanation doesn't quite make sense to me. I'm probably just being stupid. Is the reason perhaps that the -IR involves a negative value for I and therefore causes a positive potential difference?
 
Physics news on Phys.org
Without a diagram it's hard to know for sure. But in writing loop equations, you generally follow the standard that a voltage increase (like going from the - to the + terminal of a voltage source as you trace the loop) is a positive voltage term, and a voltage drop (like when you trace in the direction of the current flow through a battery) is a negative voltage term in the equation. Does that help with your question?
 
Last edited:
mmmm, not exactly. Here, imagine a simple circuit with no battery containing only a switch, a capacitor and a resistor in series. Consider that the capacitor is on the left of the resistor and the left plate is initialy (before the switch is closed) charged with a positive sign. The switch is then closed. It seems that the current should flow to the left (left for the resistor and capacitor). Because of this, if I do a loop rule it seems like the I would get +q/C-IR=0. Did that clear up my question?
 
see here: http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/RCSeries.html
 
Last edited by a moderator:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K