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Where does the 1/2 come from in the 1/2CV2 capacitor energy calculation?

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where does 1/2 come from in capacitor energy calculation

Hi,

where does 1/2 come from with the 1/2CV2 calculation for energy in a capacitor?
 

Re: where does 1/2 come from in capacitor energy calculation

Energy(work done) in charging a capacitance C to v volts is ,

∫Cv dv (between limits 0 to v)

This integration of vdv gives that as 1/2 Cv2
 
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    erikl

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Re: where does 1/2 come from in capacitor energy calculation

For a capacitor C=Q/V or Q=CV where C is the capacitance, Q is the charge on the capacitor, and V is the capacitor voltage.

Also, the energy to move a charge is E =QV where E is the energy, Q is the charge transferred, and V is the voltage it takes to transfer (move) that charge.

If you substitute from the capacitance equation, you get E = CV*V = CV².
(So where does the 1/2 come from)?

V is the final voltage on the capacitor but all the charges transferred to the capacitor did not require that final voltage to be transferred into the capacitor.
The voltage to move each charge starts at zero and linearly increases to the final voltage as the capacitor charges.
Thus the average to move the charge is 1/2 the final voltage or 1/2V, making the energy to charge the capacitor 1/2 CV², which has to equal the energy stored on the capacitor.

An interesting observation of this is that charging the capacitor from a fixed voltage through a resistor will dissipate the same amount of energy in the resistor as ends up stored on the capacitor, since the total energy transferred is CV² but the stored capacitive energy is only 1/2 CV².
Thus the charging efficiency from a resistive source is 50%.
This loss can be minimized if the capacitor is resonantly charged through an inductor with low or negligible resistance, which can give a charging efficiency of near 100%.
 
Re: where does 1/2 come from in capacitor energy calculation

Which is why there is an inductor in the path of switching DC to DC converters.

Bob
 

Re: where does 1/2 come from in capacitor energy calculation

I think of it as geometry. It's the area under a straight sloping line which forms a triangle.

The X axis is V
The Y axis is charge
C is the slope
Energy is the area under that line
That's the area of a triangle
That's the area of the rectangle bounding that triangle divided by two
Rectangle X is V, rectangle Y is C*V (because c is the slope)
Thus area is V*C*V/2 or 1/2CV^2

Of course this is a long but simple way to do a first order integral which as pointed out gives the same answer. This is also the exact same form of many other equations:
1/2MV^2 (Energy of mass M at velocity V)
1/2LI^2 (Energy of inductance L with current I)
And undoubtedly many others

Good diagram here:
https://www.quora.com/The-momentum-P-and-kinetic-energy-E-of-a-body-of-mass-M-are-related-as
 
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