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RC time Circuit; charge of capacitor as a function of time derivation

  1. Jul 11, 2014 #1
    Hey guys! New to physicsforums. I wanted to ask a more conceptual question regarding RC time Circuits. I spent some time trying to derive the equations, and I feel like I'm not setting up the problem correctly. Here's my attempt:

    Solutions according to profecssor:

    1) ##q_{charge}(t)=C\epsilon(1-e^{-\frac{t}{\tau}})##
    2) ##q_{discharge}(t)=q_{0} e^{\frac{-t}{\tau}}##

    ---
    my attempt at solution 1
    ---

    ##q##-charge on capacitor
    ##R##-Resistance encountered in circuit
    ##C##-Capacitance of capaticor
    ##\epsilon##-Electromotive Force of battery
    ##t## -time since capacitor began charging
    ##\tau## - ##RC##
    ##i## -current


    ##C=\frac{q}{\epsilon}##
    ∴##q=C\epsilon##
    ##\epsilon=iR=\frac{dq}{dt}R##
    ∴##q=RC\frac{dq}{dt}=\tau\frac{dq}{dt}##
    ∴##\frac{1}{q}\frac{dq}{dt}=\frac{1}{\tau}##
    ∴##\int_0^t \frac{1}{q}\frac{dq}{dt} dt=\int_0^t \frac{1}{\tau} dt##
    ∴##\int_{q(0)}^{q(t)} \frac{dq}{q}=\int_0^t \frac{1}{\tau} dt##
    ∴##ln(q(t))-ln(q(0))=\frac{t}{\tau}##
    ∴##ln(\frac{q(t)}{q(0)})=\frac{t}{\tau}##
    ∴##e^{\frac{t}{\tau}}=\frac{q(t)}{q(0)}##
    ∴##q(t)=q(0)e^{\frac{t}{\tau}}##
    So it's here that I'm like "I guess ##q(0)=c\epsilon## just for the sake of looking more like the professor's solution" <I realize how stupid this is... lol
    ∴##q(t)=C\epsilon e^{\frac{t}{\tau}}##

    There must be something that I'm just totally missing here. The math seems right, so I'm thinking it's the set-up that I'm missing.

    Any help is appreciated! Thank you!
     
  2. jcsd
  3. Jul 11, 2014 #2

    Orodruin

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    You have assumed that the potential drop across the resistance is the same as that across the capacitor. Is this true or can you think of something else that is more physical?
     
  4. Jul 11, 2014 #3
    Alright, so I started off on the wrong part. So, I need to derive ##V_{C}(t)=\epsilon(1-e^{\frac{-t}{\tau}})## (This was written on the paper our professor gave us, so as of now it doesn't really mean anything to me.)

    So, ##V_{C}=iR##. The current is the same as the current we've been dealing with, but what R is this equation referring to?
     
  5. Jul 12, 2014 #4

    Orodruin

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    Let us start studying the charging circuit: Draw the circuit. What will be the potential drop across the capacitor and resistor respectively? What is the potential gain across the battery? How are those related?

    For the discharging circuit: it is the same just without the battery.
     
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