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11th November 2019, 16:26 #21
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Re: Ohm's Law and Power calculation problem
As a service for the persons reading this post, I made the following calculations:
I took the first full pulse from the data file (SDS00001.CSV attached in previous thread) and it looks like this
Then I inverted the current by multiplying it with 1 and the result looks like this
Note that the current still goes negative while the voltage is positive.
Then I calculated instantaneous power P = U * I and energy ∫P
And I calculated the resistance R = U / I
Note that the resistive spike occurs at the same time as the current crosses zero.
If I in fact have turned the CT the wrong way, then I would really have a "pertuum mobile" in my hands as the output energy would be 2.45J while input is 50mJ (calculated excluding losses i.e. the final amount of energy that enters "R") i.e. 4900% efficiency. So logic dictates that I could not have turned the CT the wrong way around, hence I still believe we are talking about true negative resistance.
In general, it may also be noteworthy that if the resistor "R" is disconnected from the source it behaves just like any ordinary resistor and has a resistance of 883K ohm.Last edited by Swend; 11th November 2019 at 16:43.

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12th November 2019, 15:45 #22
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Re: Ohm's Law and Power calculation problem
As a service for the persons reading this post, I made the following calculations:
Such points are singular and must be skipped in the plot. That happens when you have a finite U(t) but the current I(t) becomes zero, you have this problem.
The reason is simple: if the U(t) or I(t) are not constants in time, they must be treated as a complex quantity. Then R(t) is also complex and has a real and imaginary parts. At the singular point, the real and imaginary parts are equal and opposite.
Such behaviours are common whenever there are reactive components.

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Yesterday, 10:20 #23
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Re: Ohm's Law and Power calculation problem

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Yesterday, 11:01 #24
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Re: Ohm's Law and Power calculation problem
Yes that's how it's calculated.
Yes, I agree in the case of singular points, but 6.7e10 as in the plot is still finite.
Your system has a natural frequency and it is damped poorly. You can approximate the system as a resistor far from this frequency (on both sides): but how far is far enough?
And what if both U(t) AND I(t), as the case is here, are not constants in time?
Just for fun:
Consider U(t)=1+2*t (a simple ramp); the current is also a simple ramp: I(t)=2+1*t; what will be R(t)??
If R(t) is constant and real, then I(t) is automatically defined for all t. But how about Re(R(t)) is constant and Im(R(t)) is also constant (but nonzero)??
Simple questions do not have simple answers. I will stop here.

Yesterday, 11:16 #25
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Re: Ohm's Law and Power calculation problem
The calculations and respective conclusions are presuming correct I and U measurements.
Particularly the current measurement is very implausible, see previous discussion, e.g. post #12.

Yesterday, 14:03 #26
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Re: Ohm's Law and Power calculation problem
Yes, but in this data series, the current is never zero, otherwise the calculating script throws a divide by zero warning, which it doesn't in this case.  so no zeros.
As I have said several times, and as it can be seen in the trace, before the pulse  both current and voltage are zero. And the pulse in the trace is the very first after power on.
Yes I can agree.
   Updated   
I'm presuming the measurements are correct until there is valid reason they are not. So far I have not heard any valid reason, "impossible" and "implausible" are not a valid reasons, they are just empty statements, so I kindly ask you to be more serious.
For which valid reason are you making that claim?
You have seen the pulse from start in #21, so I'm still waiting your sentiments.
Since it is in my interest the measurements are correct, I would like to propose that I drop by your lab at your convenience, where you can repeat the measurements with your own apparatus, and even examine the source to your personal satisfaction. And if you don't work for free  then give me your price.
Actually I would personally prefer if the current was positive all the way, I have no use for negative current.Last edited by Swend; Yesterday at 14:11.

Yesterday, 23:13 #27
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Re: Ohm's Law and Power calculation problem
I gave some hints why the current waveform is likely to be distorted by the characteristic of the transducer. There may be more problems with your circuit that are not so obvious at first sight.
In my view, the final prove that voltage and current waveforms are not consistent is the apparent "resistance" you have calculated. At this point, it's pretty clear that the measurement function needs to be validated from the very bottom.

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Today, 02:41 #28
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Re: Ohm's Law and Power calculation problem
Yes, but in this data series, the current is never zero, otherwise the calculating script throws a divide by zero warning, which it doesn't in this case.  so no zeros.
As I have said several times, and as it can be seen in the trace, before the pulse  both current and voltage are zero. And the pulse in the trace is the very first after power on
Hint: all physical measurements take time and all the results are simple averages over the measurement time period.
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