# Energy stored and returned by a capacitor

Discussion in 'General Electronics Chat' started by jag1972, Oct 10, 2011.

1. ### jag1972 Thread Starter Active Member

Feb 25, 2010
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0
I am trying to determine the Energy stored and returned by a capacitor. Energy is equal to the power multiplied by the time, E = Pt. How this is calculated depends on the information given.
If the information available was 2 sine waves, and 1 for current and the other for voltage. The two waves would have to be multiplied which is a straght forward multiplication. Once multiplied the result is the power which should be positive have twice the original frequency (current and voltage).
This Power can now be integrated with respect to time to calculate the energy.
Alternatively the voltage sine wave can be squared and divided by the resistance then integrated with respect to time or
The current sine wave can be squared and multiplied by the resistance then integrated with time.
Are my assumptions correct or is there and easier method.
In a lot of the material I have read the equation given for energy is E=0.5CV2 , however I cannot mentally relate that to what I have stated above.

2. ### #12 Expert

Nov 30, 2010
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9,430
First, there are some math whizzes on here that put me to shame, but I'll try because you've been waiting for over 6 hours.

I think you're going about it all wrong. Feel free to direct me to where you came up with this method and correct me as to what your goal is. Doing that might even get more people to contribute.

Energy in Joules = 1/2 (capacitance in farads) times voltage squared.
Voltage is only a symptom of current in a capacitor.
V = I x Time/capacitance
Voltage across the capacitor = current x time / capacitance
a milliamp-second going into a millifarad will have a symptom of 1 volt across the capacitor.
The idea of multiplying a voltage wave to arrive at energy in a capacitor escapes me.

Then there is the old, "what goes inta comes outta". Of course that is named after some famous guy I can't remember. Whatever energy is put into the capacitor will come out, minus a tiny bit for dielectric absorbtion (a form of defect) and a tiny bit for ESR (Equivalent Series Resistance) (another form of defect).

This might be the beginning of a conversation. It's up to you.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
786
I guess the current & voltage sine waves refer to the current and voltage in a capacitor in an AC circuit.

Suppose one has as the capacitor time domain voltage

$V_c(t)=V_m cos(\omega t)$

Then the corresponding steady-state time domain current in an ideal capacitor will be found from the relationship

$I_c(t)=C\frac{dV(t)}{dt}$

& ultimately

$I_c(t)=I_m cos(\omega t +\pi/2)$

where

$I_m=\frac{V_m}{X_c}=\omega C V_m$

In other words the current leads the voltage by 90°.

The instantaneous power is given by

$p(t)=V_c(t)I_c(t)=V_mI_mcos(\omega t)cos(\omega t +\pi/2)$

using the relationship

$cosAcosB=\frac{1}{2}[cos(A+B)+cos(A-B)]$

gives

$p(t)=\frac{1}{2}V_mI_mcos(2\omega t+\pi/2)$

The average power is given by

$P_{av}=\frac{1}{T}\int_0^{T}p(t)dt=0$

Which you can prove to yourself by doing the integration.

The average power is zero in an ideal capacitor with steady state ac conditions.

So you can't use that relationship to work out the energy inflow / outflow in a capacitor.

You're stuck with the 0.5CV^2 relationship for finding the stored energy.

Keep in mind that in electrostatics an interpretation of the Volt [electrical potential difference] is related to energy and charge.

In electrostatics one can think of the Volt as Joules per Coulomb. In other words the work done in moving a certain charge through a known electrical field leads one to voltage.

4. ### crutschow Expert

Mar 14, 2008
17,063
4,677
For a perfect capacitor the energy stored and returned is 1/2 CV^2 Joules, where V is the peak voltage. Thus in an AC circuit the capacitor would be charged and discharged twice per cycle. The total average power stored and returned by the capacitor would thus be 1/2 CV^2 x 2 x f = fCV^2 watts, again with V being the peak capacitor voltage.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Strictly speaking the units would be vars rather than watts. The use of watts is confusing since it implies that energy is lost in the ideal capacitor - which is impossible.

Of course one may think of a capacitor as a transient energy source - such as when a charged capacitor is discharged into a resistive load. In that case the notion of average power has little meaning because it is a transient condition. One would think of this as transient energy transfer. Certainly there is also the case of oscillatory behavior in an under-damped LCR circuit with the capacitor charged to an initial value. The energy originally stored still ends up as heat loss in the resistor R. Neither the ideal L or C consume any energy.

Last edited: Oct 11, 2011
6. ### crutschow Expert

Mar 14, 2008
17,063
4,677
I did say average power stored and returned by the capacitor, so it should be clear that no energy is dissipated.