### Author Topic: Ok lets find the capacitance of our wfc?  (Read 13381 times)

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•    • • Posts: 3741 ##### Re: Ok lets find the capacitance of our wfc?
« Reply #8 on: June 24, 2012, 18:45:42 pm »
Well it could be i'm not saying is this or that, but can  be.

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« Reply #9 on: June 25, 2012, 05:11:58 am »
Took new measurements... at the resonance the voltage is maximum at the inductor only.. at the capacitor the voltage is high at a lower frequency than resonance than at resonance and the further the frequency goes up the lesser is the woltage.

Thus the best way to find the resonance is to measure the voltage across the inductor.-
« Last Edit: June 25, 2012, 05:27:20 am by sebosfato »

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« Reply #10 on: June 25, 2012, 15:07:58 pm »
Seems to me the best way to find resonance is to find the inductance value of the coil and the capacitance of the cell and use the standard equation to calculate it.  As you stated to me recently, the reactance must be equal between the two.  Additionally, there will be a bandwidth where resonance will occur, so it may not be at one single frequency.

TS

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« Reply #11 on: June 25, 2012, 15:22:45 pm »
Took new measurements... at the resonance the voltage is maximum at the inductor only.. at the capacitor the voltage is high at a lower frequency than resonance than at resonance and the further the frequency goes up the lesser is the woltage.

Thus the best way to find the resonance is to measure the voltage across the inductor.-

Thats odd....

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« Reply #12 on: June 25, 2012, 18:35:03 pm »
Well it might seems strange, i made an excel sheet and ploted a graph, at resonace of course the reactances are the same and thats why current is maximum, however the XC is not linear like the inductive reactance, so at resonance both are the same but as the frequency is increased the higher is the XL and the XC becomes lower as the voltages should be equal to the reactance times the current, the voltages across the inductor and the capacitor become smaller but the capacitance voltage drops more than the inductor.

Also the lower is the Q factor the greater is the difference in frequency between the maximum voltage at the inductor and at the capacitor accoring to the graphs... Actually when i learned about the harmonic oscillator at the physics lecture, i learned there are two types of resonance, one is at the amplitude and the other is at the power. Maybe thats the difference...

Yet I attributed the lower voltage at the capacitor than at the inductor to the fact that although the capacitor holds the charge, it also has the greater loss in the circuit. so or the capacitance change as a function of V(t) and i think it does because i got to change the frequency to find a greater voltage if i raised the Vt applied.

Again at F=0 the capacitor has Vt the inductor has 0v

Closer to resonance the capacitor has greater voltage than the inductor, the current start to raise

At resonance the voltages should be equal but as there are losses in the capacitor it will have a lower voltage than at the inductor. current is maximum

little above the resonance the inductor voltage still greater than the capacitors, the current decreases

way above resonance the voltage at the inductor Vt and capacitor is 0  the current is 0 too

In theory this should be the behavior

In the case of water I think because of the losses, the voltage across it aways decrease from 0 frequency to max frequency. so no voltage peaks... Maybe at resonance it decrease less..

Thats why i said its easier to find the resonance measuring the voltage across the inductor...

Since stanley had the feed back on the inductor, this imply Zero current switch and imply that the frequency is adjusted to max inductor voltages.

Ts as i stated @resonance XL = XC so XL-XC=0 condition required for resonance, yes there is a bandwidth not much wide.

What i'm saying is the waters capacitance is not the calculated value because it's a lossy dielectric, this experiment thus allow you to empirically determine the capacitance observing the frequency where resonance happens.

The first graph has 100 ohm resistance and the other 20 ohms

C = 500n
L = 1mh
Fres=7118

« Last Edit: June 25, 2012, 19:16:28 pm by sebosfato »

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•    • • Posts: 3741 ##### Re: Ok lets find the capacitance of our wfc?
« Reply #13 on: June 25, 2012, 19:53:31 pm »
If you try this simple circuit you see that the capacitor having a resistance in parallel with it shows a smaller voltage than the inductor.

Now i understand why. The Voltage across the inductor will be = to the current passing thru the capacitance + the current passing thru the resistive path. while the capacitance only the capacitive current...

\$ 1 3.4000000000000003E-7 61.195984441989225 50 5.0 43
r 352 128 560 128 0 1.0
s 560 128 624 128 0 1 false
c 352 400 560 400 0 5.000000000000001E-7 -105.48842737546646
l 560 128 560 400 0 0.0020 3.6875247967452998
v 352 400 352 128 0 2 5000.0 50.0 0.0 0.0 0.5
r 432 448 496 448 0 250.0
w 432 448 352 400 0
w 496 448 560 400 0
o 3 64 0 35 320.0 6.4 0 -1
o 2 64 0 35 320.0 6.4 1 -1
o 0 64 0 35 5.0 6.4 2 -1
o 4 64 0 35 80.0 6.4 3 -1

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« Reply #14 on: June 25, 2012, 22:09:16 pm »
Just did other tests, and i'm not going to share much of the parameters to force you doing this experiment

I tried different inductors, using and not using cores, and got different readings for the capacitances in respect to frequency...

The lowest capacitance I found were at 50khz around 350nf +- 20nf

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•   • Posts: 431 ##### Re: Ok lets find the capacitance of our wfc?
« Reply #15 on: June 25, 2012, 22:23:57 pm »
When you did your capacitance calculations for your water cells, did you factor in the dielectric of water into the equation?  If factoring in the dielectric into the equations still leads to such a massive error, then either the equation is wrong, or the water dielectric value being assumed is wrong.  Either way, it would be more appropriate to correct the equation and dielectric factors and help everyone to correctly calculate the dielectric values.  It would save time and help everyone figure out this riddle.

TS