Standing Waves in Generators

[i_ron]

Sounds funny, forgive me but english is not my native language.
What would the expression "there's an awful lot of cat in that attempt" exactly indicate?

To me it could also be a reference to the Ivor Catt anomaly

http://www.ivorcatt.com/28anom.htm

Ron

[Cap-Z-ro]

Well, thats certainly a read and a half...thanks for posting that up i_ron.

Regards...

[plengo]

Excellent conversation fellas !

[gyulasun]

....
In fig 2 in the pic what I get is that both caps grow in voltage fairly fast, after about 30 seconds of run both caps are at 30 to 40 volts not just one, and they seem to get there faster than in what I am assuming is the conventional way of hooking them up.

The caps are rated at 14000mfd 30V, so I take 30 seconds of 12V 0.1A and charge both caps to 35V so that is about 7 joules or roughly 14 watt seconds of power so it is nothing overly interesting except that I am having a hard time figuring out WHY both caps charge up so much and that is without the common rail being used, or even if I use the common rail and install a cap.

Hi webby1,

In Fig 2, when you close then open the switch, the big voltage spike from the collapsing magnetic field would charge up BOTH capacitors via the right hand side diode because the two caps are in series from the spike current point of view.  When you open the switch the voltage spike has a polarity between the coil ends as follows: the coil end connected directly to the switch will be positive with respect to the other coil end (this latter is also the battery positive)  and current can flow via the right hand side diode and via the two caps and towards battery negative, to close the circuit for the voltage spike.  And in case part labeled A is still connected AND has a certain impedance or resistance, then the spike current can of course flow via part A towards the other coil end (which is connected to battery positive).  The higher the inner impedance or resistance of part A, the more energy can charge up the capacitors, in case part A is a dead short, then the left hand side cap is charged up to the battery voltage (from the battery because it is wired in parallel in that case) and the higher the inner resistance or impedance of part A, the more energy remains for charging the series caps.  In this latter case the battery also receives a certain (remaining) energy from the spike because it is included in the voltage spike closed circuit.

Hope this helps?

Gyula

[gyulasun]

Hi,

A correction to your sentence: "...in fact the cap becomes a 14000mfd 60V cap,"

I would say: the cap becomes a 7000mfd 60V cap  (because two 14000mfd in series gives 7000mfd).  Agree?

And this explains why the charge up time quicker versus the case in your Fig 1.

Gyula