Homework Help: Discharge of a capacitor

1. Jun 3, 2010

Daniiel

I have a quick question.
part of one of my questions asks
"Find the rate at which thermal energy is produced in a resistor"
i havn't included the whole question because i dont think it matters, its just like the last part of it
so we have
http://edugen.wiley.com/edugen/courses/crs1650/art/math/halliday8019c27/math148.gif
thats the discharge of energy out of the capacitor
so
would the inverse of that be the rate at which the energy is moved into the resistor
like
1/u = 2ce/q (i left out alot of things but thatsl ike the base of it)

orr
do you juts switch the e^- to e^+
because e^- is decay and e^+ is growth

yeh im just abit confused which is right, or if either is right haha.
thanks

2. Jun 3, 2010

kuruman

It does matter. If the resistor is connected to the battery, thermal energy is constant with respect to time; if it is connected to a discharging capacitor (as seems to be the case here) then the thermal energy will be a function of time.
No. That's the energy stored in the capacitor at any time t.
None of the above. The rate at which energy is dissipated in the resistor (i.e. power dissipated) is given by
P = i2R, where i is the current in the resistor at a given time t. That's what you need to find an expression for.

3. Jun 3, 2010