# Ohm's Law and Power calculation problem

1. ## Ohm's Law and Power calculation problem

Hi friends,

I understand there are three ways of calculating power

(UI) P = U * I
(UUR) P = Uē / R
(IIR) P = Iē * R

When I measure voltage, current and resistance in my circuit, and plug the values into the above formulas - I get three very different results. So the question is which one is correct?

(UI) P = 809.471 W
(UUR) P = 25W
(IIR) P = -3.731 W

All above are average power.

This is my circuit

This is the voltage and current measurements. The resistance is 1M ohm both nominal and measured with a DMM. I also checked the voltage divider, it has a flat frequency response up to about 20MHz.

And this is the calculated instantaneous power using the above three formulas.

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

Hi,

* DC voltage and DC current
* instantaneous voltages and current (V and I at the same time)
* and some other cases

But they donīt necessarily apply for pulsed or any other waveforms.

***********
(IIR) P = -3.731 W
This power value is impossible for real circuits.
* you donīt have negative resistance
* and for any value of current the result can never be negative. (I x I is always positive)

***********
Voltage and current have opposite direction, this is impossible on a real resistor. Check the polarity / sign.

Also with a resistive load: V = I x R, but this seems not to be the case with your circuit.

***********
It seems you messure V and I.
But how do you measure them?
* Instantaneous
* RMS (over which time?)
* average (over wihich time?)
All these values may be correct or false depending what you want to measure/calculate. You have to choose the correct calculation scheme.
***

The picture shows "DC pulse".
I expect a square wave with a fixed ON time and ON voltage (non fluctuating) and a fixed OFF time and OFF voltage (non fluctuating)

But this seems not to be that case with your circuit. Please clarify. Show the voltage waveform at the voltage source.

***

Klaus

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

In addition to what klaus has said, P is not UUR. P is UU/R.

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

Hi Klaus

First I made a typo, I mixed the UI value up with IIR value, here are the corrected values.

(UI) P = -3.731 W
(UUR) P = 25W
(IIR) P = 809.471 W

Originally Posted by KlausST
Hi,

* DC voltage and DC current
* instantaneous voltages and current (V and I at the same time)
* and some other cases

But they donīt necessarily apply for pulsed or any other waveforms.
Yes I agree, so maybe I should rephrase my question to "Which formulas should I apply to the measured waveform"? Because I'm not sure.

Originally Posted by KlausST

This power value is impossible for real circuits.
* you donīt have negative resistance
* and for any value of current the result can never be negative. (I x I is always positive)
Yes for IIR, but for UI the result is negative i.e. IF the current really is negative.

Originally Posted by KlausST
Voltage and current have opposite direction, this is impossible on a real resistor. Check the polarity / sign.
I can assure you it is a real standard physical 1 Mega resistor, I can hold it in my hand and measure the resistance with a DMM. The only possible error could be if I placed the CT (current transformer) the opposite way, but I'm pretty sure it's correct, the Polarity Arrow points towards positive.

Originally Posted by KlausST
Also with a resistive load: V = I x R, but this seemns dīnot to be the case with your circuit.
Yes, I also noticed that.

Originally Posted by KlausST
It seems you messure V and I.
But how do you measure them?
* Instantaneous
* RMS (over which time?)
* average (over wihich time?)
I measure them with a storage scope, the scope probes x10 are attached directly to the ->> outputs on the schematic.

Originally Posted by KlausST
All these values may be correct or false depending what you want to measure/calculate. You have to choose the correct calculation scheme.
I would like to calculate the instantaneous power, which calculation scheme do you suggest?

Originally Posted by KlausST
The picture shows "DC pulse".
I wrote "DC pulse" because the traces looks like a DC pulse to me.

Originally Posted by KlausST
I expect a square wave with a fixed ON time and ON voltage (non fluctuating) and a fixed OFF time and OFF voltage (non fluctuating)

But this seems not to be that case with your circuit. Please clarify. Show the voltage waveform at the voltage source.
The source will produce zero volts and zero amps if there is no load resistor, so what you see in the oscillogram is the actual "pulse".

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Originally Posted by Akanimo
In addition to what klaus has said, P is not UUR. P is UU/R.
"UUR" is just a more meaningful name than if I named them I, II and III.

As you can see in my OP "(UUR) P = Uē / R"

•

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

Hi,

my question to "Which formulas should I apply to the measured waveform"?
The waveorm is one problem.
* it behaves like a "complex energy source".

