Quick RC circuit question

Ja4Coltrane

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 doesnt 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 doesnt 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?

berkeman

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?

Ja4Coltrane

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?

denverdoc

see here: http://www.ac.wwu.edu/~vawter/Physic.../RCSeries.html