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

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.

Correction: I meant to say, "where t=RC," not "where RC=1"

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

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.

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

--- 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

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

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?

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.