Yes for IIR, but for UI the result is negative i.e. IF the current really is negative.
--> correct the sign.

, I can hold it in my hand and measure the resistance with a DMM
Let's say you have an ideal 1M resistor --> your DMM easures 1M
Now add a parallel capacitor --> your DMM still measures 1M (but it will behave totally different at a pulsed source)
Now add a series inductance --> and still your DMM measures 1M, (but it will behave even worse at a pulsed source)
Because it measures with DC, nit AC. It is no RLC meter!

the Polarity Arrow points towards positive.
It is a current measurement device, thus the arrow shows the technical current direction (from positive voltage to negative voltage)

I would like to calculate the instantaneous power, which calculation scheme do you suggest?
For non ohmic loads you need to use: P(t) = V(t) x I (t)

I wrote "DC pulse" because the traces looks like a DC pulse to me.
For me it does not seem to be a DC pulse...but we don't see the source voltage at all...
I expect a DC pulse to have straight horizontal lines only.

Klaus

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

You can calculate instantaneous power at a single point in time - then integrate, then average over larger portions of time

but the simple formulae you listed only apply when the input components are non time varying - or single phase AC all in phase ....

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

Originally Posted by Swend
...
"UUR" is just a more meaningful name than if I named them I, II and III.

As you can see in my OP "(UUR) P = Uē / R"
Sorry, I missed that.

If I may ask, why are you measuring voltage at that node? If you want to filter, I think you should divide before you filter rather than have those caps connected like that.

If your load is a resistor, then you cannot have current leaving the resistor when the voltage across the resistor is +ve.

Also, whatever you are using to measure the current, make sure that it is properly connected.

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

Hi Klaus.

First let me make my intentions very clear:

What I have recorded is NOT the desired result, so I need to establish a baseline to move away from the current result. I need help/suggestions to calculate and establish such a baseline.

I have recorded what I have recorded, if I made measurement and/or calculations mistakes, then I would like to know about it.

Originally Posted by KlausST
Could you please elaborate on that claim?

Originally Posted by KlausST
It is a physical 1 Mega ohm standard high voltage glass resistor. So physically it is a resistor.

Originally Posted by KlausST
* it behaves like a "complex energy source".
Yes I can agree with that.

Originally Posted by KlausST
--> correct the sign.
If I correct the sign, then it is not what I have measured and recorded.

Originally Posted by KlausST
Let's say you have an ideal 1M resistor --> your DMM easures 1M
Now add a parallel capacitor --> your DMM still measures 1M (but it will behave totally different at a pulsed source)
Now add a series inductance --> and still your DMM measures 1M, (but it will behave even worse at a pulsed source)
Because it measures with DC, nit AC. It is no RLC meter!
So you are saying that the resistor's combined R,L and C is producing the undesired measurement result?

Originally Posted by KlausST
It is a current measurement device, thus the arrow shows the technical current direction (from positive voltage to negative voltage)
No, the arrow indicates electron flow i.e. from negative to positive, check the definition of the "polarity Arrow" that I linked to.

Originally Posted by KlausST
For non ohmic loads you need to use: P(t) = V(t) x I (t)
And the result of that calculation you can see in the second graph in my first post. I believe that calculation also holds true for ohmic loads.

Originally Posted by KlausST
For me it does not seem to be a DC pulse...
Then forget that I said "DC pulse", you can call it what you prefer, I will refer to it as "the trace".

Originally Posted by KlausST
but we don't see the source voltage at all...
I can't show you something that I don't have. The load completes the circuit, if there is no load, there is no circuit, if there is no circuit, there is no current, if there is no current there is no voltage. That's the nature of the source.

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Originally Posted by Easy peasy
You can calculate instantaneous power at a single point in time - then integrate, then average over larger portions of time

but the simple formulae you listed only apply when the input components are non time varying - or single phase AC all in phase ....
Sorry, that's my mistake, I made the assumption that it was clear that I must have recorded the results with a DSO, and that a DSO produces instantaneous voltage readings which is stored in a file. Then I make the simple calculations on each point in time. Or as KlausST expressed it P(t) = U(t) x I (t) or P(t) = U(t)ē / R or P(t) = I(t)ē * R.

These results can be seen in the second graph of my first post.

Then I average the P(t) over the entire "trace" duration to get average power for all three methods:

(UI) P = -3.731 W
(UUR) P = 25W
(IIR) P = 809.471 W

However I do not integrate the P(t), I understand that an integration of P would yield E.

