a question i cant solve

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

  1. needshelp

    Thread Starter New Member

    Mar 23, 2004
    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
    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
    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
    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
    Ok, I see what you have done and although I haven't checked through with the original question I'm sure its right ;)