Energy stored in a battery, formula?

Consider schematic below.

simulate this circuit – Schematic created using CircuitLab

With a DC source we can use this arrangement to find the force between the plates of the capacitor. We can do this by finding the mechanical work done pulling the plates apart, the change in the energy stored in the battery and the change in energy stored on the capacitor. I have been told (by more then one professor) that the energy stored in the battery is of the form $QV$, where $Q$ is the charge on the capacitor (the sign may be negative). But I think this is wrong, this formula says that when $Q$ is $0$ (i.e. when the capacitor is first connected) the energy stored in the battery is $0$. This to me is nonsense. I think the energy on the battery would be better expressed as $E_0-QV$ where $E_0$. Is my formula better? And in the general case what is the energy stored in the battery?

• The energy stored in the capacitor is $$0.5 CV^2$$, which has nothing to do with the energy in the battery. Mar 19, 2015 at 9:31
• I think it sounds like you're confusing concepts - the concept of the energy in a cell (a total amount), and the energy flowing into the capacitor (an instantaneous amount). To link the two concepts together you have to add (or remove) the factor of "time". Mar 19, 2015 at 9:55
• @pjc50 Why 0.5? What does that constant mean? Sep 29, 2016 at 15:23
• It's from integration - see en.wikipedia.org/wiki/Capacitor "Energy of Electric Field" Sep 29, 2016 at 16:02
• As you begin to discharge the capacitor, the voltage starts dropping. Think of the capacitor as a triangle with one 90 deg angle: one side of that angle is the charge and the other side is the voltage. The area is the energy, E = 0.5 * Q * U, Q = U * C Feb 8, 2017 at 5:08

Total Energy stored in the capacitor, = QV/2 = 0.5 CV^2

where,

Q = amount of charge stored when the whole battery voltage appears across the capacitor.

V= voltage on the capacitor proportional to the charge.

Then, energy stored in the battery = QV

Half of that energy is dissipated in heat in the resistance of the charging pathway, and only QV/2 is finally stored on the capacitor.

We cannot say that Q=0 because as soon as you connect the battery in series with the parallel plate capacitor, it starts charging the capacitor. But if you say that Q=0 , this implies that there is no battery.

Initial charge, Q = 0 means that time before you connect the battery, say at t=0, the charge on the capacitor is zero.

The energy stored in the battery (i.e. it's capacity) is expressed in Wh (watt hours.) To calculate the energy yourself then you need a battery and a constant current drawing load. The curve of power consumed from the battery over this time has to be integrated. That will give you the energy stored in the battery, and drawing the voltage to time will get you a discharge curve. You can lookup some battery datasheets like the Sanyo Eneloop battery I have taken this discharge curve from.

The energy stored can simply be given as

V*Ah = Wh

where Ah is the charge stored in the battery.

Without using integrals, for simple understanding purposes say, a battery has 2Ah rated at 1.5V such as the Eneloop ones, then the energy stored is around 3Wh.

If I have a resistor of 1 ohm connected across this battery and by neglecting internal resistance of the battery I would be drawing 1.5A of current. Since open circuit voltage changes with time as shown in the curve we may need to decrease the resistance now and then to make the current a constant so thats why we need a constant current drawing load for accurate results.

Nevertheless, now the resistor is drawing 1.5A and at 1.5V its 2.25W of power. Now the energy is 3Wh the battery will completely discharge into power by

3Wh/2.25W = 1.33 hours.

This is a fast and easy way used to calculate amount of energy left in a battery, in the industry a better way to measure is by using its SOC (state of charge) for which numerous papers are available for reference.

Case is DC source with capacitor:

Since $$i=c\frac{dv}{dt}$$ As voltage change $dv = 0$ thus $i = 0$.

And the circuit acts like an open circuit to DC.

But here note is: If battery (DC voltage) is connected across a capacitor, the capacitor charges.

I don't think QV of the capacitor shows how much energy is in the battery at all. A capacitor hooked up to a battery will reach charge Q when it reaches the voltage of the battery. Voltage of the capacitor increases until it reaches V of the battery then stops. There could be much more charge that can be used in the battery.

I think you are mixing battery and capacitor together- they are not the same thing. A battery is an electrical energy source, the capacitor is an energy storage load.

If you charge your capacitor and want to use it as "a battery", then your equation works for answering how much energy has been used up, or how much charge/voltage is left.

           Eo-QV = Enow = Qnow*Vnow


With that being said, a battery's energy source is chemical in nature, while a capacitor is solely electrical. The above equation in a battery is a crude estimation

The thing to remember is the voltage drop behavior of a battery is given in a figure in a previous answer, and for the capacitor the voltage drop is inverse exponential.