Stubblefield coils (bifilar) and speculations

[DrStiffler]

@ Dr. Stiffler:

First, your input is ALWAYS welcome here on this topic.  As you know, I have been following your work on the SEC with great interest. (and great ignorance on my part)

Second, I have to second Jeanna's question about how to properly measure E and I at the same time?  I have checked E first usually, then set the meter to mA to measure I.  So, with  say about .8 E and 80 mA's, P would equal .646?

@ Jeanna:

One thing I do that might help is to make up some jumper leads with alligator clips soldered to the ends and I just clip them on and off my cu and fe wires because as you say, they can get very brittle very quickly. (I found this out myself by snapping one off already)  This way, I don need to position the original wires much at all, I just move the jumper wires wherever I want them to be.  My jumpers are a pretty heavy gauge so I don't think/hope I am not adding much to the resistance.

Bill

@Bill
This will not work.

You need to have a known load and you can measure the voltage across the load and get watts by w=I^2 X R where the I is the current in amps and R is the resistance of the load. Measuring the voltage across an unloaded coil followed by connecting a current meter across that coil is meaningless in result. If you want to connect a meter in current mode across the coil then connect across the current meter another meter to measure the voltage drop across the meter measuring the current. Then calculate....

Unless your meter is the correct impedance for the coil, you are seeing much lower voltage when reading the current and high voltage in voltage mode because the coil is across such a high impedance.

[resonanceman]

@ Jean

Unless your meter is the correct impedance for the coil, you are seeing much lower voltage when reading the current and high voltage in voltage mode because the coil is across such a high impedance.

RStiffler

I had not thought of that .

Given that   most of the time all we really need to know  is  how  our coils compare with   our other  coils and  similar  coils that others here have made ............isn't a simple   voltage and   current reading  good enough ?

gary

[jeanna]

Does this make sense or have I just thrown mud in the water???

Umm, I am not sure.

I will do what you suggest, then I will ask my next question.

Howsthat?

jeanna

[jeanna]

Does this make sense or have I just thrown mud in the water???

So, I have 4 coils in the ground that I check for V , A  every day.

They all have a different resonant frequency, for sure.

Are you saying that although I can compare my daily results, I have no absolute result unless I can get the coil resonance in accord with the meter resonance?

And if that is what you are saying, how will I know?

thank you

jeanna

[DrStiffler]

@ Jean

Unless your meter is the correct impedance for the coil, you are seeing much lower voltage when reading the current and high voltage in voltage mode because the coil is across such a high impedance.

RStiffler

I had not thought of that .

Given that   most of the time all we really need to know  is  how  our coils compare with   our other  coils and  similar  coils that others here have made ............isn't a simple   voltage and   current reading  good enough ?

gary

No. Not in my mind. How can you say a bottle of milk is good by looking at it? What I mean is a coil of large gauge wire may indeed be able to supply  large amounts of current at a lower voltage, where a smaller wire gauge and more turns will be a higher voltage and lower current. But how can you know the capacity without loading it? Does a voltage comparison make then equal in performance, no, not at all.

So if we have a two 2U6 (9volt transistor batteries connected in series ~100mAHr) which would measure ~18V and three 6V golf cart batteries connected in series ~18v and measure only the voltage, would you be able from the voltage alone be able to tell which had the greater capacity?