# A confusion about Energy and Capacitors

1. May 31, 2009

### TheWind777

I have some questions about the energy stored in capacitors.

Looking in Physics books, there are two equations which govern the amount of energy stored in capacitors:

Energy in Joules (Watt*Seconds) = .5 * Q^2/C

Where Q is the charge in couloumbs and C is the capacitance in Farads.

Energy in Joules (Watt*Seconds) = .5 * V^2 * C

Where V is the voltage.

Now, if you were to take a 100F capacitor and charge it to 2.5 volts, the energy would be .5 * 2.5*2.5 * 100 = .5 * 6.5 * 100 = 312.5 Joules

If you take the very same capacitor and double the voltage. This time you charge it to 5 volts...

energy would be .5 * 5*5 * 100 = .5 * 25 * 100 = 1250 Joules

So, if you raise the voltage by 2, you get 4 times the stored energy?

What kind of magic is that? How can you get a four-times increase of energy just by doubling the voltage? Wouldn't that be cold fusion or perpetual motion if I could do that?

In theory, if you could get a four times increase in energy just by increasing the voltage twice, couldn't you increase the voltage to 10,000 volts, put a bunch of high-capacitance capacitors together and store 500,000,000 joules of energy?

If that were the case, we could harness the power of chain letter math.

Something has to give if you increase the voltage on the same capacitance. You can't get four times the amount of energy out just by raising the voltage twice, can you?

And, if so, where did that power of two come from anyways? What combination of two other equations gives you the V squared?

The secret can't be with charge, because charge is also linear. You raise the voltage by two, the charge increases by two.

Secret can't be the capacitance, because you're using the very same capacitor in both cases.

Equation that governs charge, and capacitors.

Q = CV

Or, swapped around...

V = Q/C

So, what am I missing? What goes down to compensate as the apparent energy goes up? The capacitance is staying the same. The voltage is doubling. The charge is doubling. where does the 'four times more energy' come from?

I can envision that possibly the answer might be time. Could the extra energy be allowable because time is the secret missing variable? Could the time it takes to charge it up that extra amount be the answer to the hidden 'where did the energy come from?.

If not, then what compensates by dropping to balance out that four-times-the-energy value?

You can't really get four times the energy just by doubling the voltage, can you?

And, if you can get four times each time you double the voltage, why can't you just charge up capacitors to a couple hundred volts, then drop them back down to a lower voltage when you want to use them?

Last edited: May 31, 2009
2. May 31, 2009

You can.Something similar happens with a resistive circuit,doubling the voltage doubles the current and quadruples the power whereas with a capacitor if the voltage doubles the charge doubles.The energy is not proportional to the voltage it is proportional to the voltage times the charge or the voltage squared(E=0.5QV or one of the equations you presented above)

3. May 31, 2009

### Bob S

Suppose I said that the kinetic energy of a moving car were (was?)
E = (1/2)m v2 Would you believe this?
So, if you raise the velocity by 2, you get 4 times the stored energy?

What kind of magic is that? How can you get a four-times increase of energy just by doubling the velocity? Wouldn't that be cold fusion or perpetual motion if I could do that?

4. May 31, 2009

### TurtleMeister

Also, connecting two capacitors together will not give you four times the energy. If you connect two equally charged capacitors of the same capacitance in series you will double the voltage and cut the capacitance in half .5 * (v*2)^2 * (c/2) giving you double the energy (not four time the energy). If you connect the same capacitors in parallel you will double the capacitance .5 * v^2 * (c*2) giving you double the energy again. So there's no way to get more energy than you put in.

5. May 31, 2009

### mikelepore

Derivation of capacitor energy E = (1/2) C V^2

Charge a capacitor
from initial voltage 0 at initial time t=t_i
to final voltage V at final time t_f

Instantaneous voltage v
Instantaneous current i = C dv/dt
Instantaneous power p = vi

Substitute i expression into p expression
p = Cv dv/dt
Energy E = integral from t=t_i to t=t_f of p dt
Substitute p expression into E expression
E = integral Cv (dv/dt) dt
Cancel dt/dt
E = C integral v dv
E = (1/2) C V^2

Last edited: May 31, 2009
6. May 31, 2009

### LewisEE

We have to keep in mind where the energy that is ultimately stored in the capacitor's electric field initially comes from: the battery. Just as Bob S hinted at with the car, there is more to it than simply doubling the voltage. Doubling the voltage implies that you are now packing more charge onto the capacitor, which requires work. So, it's not a free lunch. ALL energy that gets stored in a capacitor must be supplied by the battery.

7. Jun 1, 2009

### dE_logics

Notice that the energy stored is not a leaner function of V in a capacitor, its a square relation...v2...so if you double the V, you get 4 times the voltage stored.

I seriously wish we could...but...till now, I think the maximum capacitance for commercial purposes is like.......20F...and if you increase the V, the output will too be high (though I have major un-resolved ambiguities with this).

This is the energy 'stored'...you gotta give the same amount of energy as the input.

The charge accumulation is directly proportional to V...so if we doubles, change accumulations too doubles.

But the energy stored per electron will too double; causing a 4x effect.

Same reason...its more of maths.

Last edited: Jun 2, 2009
8. Jun 5, 2009

### Bob S

Here is a quantitative measurement of the energy stored on (in) a capacitor. Suppose I have a capacitor C with a voltage V0 on it. I attach it to a resistor R and discharge it. How much energy is dissipated in the resistor? The initial voltage is V(t) = V0, and current I(t)= V0/R. The voltage and current both decay as exp(-t/RC). So the total energy dissipated in the resistor is
E = integral0inf [I(t) V(t)]dt = int[V02/R exp[-2t/RC)]dt = (V02/R)(RC/2) = (1/2)CV02.
Using Q=CV you can eliminate eiher C or V0.