Fun Capacitor Circuit

[hartiberlin]

1)Take 3 identical caps 2) lets call them C1,C2 and C3   3)charge C1 to 12 volts 4)connect C1 neg to C2 neg 5) connect C2 pos to C3 neg 6) finally connect C3 pos back over to C1 pos
You will end up with approx. 8 volts in C1 and 4 volts in each C2 and C3  - thats it !

Best,

Okay, you end up with 3 x 22.000uF caps in series which have due to the series circuit now
about 7333 uF with a total voltage of 16 Volts= 8 + 4 + 4 Volts.
So does this 16 Volts at 7.333 uF run your motor longer than the 12 Volts at 22.000 uF ?
If yes, it is probably due to the motor you are using, that it has a better efficiency at
higher voltages or something like this...
If you would use just a load resistor you would see, that it will get hotter
at the 12 Volts at 22.000 uF cap.

[NerzhDishual]

Salve a tutti

Very sorry.

But, I am just wondering.

According to pg46 descriptions and my following pics (should these pics being accurate).

In that case, the most important question could be: how can you charge a cap with a
single wire?

Or, I am mistaken something?

Best

[hartiberlin]

@NerzhDishual 

he is just placing C2+C3 in parallel with C1.
Then C1 discharges until it has 8 Volts and also C2+C3 have
8 Volts, so each C2= 4 Volts and C3= 4 Volts.
Then he puts alls 3 caps in series getting 16 Volts but at a lower
total capacity...
So I don?t think, he can run the motor longer on the 3 caps in series...
at least not in normal theory....

[pg46]

Thanks NerzhDishual -

Very nice diagrams, thanks! Yes indeed, that's the way I connected them.
Please note that the last connection from C3 pos back to C1 pos need only be momentary. Thanks also for your comments.

hartiberlin- You can see by the connections that my circuit is neither in series or parallel or else they are both! or maybe you could say that C1 and C3 are hooked in parallel but with a cap(C2) "in-line" between them
I do not connect any caps in series after my circuit as although that would be more volts, there would be less capacitance and so it will not run a motor any longer than the original C1 cap charged to 12 volts. So, before and after my circuit I do all testing on caps individually or else in parallel only.

In this group of tests and examples I used all the same caps, 35volt 22,000uf caps and a single DC motor for any discharge timings.

SET UP #1 - I charge C1 to 12 Volts, 22,000uf. If I connect C1 in parallel to empty C2 and C3 caps, I will end up with 4 volts and 66,000uf.  When I discharge this through my motor, it will take 30 seconds to empty the charge.
SETUP #2, if I charge C1 to 12 volts at 22,000uf but then this time I run the pg46 circuit with C2 and C3 , I can get 1 cap at approx. 8 volts(C1) and two caps(C2&C3)at 4 volts each. So, then if I introduce another cap C4 and reconnect the 4 caps in parallel I will end up with 4 volts @ 88,000uf. Now, when I discharge this through the motor, this time it will take 40 seconds to empty the charge, approx. a 30% gain over the "normal" circuit with the same 12 volt input charge.
The reason I use 4 volts to time the discharges is because "4" is a common denominator in both the original 12 volt charge and the resulting voltage from the pg46 ciruit(16v) The motor will run faster or slower depending on the voltage so I wanted to make sure I did the timings all using the same voltage.
Another way I can say all this is that in setup#1, I can produce 3 caps of 4 volts each from a single 12 volt charge. With my circuit I can produce 4 caps at 4 volts each with the same single 12 volt charge.
If one thinks of capacitors as batteries, one person might offer 3- 4 volt batteries from a single 12 volt charge whereas with my circuit one could offer 4 - 4volt batteries from a single 12 volt charge

Best,

[hartiberlin]

Okay, PG46,
now I see, what you are doing...
But you did not test,how long the motor will
run on the 12 Volts C1 cap only ?

The only thing you do with your circuit is, you
reduce the losses a bit..
The 12 Volt Cap has 1584 units energy,
the 3 caps at 4 Volts have 528 units energy
and the
4 caps at 4 volts have 704 units of energy, so
you reduce the losses from about 66 % to just 50 % ,
but in total you have lost about 50 % of the used input energy all in all.

You put 1584 units of energy into the  C1 cap, when you had charged it up to 12 Volts
and end up with 4 caps at 4 Volts , that is just left only 704 units of energy !

So just let the motor run on the 12 Volts C1 cap and it will
probably run for about 50 seconds or more...
so the energy equation is still valid with Wcap= 0.5 x C x V^2