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

On Fri, 04 Nov 2005 13:32:03 -0800, jalbers@xxxxxxx wrote:

> 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

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

.

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