# Ohm's Law and Power calculation problem

1. ## 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.  Reply With Quote

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2. ## Re: Ohm's Law and Power calculation problem

As a service for the persons reading this post, I made the following calculations:
Please explain: if you are using R(t)=U(t)/I(t) to calculate the resistance, I have no problem. But the 4th graph is misleading, because you are trying to plot a point where the R becomes infinite.

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.  Reply With Quote

3. ## Re: Ohm's Law and Power calculation problem Originally Posted by c_mitra Please explain: if you are using R(t)=U(t)/I(t) to calculate the resistance, I have no problem.
Yes that's how it's calculated. Originally Posted by c_mitra But the 4th graph is misleading, because you are trying to plot a point where the R becomes infinite.

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.
Yes, I agree in the case of singular points, but 6.7e10 as in the plot is still finite. Originally Posted by c_mitra 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.
And what if both U(t) AND I(t), as the case is here, are not constants in time?  Reply With Quote

4. ## Re: Ohm's Law and Power calculation problem

Yes that's how it's calculated.
What you have done is called point-wise division. This may not be same R(t)=U(t)/I(t) under some conditions. Particularly near singularities. I won't go into these details. Recall that division by zero is not defined.

Yes, I agree in the case of singular points, but 6.7e10 as in the plot is still finite.
In computer, we can define a zero but not infinity because infinity is not a number. You can see what I mean if you convert the y-axis scale to a log scale.

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?
You must know the waveform: in other words, the voltage a moment back has an effect on the current at the present moment (if your circuit is reactive). Because of Ohm's law, only two of the three (U,I and R) are independent.

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 non-zero)??

Simple questions do not have simple answers. I will stop here.  Reply With Quote

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5. ## 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.  Reply With Quote

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6. ## Re: Ohm's Law and Power calculation problem Originally Posted by c_mitra What you have done is called point-wise division. This may not be same R(t)=U(t)/I(t) under some conditions. Particularly near singularities. I won't go into these details. Recall that division by zero is not defined.
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. Originally Posted by c_mitra You must know the waveform: in other words, the voltage a moment back has an effect on the current at the present moment (if your circuit is reactive). Because of Ohm's law, only two of the three (U,I and R) are independent.
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. Originally Posted by c_mitra Simple questions do not have simple answers. I will stop here.
Yes I can agree.

- - - Updated - - - Originally Posted by FvM The calculations and respective conclusions are presuming correct I and U measurements.
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. Originally Posted by FvM Particularly the current measurement is very implausible, see previous discussion, e.g. post #12.
For which valid reason are you making that claim? Originally Posted by FvM I agree with KlausST that the current measurement channel seems to be inverted. I' d think about possible alternative explanations if I see the pulse response form the start.
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.  Reply With Quote

7. ## 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.  Reply With Quote

8. ## 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.
It need not be exactly zero; in floating point asthmatics, you test for zero with a small number.

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
when both current and voltage are zero, what is the calculated value of the resistance?

Hint: all physical measurements take time and all the results are simple averages over the measurement time period.  Reply With Quote

9. ## Re: Ohm's Law and Power calculation problem Originally Posted by FvM 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.
So you are stating a personal opinion, and because of that opinion it's clear to you that measurements are wrong and therefore you can't be bothered to look at the facts even for all the money in the world.

It makes me think of this Leo Tolstoy quote:

I know that most men, including those at ease with problems of the greatest
complexity, can seldom accept even the simplest and most obvious truth if it
be such as would oblige them to admit the falsity of conclusions which they
have delighted in explaining to colleagues, which they have proudly taught to
others, and which they have woven, thread by thread, into the fabric of their lives.

- - - Updated - - - Originally Posted by c_mitra when both current and voltage are zero, what is the calculated value of the resistance?
Then you have a serious singularity problem. Originally Posted by c_mitra Hint: all physical measurements take time and all the results are simple averages over the measurement time period.
How can a single point in time measurement be an average?

Anyway I'm getting fed up by this thread, it's generating NOTHING of value. I will declare it as [SOLVED].

For those who have a sincere technical interest in the subject, please feel free to contact me by PM, this is apparently not a subject for public consumption.  Reply With Quote

10. ## Re: Ohm's Law and Power calculation problem

How can a single point in time measurement be an average?
That is the point I am trying to convey: it is not possible to measure any quantity in ZERO time; all physical measurements take time to perform.

Then you have a serious singularity problem.
No, it is still defined. Consider a sine wave applied to a pure resistor; the current and voltage graphs will be same (except for a scale factor). The resistance value will still be known when the voltage is zero (say at t=0).

if U(t) =a sin (alpha t) and I(t) = b sin (alpha t), R(t) = a/b but at t=0, it is 0/0. You need to use l'Hospital rule to get R(t=0)=a/b.

I too give up.

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11. ## Re: Ohm's Law and Power calculation problem Originally Posted by c_mitra You need to use l'Hospital rule to get R(t=0)=a/b
Well, that trick I didn't know, so thank you for that.

Then it's only fitting that I amend my statement to "it's [this post] generating VERY LITTLE of value"  Reply With Quote

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