Battery charging a Capacitor

Question

Ok I know there is a similar question already on here, but mine takes it just a step further. What I would really love to know is exactly how much energy is being transferred from a battery into a capacitor. I am under the assumption that it is 100% of the energy and can not for the life of me understand how it could possibly be only half of the energy, it makes no sense to me and I refuse to believe that.

This is something similar I believe to the 2 capacitor problem in which one fully charged capacitor is directly connected to another equivalent capacitor that is completely discharged to 0V. They both end up being the same voltage of course and therefore equalized, but then it is said that there is only half the energy now left in the 2 capacitor system. How can this be?

Yes it is true that no matter which way you now connect the 2 caps together, which there are only 2 ways, series or parallel, then the total power is only half but at the same time I understand that I have only discharged the first capacitor by half while charging up the other one by half, logically this is still 100% of the energy, but now only divided in half.

So my main question again is exactly how much power is being transferred from a battery to a capacitor that begins at 0V? I understand also that there would be a tiny minuscule resistive loss through the wire, but really it's not enough to say anything about.

Even with a resistor in between the battery and capacitor, I understand there is a slight loss of power through heat dissipation in the resistor, but again what would that have to do with the power leaving the battery and then entering into the capacitor, it would only have less current and would transfer more slowly because of the resistance. Still whatever power is leaving the battery would be the exact same power entering the capacitor despite the resistor.

Am I anywhere close to understanding this transferring of power from battery to capacitor, even with a resistor between the two?

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Great answers down below. It made me think a little bit and I realized it really is destroying half the energy. Not really destroying it but simply wasting it away in heat dissipation, and also LENZ LAW opposing fields in the tiny inductance of the wires connecting the battery to the capacitor, at least I think it is. I'm no expert.

However with much more thought I thought how could one charge a capacitor from a battery more efficiently and came up with a simple solution. I did this experiment and it works great and I can charge a capacitor from a battery at roughly 84% efficiency.

I got a 555 timer and made a simple DC pulse circuit and a coil I had wound with about 100 Ohms of resistance. I then pulsed that coil into resonance in which the input current became very small and the output in the form of a collapsing electromagnetic field was very high in the form of a kickback spike of high voltage, many times more than that of the battery and rerouted it through a simple diode, and Voila, 84% transfer of energy from the battery to the capacitor, case solved as far as I'm concerned, would love see somebody beat that efficiency.

Answer 1

I understand also that there would be a tiny minuscule resistive loss through the wire, but really it's not enough to say anything about.

On the contrary, it's crucial.

Assuming an ideal voltage source (can supply unlimited current) of voltage $V_S$, an ideal resistor of resistance $R$, and an ideal uncharged capacitor of capacitance $C$, are suddenly connected in series, e.g., with an ideal switch, the resulting series current is given by

$$i(t) = \frac{V_S}{R}e^{\frac{-t}{RC}}\;,\; t \ge 0$$

The power delivered by the source is

$$p_S(t) = V_Si(t) = \frac{V^2_S}{R}e^{\frac{-t}{RC}}\;,\; t \ge 0$$

The power delivered to the resistor is

$$p_R(t) = i^2(t)R = \frac{V^2_S}{R}e^{\frac{-2t}{RC}}\;,\; t \ge 0$$

The power delivered to the capacitor is

$$p_C(t) = p_S(t) - p_r(t) = \frac{V^2_S}{R}(e^{\frac{-t}{RC}} - e^{\frac{-2t}{RC}})\;,\; t \ge 0$$

It follows the energy delivered by the source is

$$U_S = \int_0^\infty p_S(t) dt = \frac{V^2_S}{R}\left(RC\right) = CV_S^2$$

The energy delivered to the resistor is

$$U_R = \int_0^\infty p_R(t) dt = \frac{V^2_S}{R}\left(\frac{RC}{2}\right) = \frac{CV_S^2}{2}$$

It follows that the energy delivered to the capacitor is

$$U_C = \int_0^\infty p_C(t) dt = \frac{V^2_S}{R}\left(RC - \frac{RC}{2} \right) = \frac{CV_S^2}{2}$$

Note that the resistance $R$ is not a factor in the results for the energy!

However, we can't set $R = 0$ otherwise the current and powers are undefined.

But, we can set $R$ arbitrarily small without changing the result that the resistance dissipates half of the energy (delivered by the source) while the capacitor stores the other half.

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In the physical case, there is unavoidable inductance and radiation resistance such that, as the 'ordinary' resistance is reduced, inductive and radiation effects dominate.

Answer 2

When you connect a battery to a capacitor, a "real" circuit has at least four components in series:

• the voltage source (battery)
• the capacitor
• series resistance
• series inductance

Any wire has inductance, since current flowing through it will induce a magnetic field. It is possible to have wires without resistance - we call them superconductors. So let's look at a circuit that has no resistance, and just the other three.

When you close the circuit, current will flow. The inductance will resist the rate of change of current, but it cannot completely stop it. The capacitor will charge up. When the voltage on the capacitor has reached the same voltage as the battery, current is still flowing (because of the inductance). It turns out that the energy stored in the capacitor is exactly the same as the energy stored in the inductor.

What happens next is that the inductor will force more current into the capacitor until the current is zero. At that point the voltage on the capacitor is higher than the voltage on the battery, and the capacitor will discharge through the inductor. What you end up with is an oscillating system, with frequency $\omega = \frac{1}{\sqrt{LC}}$.

In a real system, there is a little bit of resistance (even if the wires are superconducting, the battery will have some internal resistance. So will the plates of the capacitor.) That resistance will over time dissipate the additional energy until the oscillations stop and the capacitor voltage settles at V. It might take a long time if resistance is small - but it will happen.