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Homework Help: Discharge of a capacitor

  1. Jun 3, 2010 #1
    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. jcsd
  3. Jun 3, 2010 #2

    kuruman

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    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.
     
  4. Jun 3, 2010 #3
    what about
    P(t)= V(t) I(t)
    is that alright aswell?
     
  5. Jun 3, 2010 #4

    kuruman

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    Yes it is, but you will have to find two functions of time, V(t) and I(t). If you use P = I2R, you will need only I(t).
     
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