Quote from: Cap-Z-ro on April 17, 2011, 11:11:58 AM
100 feet of wire stretched horizontally in the air 100 up will accumulate a charge...is that not by definition a capacitive property ?

The varying values of resisters is defined as a resistive property, is it not ?

You appear to be engaging in games of semantics, are you not ?

Regards...

You appear to be mixing up and confusing two very different phenomena.

An elevated conductor such as you mentioned acquires a charge separation because it is immersed in the earth's electric field and has become one plate of an "earth capacitor". Measured with respect to earth ground there will be a potential difference.

What we are discussing in this thread is the use of a piece of wire to conduct electron flow in a closed circuit for charging one capacitor from another.

Two very different concepts actually.

.99

Hi P.99

Yes i aggree with you, but it is not what i meant in my post.

What i meant is that the virtual circuit would be 2  NORMAL caps (that is with some resistance)  directly connected  (without diode and inductor) by a  hypothetical zero resistance Wire.

The idea to check here is .
When we have to transfer the energy from a cap to an other cap of same capacity  (or not)  even if we would have the best transfer wiring (zero resistance), we will lose anyway 50 % of energy.

So if we get exactly 50 % of energy lost after the transfer,  in this case, this direct transfer was 100 % efficient,     as it seems to be impossible to get more , without using a diode  and an inductor in the circuit.

So to resume,

2 normal caps directly connected with an hypothetical " zero resistance" wire   have  the best direct transfer possible efficiency (100% ) and this high end transfer will result in 50 % energy lost in the best case.

So if the wire is somehow resistant,which is always the case in real life,  the direct transfer will not be 100 % efficient , but perhaps 98 %  . That is we will always and anyway lose  more than 50% of energy in a direct transfer in our terrific real life.

Does it make sense ?

Thank's for your patience

Laurent

If that wire is coiled its capacitance increases to the point it is recognized as a capacitive inductor...isn't that correct ?

Regards...

 

Quote from: Cap-Z-ro on April 17, 2011, 11:34:07 AM
If that wire is coiled its capacitance increases to the point it is recognized as a capacitive inductor...isn't that correct ?

Regards...

The capacitance will remain constant because the plate surface area and distance to the earth has not changed.

The conductive plate made up by the wire is mostly just a plate, regardless of it's configuration. It is all about surface area.

All capacitors have some integral series inductance as well, but this is many magnitudes lower than the capacitance value. Placing the "lead wire" at one end of this coiled capacitor plate may have the effect of very slightly increasing this stray inductance.

.99

Quote from: woopy on April 17, 2011, 11:33:26 AM

Hi P.99

Yes i aggree with you, but it is not what i meant in my post.

What i meant is that the virtual circuit would be 2  NORMAL caps (that is with some resistance)  directly connected  (without diode and inductor) by a  hypothetical zero resistance Wire.

The idea to check here is .
When we have to transfer the energy from a cap to an other cap of same capacity  (or not)  even if we would have the best transfer wiring (zero resistance), we will lose anyway 50 % of energy.

Yes, that is correct.

Quote
So if we get exactly 50 % of energy lost after the transfer,  in this case, this direct transfer was 100 % efficient,     as it seems to be impossible to get more , without using a diode  and an inductor in the circuit.

So to resume,

2 normal caps directly connected with an hypothetical " zero resistance" wire   have  the best direct transfer possible efficiency (100% ) and this high end transfer will result in 50 % energy lost in the best case.

So if the wire is somehow resistant,which is always the case in real life,  the direct transfer will not be 100 % efficient , but perhaps 98 %  . That is we will always and anyway lose  more than 50% of energy in a direct transfer in our terrific real life.

Does it make sense ?

Thank's for your patience

Laurent

I know what you are saying, however we should be careful when distinguishing between % efficiency of transfer and % of energy transferred.

With two normal capacitors the energy transfer efficiency is 50%, regardless if the resistance is in the capacitors or the piece of wire joining them (they can be considered together).

The amount of energy transferred to C2 is 25%. The amount left in C1 is 25%. The amount burned up in the wiring is 50%.

So in summary, we can not say that this "normal" energy transfer (with the 50% loss) is 100% efficient, because it is only 50% efficient. In fact it cost us 50% of the total energy just to move 25% from one place to the other. In that sense, you might view it as being even less than 50% efficient.

.99