### Author Topic: Ok lets find the capacitance of our wfc?  (Read 14189 times)

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##### Re: Ok lets find the capacitance of our wfc?
« Reply #16 on: June 25, 2012, 23:20:25 pm »
The value of 78.5 is obtained in a different way, it is determined differently...

The problem i see for this is that i think the dielectric constant or the water capacitor behaves different according to applied voltage and frequency.

So the best way is to measure the capacitance for different shapes, sizes, voltages and frequencies, and plot graphs with the results.

I'm using square wave, probably would be better a sine wave...

Yes i used the dielectric constant in the calculation and the calculation is correct because i can measure the capacitance without water inside... but the capacitance don't get anyway close to the capacitance measured on this experiment.

We should be able to create thru calculation a dielectric constant function of voltage and frequency.

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##### Re: Ok lets find the capacitance of our wfc?
« Reply #17 on: June 26, 2012, 01:04:42 am »
Incidentally, when using distilled water, my calculations have usually been within 200Hz of accurate when manually tuning to resonance.

TS

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##### Another method to determine the capacitance of the wfc
« Reply #18 on: June 26, 2012, 07:42:00 am »
Its know that a RC capacitance resistance series circuit, has a time constant t=R*C

The time constant is the time it takes for a capacitor to charge from zero to 63,2% of the applied voltage. 0,632 * Vo

So If any of you have an oscilloscope and could try some measurements would be a nice.

Some one could Even create a comparator circuit regulated for this parameter and measure the output pulse length.  Its easy you just need an opamp and a two resistors divider circuit or a trim pot for the reference. Sharing a common ground the opamp and the cell and having the resistor between the positive and the wfc capacitor the up side of the capacitor is connected to the other input of the opamp... switch on and freeze the scope shot. another way is to record the signal as audio  *.*.wav to the computer... and analyze the pulse length with any good audio editing program like sound forge.

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##### thinking a little better
« Reply #19 on: June 26, 2012, 07:49:29 am »
Let take out the juice.

just need a battery a resistor a switch a wfc and a potentiometer and a pc. connect the resistor to the cell to the car battery to the switch now across the cell connect the trimpot, and send the signal to the audio in port of the computer... press record in any audio recording software and switch on.

Than inside sound forge or the like measure where the voltage arrived at 63,2%. thats the time constant.

Try different resistors

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##### How to measure the resistance of the cell
« Reply #20 on: June 26, 2012, 07:57:23 am »
Connect a battery a resistor and the cell. Measure the current and voltage across the resistor and the cell

The resistance might be a function of voltage.

This is kind of how you measure the internal resistance of a battery (polarization)
You measure the open circuit voltage than connect the battery to a resistor measure again the voltage across the battery and the current. The voltage drop Vd=I*Ri is the current times the internal resistance thus Ri=Vd/I

At the end they are both electrochemical components.

So consider its internal resistances.

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##### Polarization
« Reply #21 on: June 27, 2012, 13:32:08 pm »
If I remember well the relative dielectric constant is determined as the ratio between applied voltage and the induced contrary voltage inside the capacitor and is measured by charging the capacitor and subsequently disconnecting it from the source while still measuring the voltage across the capacitor.

From the document i attached there is an example for the hydrochloric acid which has a dielectric constant of 4.12 acording to the reference...

In the example is not really clear, because they say E0=1,5v and Enet=1,15v being Enet=E0-E1 solving E1 becomes 0.35v

Than they say that K=V0/V

however 1,5/1,15=1,3 and the value i found for hydrochloric acid dielectric at the reference is 4,12

so i found that 1,5-1,15=0,35 that would be the E1 the induced contrary voltage than if i divide 1,5/0,35 than i find a value that is in agreement with the reference 4,3.

For example you charge a water capacitor to 100v And its dielectric constant is 81k Thus the induced contrary voltage equals 100v/81k=1,234v So if a capacitor was charged up to 100v and is disconnected the measured voltage is 100v-1,23v=98,76v

If the document was right:

If it was like they say in the document the calculation for water would be 100v/81=1,23v however this time the 1,23v is the remained Enet so the E1 would be 98,76v

If it were vacuum which has 1 as dielectric constant, the induced contrary voltage should be zero than i would guess, so 100/1=100 so Enet=100 and thereto E1 polarization = 0. Which seems to be the case...

So why the contradiction on the hydrochloric acid? is the document wrong only there?

reference:
http://www.asiinstr.com/technical/Dielectric%20Constants.htm#Section%20H

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##### Voltage difference
« Reply #22 on: June 27, 2012, 13:38:24 pm »
I was thinking about it to try to explain to myself why the voltage across the capacitor has no peak at resonance while resonance still occur clearly at the inductor.

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##### Re: Ok lets find the capacitance of our wfc?
« Reply #23 on: June 27, 2012, 13:57:26 pm »
Mystery solved...

At least in part.

The document was right, although the example of the hydrochloric acid is wrong in numbers.

Thats because a dielectric constant affects the ratio of voltage to charge since Q=CV If the capacitance increase the voltage reduces... This reduction is actually this contrary induced voltage (polarization)

So for water if you apply 100v there is 810v induced contrary thereto reducing voltage per unity charge ("increasing capacitance") thus the electric field so the electric field applied is 100v after the capacitor is disconnected from the source.

Or this is correct or the assumption made at the document is wrong... Maybe they wanted to mean that if the dielectric is inserted than the effective electric field reduces by a factor K instead of turning on the switch and turning off while still measuring the voltage.

Because if this is true and the dielectric strength of water is 2,946MV/cm it would become clear that applying 3,637kv the water would breakdown.

reference
http://www.tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/exercise/s3_2_1.html#_dum_9
« Last Edit: June 27, 2012, 16:35:03 pm by sebosfato »