# Capacitor math anomaly

I may have a mathematical anomaly concerning the energy transfer between two capacitors.

Here is the experiment:

CAPACITOR SPECS COULOMBS JOULES
CAPACITOR 1 aka C1
C1 11200UF@19V
@ 19V 212.8m 2.02 @17.48V 195.77m 1.71
TOTAL DISCHARGE 17.03m .31

CAPACITOR 2 aka C2
C2 1000UF@100V
TOTAL CHARGE/ C2@ 17.48V 17.48m .16

Capacitor calculator: http://highfields-arc.6te.net/constructors/olcalcs/caclchrg.htm [Broken]

Note: All non-polarized capacitors (polypropylene dielectric- ESR rating of .1582ohms) used in test are of high quality and were well within their tolerances and should maintain at least 90 to 95% efficiency through out all testing cycles.

Wire leads used to connect components to circuit are the actual leads on the capacitors, installed by the manufacturer.

Test:

C1 (19v starting voltage) discharges directly into C2. Both capacitors equalize to an identical voltage of 17.48v within 1 second.

Note: According to the math, C1 discharged 17.03m coulombs @ .31 joules during this cycle directly into C2, which can only hold .16 joules @17.48v.

What happened to the remaining .15 joules during this energy transfer?

This one has me wondering!!!!

Thanks for any help guys!!!!

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## Answers and Replies

The conserved quantity is Coulombs. Work out the energy balance with resistive wires (include the energy loss in the wires), and then take the limit as the wire resistance goes to zero.

Bob S

sophiecentaur
Gold Member
2020 Award
If you discharge a perfect capacitor through wire of no resistance into another capacitor you will produce a resonant circuit that will 'ring' for ever - energy sloshing from one capacitor into the other and back because of the inherent inductance in the construction. The only thing which will eventually cause the oscillations to decay will be radiation of em into space.

Another problem might help illustrate this further. Suppose we have a 1 kilo cannonball traveling at 10 m/s. It has 10 units of momentum and 50 Joules of kinetic energy. Let's say this cannonball embeds into a 10 kilo block of wood that is resting on a frictionless surface. Because it embedded, all the momentum is now carried by combined mass of the two objects. The momentum is 10 and the new mass is 11 kilos. That means the new speed is 10/11 ~ 0.9 m/s. That means that it now has ~4.5 Joules of kinetic energy. 45.5 Joules have disappeared, maybe.

That's a substantial energy loss. If you actually did this experiment (or something similar with pendulums) the place that the energy goes will become obvious. It turns into sound waves. Remember that energy can transform. Sometimes, the only things that we can be sure haven't left confines of our problem will be things like momentum, charge, current, etc.

Thanks Guys
excellent info.
Do any of you know anything about the "capacitor paradox".
After a little research, found this... Could this be what I'm up against?

If I'm breaking off the original question to much, let me know and I will start a new thread if needed.

sophiecentaur
Gold Member
2020 Award
Yes. The paradox is only there if you neglect the real world losses - resistance and radiation.

When performing the above experiment with larger Capacitors and assorted loads/resistors and a calorimeter--
I have observed the following results:
For example,
when discharging C2 (17.48v to 0v) through load to ground, consistently produced around 95-98% as much heat when compared to
C1 (19v to 17.48v) discharging through load to ground

I looked at this as C2 (at 17.48v) storing approx. 98% of energy discharged from C1.
Is this correct? Just want to be sure..

sophiecentaur
Gold Member
2020 Award
The energy stored will be half C Vsquared. Do the sums for before and after conditions and see what you'd expect to get.
You haven't quoted capacitor values so you'll have to work yourself, it out using those values. You can be pretty sure that your measurements will agree with theory - minus a bit for losses that your calorimeter hasn't caught. Calorimetry is unreliable unless you have high class kit.

uart
Here is the experiment:

CAPACITOR SPECS COULOMBS JOULES
CAPACITOR 1 aka C1
C1 11200UF@19V
@ 19V 212.8m 2.02 @17.48V 195.77m 1.71
TOTAL DISCHARGE 17.03m .31

CAPACITOR 2 aka C2
C2 1000UF@100V
TOTAL CHARGE/ C2@ 17.48V 17.48m .16

Capacitor calculator: http://highfields-arc.6te.net/constructors/olcalcs/caclchrg.htm [Broken]

Note: All non-polarized capacitors (polypropylene dielectric- ESR rating of .1582ohms) used in test are of high quality and were well within their tolerances and should maintain at least 90 to 95% efficiency through out all testing cycles.

Brad are you sure about that, 11200 uF non polarized polypropylene capacitor. That is an enormous capacitance for that type of capacitor. The largest polypropylene caps I’m aware of are motor start capacitors and they usually go up to about 100uF. I would have expected that you’d only find 11200uF in an electrolytic capacitor. Where did you get such a large polypropylene non polarized capacitor and physically how large is it?

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Let's look at this problem from the standpoint of conservation of energy, first with a series resistance R between the two capacitors when C1 (11,200 uF @ 19 volts) is discharged into C2 (1000 uF @ 0 volts). The series equivalent capacitance of C1 and C2 is 918 uF, which I will call C3. Initially, the voltage between the two capacitors is 19 volts, and the initial power dissipated in R = 192/R. The total energy E dissipated in R during the complete discharge of C1 into C2 is

E = ∫oV(t)2/R dt =192oe-2t/τ/R dt = 192·τ/2·R

where V(t) = 19 e-t/τ and τ = R·C3

So E = 192R·C3/2·R = 192·C3/2 which is independent of R !!!

So the total dissipated energy is 192 x 918 x 10-6/2 = 0.165 Joules, independent of R. So even when R -> zero, there is energy dissipated during the charge transferred from C1 to C2. Charge (number of electrons) is conserved.

Bob S

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