Why Does Charging a Capacitor Result in Half the Energy Loss?

[gangsta316]

Why is it E = \frac{1}{2} C V^2? I mean why is it half? It seems to arise from integration but can't we say that Q = CV and then multiply both sides by V to give QV = CV^2? The left hand side is equal to work done because voltage is defined by work done/charge.

 

[Doc Al]

The problem with using W = QV is that it's only true if the voltage is fixed. But when you charge the capacitor, the voltage across it is not fixed. You do need to integrate.

 

[robphy]

Following up on Doc Al's post,
it might be helpful to write the energy-stored as
E=\frac{1}{2}\frac{1}{C}Q^2=\frac{1}{2}\left( \frac{1}{C} Q \right) Q
... then note the analogy with a spring (considered in a mechanics course):
E=\frac{1}{2}kx^2=\frac{1}{2}\left( k x \right) x.
Now, ask why isn't the energy-stored in a spring F x =kx^2?
(How would this question be answered in your mechanics course?)

 

[Delphi51]

You have to integrate charge x energy/charge as the capacitor charges up because the value of Q is changing. But it is changing steadily, so no knowledge of calculus is needed:

The integral or the sum of all the little Q x V steps is the area under the triangular graph.
The area of the triangle is A = 0.5 base x height = 0.5 QV = 0.5 CV^2
The factor of one half comes from the triangular shape of the charging graph and is due to the average Q during charging being 0.5 times the final Q.

 

[gangsta316]

http://www.s-cool.co.uk/alevel/physics/capacitors/time-constant-and-energy-stored-in-capacitors.html

Note: the energy used by the cell to charge the capacitor, W = QV, but the energy stored on the capacitor = 1/2 QV. So half the energy is lost in the circuit as heat energy as the capacitor is changed.

As capacitors are able to store energy, they can be used in back-up systems in electrical devices, such as computers.

Why is it exactly half of the energy which is lost as heat? This sounds like an approximation but it should be exact according to the graphs and calculus.

 

[Delphi51]

Why is it exactly half of the energy which is lost as heat?

A fascinating observation, Gangsta! It never occurred to me that energy is lost in charging a capacitor with a battery. I guess the energy is lost in the internal resistance of the battery. If you think of the battery as an ideal voltage source in series with a resistance, initially (when the capacitor has no charge and no voltage) 100% of the energy flowing is lost in the capacitor. Later when the capacitor is fully charged, no current flows and no energy is lost. It happens smoothly with V rising linearly on the capacitor, so the same math applies and the average efficiency is 50%.

 

[Phrak]

A fascinating observation, Gangsta! It never occurred to me that energy is lost in charging a capacitor with a battery. I guess the energy is lost in the internal resistance of the battery. If you think of the battery as an ideal voltage source in series with a resistance, initially (when the capacitor has no charge and no voltage) 100% of the energy flowing is lost in the capacitor. Later when the capacitor is fully charged, no current flows and no energy is lost. It happens smoothly with V rising linearly on the capacitor, so the same math applies and the average efficiency is 50%.

That's essentially correct. No matter what the series resistance, the energy lost is (1/2)CV^2. A current source can be used to charge a capacitor efficiently.