Ok lets find the capacitance of our wfc?

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It seems nice, I wish you luck. Thanks for the link 

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Another mean for discovering the capacitance is to construct a relaxation oscillator with an opamp. The frequency output ^-1 will give you the period of the RC product.
The opamp positive input must be set to 63% of the supply voltage...

As the water is leaky a mean to provide the current needed is to use a low R value. Thus a simple opamp relaxation is not enough, you need a Half bridge to boost the opamp current.

I believe this will give out the best results... if correctly used.

The opamp must be supplied with the same voltage or so going into the bridge.... So one could use two batteries, a full wave bridge rectifier from a variac and using a ground connection to split the DC output into positive and negative between two big capacitors at the dc output...

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$ 1 9.2E-9 37.11724081536377 61 10.0 50
a 464 384 640 384 0 15.0 -15.0 1000000.0
w 464 368 464 320 0
r 464 320 576 320 0 100.0
w 576 320 608 320 0
w 784 448 576 448 0
r 576 448 464 448 0 62000.0
w 464 400 464 448 0
r 464 448 368 448 0 20000.0
g 368 448 368 480 0
c 464 320 368 320 0 1.0E-7 0.6670691661928481
g 368 320 368 352 0
O 640 384 704 384 0
r 640 528 640 384 0 0.1
w 784 336 608 320 0
O 928 528 992 528 0
R 864 448 864 496 0 0 40.0 -15.0 0.0 0.0 0.5
R 864 224 864 176 0 0 40.0 15.0 0.0 0.0 0.5
w 864 224 864 272 0
w 864 400 864 448 0
w 864 336 864 368 0
w 864 304 864 336 0
t 928 384 864 384 0 -1 29.999632015926228 0.6985570474976956 20.0
t 928 288 864 288 0 1 -3.67984073772476E-4 0.6985570474976956 20.0
w 784 336 864 336 0
r 464 256 368 256 0 1000.0
w 368 256 368 320 0
w 464 256 464 320 0
w 640 528 928 528 0
w 928 528 928 384 0
w 928 384 928 352 0
w 928 352 928 288 0
w 784 336 784 448 0
o 9 64 0 35 5.0 1.6 0 -1
o 11 64 0 42 20.0 9.765625E-5 1 -1
o 12 64 0 35 0.009765625 0.1 2 -1


Here i used the voltage divider to tune the threshold to satisfy the RC time constant frequency for a known capacitor. Than the only limitation is that as the R in series is = to the R in parallel with water the frequency will become a little smaller, this off set can be interpreted as the ratio between the resistances. In this case the precision will be within around 7%... As the series resistance is preset much smaller than water leak the precision goes up to 1% or less...

 

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So using test 4 from the link in my previous post, I have determined that the actual capacitance of my one test cells is 40nf where I originally calculated it to be 25nf.  What is more (and possibly more important) I have also determined that the water cell's internal resistance is 3 ohms.

This also makes the dielectric value of the water I am presently using to be 125. Either that or I am using the wrong constant for permittivity.

In case anyone is interested.   

TS

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So using test 4 from the link in my previous post, I have determined that the actual capacitance of my one test cells is 40nf where I originally calculated it to be 25nf.  What is more (and possibly more important) I have also determined that the water cell's internal resistance is 3 ohms.

In case anyone is interested.   

TS

Presumably, to use the fact that the resistance is 3 ohms, to make use of this, I would need to create a pulse that is short enough that 3 ohms is enough to prevent dielectric breakdown by current.

TS

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Hi TS

What are the area and separation of your electrodes, if were tubes what are the length and diameters?

How did you made the test? what were the values of the components?

if you connect a battery with 12v across the water the current is 4 amps ?

Is that pure water? tap? distilled?

Thanks

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I'm trying some variations to confirm my results.  I'll provide more details soon.

TS

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Make a tables, take measurements at different frequencies, different voltages and components...

than you can compare and be more sure of your measurement...

good luck