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Work required to charge a capacitor to 2 RC time constant level

Discussion in 'Electronic Basics' started by [email protected], Nov 4, 2005.

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  1. Guest

    I know that the work required to fully charge a capacitor is given as:
    1/2 C V^2

    How much work is required to charge a capacitor to approximately 63% or
    2 RC time constants? I know that it is going to be the area under the
    curve: Vsource [1-exp(-t/RC)] from zero to 2RC , but it has been a long
    time since I have done any integration. I am looking for a ball park
    estimation if the exact answer is too difficult to derive. Any help
    would be greatly appreciated. Thanks
     
  2. kell

    kell Guest

    Wait, isn't 63% just one time constant? I'm pretty sure it is. Plug
    it into a calculator, or even just do a napkin calculation using 2.7 as
    an approximation for e. The quantity inside the brackets of the
    exponential formula is about 17/27, or 63% for RC=1.
    At any rate, the amount of work would equal the amount of energy
    stored, which as you said is 1/2 C V^2, where V is just the voltage on
    the cap. You don't have to do any calculus if you can calculate the
    voltage on the cap. But calculate the voltage as a function of time
    correctly.
     
  3. kell

    kell Guest

    Correction: I meant to say, "where t=RC," not "where RC=1"
     
  4. John  Larkin

    John Larkin Guest

    If V is the final charged voltage, the energy to get the cap to 0.632
    of V is just

    e = 1/2 * C * (0.632*V)^2.

    which is 0.4 times as much energy as the whole shebang.

    Of course, that doesn't count the energy lost in a charging resistor,
    if any.


    John
     
  5. kell

    kell Guest

    Okay, I just now did a calculation to give the amount of energy
    required
    INCLUDING heat lost in the charging resistor.
    Current through the rc combo is I = (V/R)e^(-t/RC)
    Power = VI = (V^2/R)e^(-t/RC)
    Energy = integral of Power with respect to time = -V^2 C e^(-t/RC) + K
    set the constant K so that the integral is zero when t = 0
    E = V^2 C (1 - e^(-t/RC))

    Hope I did that right.
     
  6. Bob Monsen

    Bob Monsen Guest

    Actually, the energy required to charge a capacitor to V is C*V^2.
    However, 1/2 of that ends up dissipated by the resistance between the
    voltage source and the cap.

    So, the energy required to charge the cap is the energy dissipated by the
    resistance + the energy stored on the cap at the end. Integrate the power
    through the resistance as a function of t from 0 to 2*R*C, and add that to
    the final energy on the cap.

    The energy dissipated by the resistor between t=0 and t=2*r*c is

    c v (2 exp(-2) - exp(-4) + 1)

    You can figure out the rest.

    The old 'two cap' problem illustrates the issue: if you have
    two equal value caps, one of which is charged to V, and the other of which
    is uncharged, and you connect them up, the initial energy in the cap was
    1/2 * V^2 * C, and the final energy is the energy in both caps, which
    is 2 * the energy stored in one, is 1/2 *(V/2)^2 * C. Thus, the
    energy in the final system is 1/2 the energy in the system originally. The
    question is where did the energy go, but the answer is obvious now...

    --- Regards,
    Bob Monsen

    The chief aim of all investigations of the external world should be to
    discover the rational order and harmony which has been imposed on it by God
    and which He revealed to us in the language of mathematics.
    - Johannes Kepler
     
  7. Guest

    I just want to make sure that I am interpreting what you are saying
    correctly. The total work that the battery is doing to fully charge
    the capacitor with a series resistor is going to be CV^2 not 1/2 CV^2
    where V is the voltage across the capacitor or in this case will be the
    same as the battery.

    The total work that the battery is doing to charge the capacitor to
    86.5% of the battery voltage (2 RC time constants) with a series
    resistor is going to be C(V*.865)^2 not 1/2 C(V*.865)^2 where V is the
    voltage across the capacitor or in this case will be 86.5% of the
    battery voltage.

    I think I get a slightly different result when calculating the energy
    dissipated by the resistor between t=0 and t=2*r*c

    Integral(0, 2RC, Vbattery^2/R exp(-2t/RC) dt = (Vbattery^2 C)/2
    (1-exp(-4)) which works out to be about .4908421806 Vbattery^2 C

    So according to this the total work done by the battery after 2 RC
    constants would be .4908 Vbattery^2 C + .5 (Vbattery*.865)^2 C

    Is this correct?
     
  8. Bob Monsen

    Bob Monsen Guest

    Actually, I got the integration wrong; I was using the voltage across the
    cap, rather than the voltage across the resistor.

    I believe you are correct about the energy at t=2rc.

    ---
    Regards,
    Bob Monsen

    Beyond the natural numbers, addition, multiplication, and mathematical
    induction are intuitively clear.
    - Luitzen Brouwer (1881-1966) (intuitionist)
     
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