a question i cant solve

Discussion in 'Homework Help' started by needshelp, Mar 23, 2004.

  1. needshelp

    Thread Starter New Member

    Mar 23, 2004
    2
    0
    here is the question..if anyone can solve it id be grateful:

    Consider a battery of voltage V which charges a capacitor of capcitance C through a resistor of resistance R. Derive an expression for the work done by the battery in charging the capacitor. Careful: in the process there is both energy dissipated in the resistor and stored in the capacitor. [Hint: Find the power output of the battery as a function of time, and integrate it to find the work done.]

    thanks alot :eek:
     
  2. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    hi
    Work done= (power dissapated across resistor*time)+(energy stored in capacitor)

    Current through the circuit is given by
    i=[E e^(-t/RC)]/R
    where E is the emf of the source
    Power dissapted across the resistor is (i^2)R
    The energy stored in a capacitor is given by 0.5(dqV)
    where dq is the charge on the capacitor at any instant of time and V is the voltage drop across the capcitor at the same instant of time.
    We know that dq=i dt and V=E[1-(e^-t/RC)]

    In dt time energy dissapated accross the resistor is = (i^2)R dt

    Therefore the expression for work done by the battery in time dt is
    dw = (i^2)R dt + 0.5dqv
    dw = {[E e^(-t/RC)]/R}^2 (R dt) +{[E e^(-t/RC)]/R }{E[1-(e^-t/RC)]}dt

    integrating this expression give the work done
    some simplification of the expressions is necessary before integrating which i am avoiding in order to save on steps ( btw its your homework!!!!)

    hope this helps you
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    145
    Good answer haditya, however one small point:

    The energy stored in the capacitor is given by 0.5(qV)

    Correct me if you think I'm wrong, but if I remember:

    Energy stored W = ∫ vi dt = ∫ v dq

    i.e. dq = i dt

    From v = qC

    ∫ q/C dq from the limits 0 to Q gives
    W = 0.5 Q²/C = 0.5 CV²
     
  4. haditya

    Senior Member

    Jan 19, 2004
    220
    0
    Yes.. what you are saying is true... what i have done is the following:-

    Energy stored in the cap is dqv, where dq and v are the charge stored on the cap and "potential diff across the cap" respectively at an arbitrary time t
    Then i have written the voltage v and charge q as a function of t so as to facilitate integration with respect to time
     
  5. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    145
    Ok, I see what you have done and although I haven't checked through with the original question I'm sure its right ;)
     
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