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Originally Posted by Akanimo
If I may ask, why are you measuring voltage at that node?
R2+R3+C1+C3 forms a voltage divider, and that is the correct node to record the divided voltage.

Originally Posted by Akanimo
If your load is a resistor, then you cannot have current leaving the resistor when the voltage across the resistor is +ve.

Originally Posted by Akanimo
Also, whatever you are using to measure the current, make sure that it is properly connected.
Yes, that's a good idea.

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

Hi,
Could you please elaborate on that claim?
You talk about R, an ohmic load. But an ohmic load has a constant relationship of V and I.
But in your case it is not constant, it varies with time.

You have measured V and I.
You calculated V x I to get P.

So please calculate V/I of your measured values. To me it seems the result is not constant "1M Ohms".
Thus it is no ohmic load and can't be treated (calculated) according Ohmīs law.

I donīt know whether the measurement values are wrong or the load is a complex load ... they just donīt match.
For sure I may be mistaken.
******
It is a physical 1 Mega ohm standard high voltage glass resistor. So physically it is a resistor.
As already explained... Mabe it is 1M OHms at DC. But your pulse measurement show other values.

I try to explain: Maybe you buy a rubber rope. You buy 1m length. (without force)
But now you use the same rope but with force. Your measurement says now it is 1.2m.
... and now you ignore the measurement and say "Iīm sure I bought 1m"

--> The length of a rubber rope depends on force.
--> The impedance of a "resistor" depends on frequency (signal waveform).
Use a DMM and it shows 1MOhms.
Use an RLC measuremnt tool and it shows something different.
******

No, the arrow indicates electron flow i.e. from negative to positive, check the definition of the "polarity Arrow" that I linked to.
I (just for me speaking) see no need for further discussion. V(R) x I(R) is always positive on a true resistor.

******
I can't show you something that I don't have. The load completes the circuit, if there is no load, there is no circuit, if there is no circuit, there is no current, if there is no current there is no voltage. That's the nature of the source.
What "load" do you need to measure the voltage of an AC outlet? ... or a car battery?

--> Iīm very sure I did measure the battery voltage without connecting a dedicated load.

******
Then forget that I said "DC pulse", you can call it what you prefer, I will refer to it as "the trace".
In your measurement results I can not see a true R. I see a complex load. Thus the results will depend on waveform. You are free to ignore this.

******

Mathematics:

P = V x I
and R = V / I (rearranged: V = R x I, I = V / R)

Now take: P = V x I .... and replace "I" with "V/R" according above formula "I = V/R".
The result is P = V x (V / R) and solve this: P = V x V / R = V^2 / R.

This means P = V x I and P = V^2 / R
In your case this is not true.
There are just three possible reasons:
* The measurement of I is wrong
* The measurement of V is wrong
* or "R" is not a true ohmic "R". Maybe it is non linear, maybe it is complex...

I donīt know which reason generates the problem. I can only guess.

Klaus

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

What is the essnce of C1?

•

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

Hi Klaus

Originally Posted by KlausST
The simple and only reason I talk about R as an ohmic load is:

Originally Posted by KlausST
But an ohmic load has a constant relationship of V and I.
But in your case it is not constant, it varies with time.
I did not say it behaved like an ohmic load, because according to the measurements it definitely doesn't, which is the only reason I'm posting this. I was initially confused and in doubt on how to handle this and make a baseline.

Originally Posted by KlausST
You have measured V and I.
You calculated V x I to get P.

So please calculate V/I of your measured values. To me it seems the result is not constant "1M Ohms".
Please see the requested result hereunder

Originally Posted by KlausST
Thus it is no ohmic load and can't be treated (calculated) according Ohmīs law.
OK, so what do you suggest I treat it like then?

Originally Posted by KlausST
Use a DMM and it shows 1MOhms.
Use an RLC measuremnt tool and it shows something different.
OK. I don't have a RLC-meter, I have a scope, a function generator and a handful of various components. Please suggest a set-up that would give a satisfactory measurement result, and I will perform the measurement accordingly.

Originally Posted by KlausST
I (just for me speaking) see no need for further discussion. V(R) x I(R) is always positive on a true resistor.
Sure I agree, no need to debate that. So we can conclude that my load is not a true resistor, but where do I go from here?

******

Originally Posted by KlausST
What "load" do you need to measure the voltage of an AC outlet? ... or a car battery?

--> Iīm very sure I did measure the battery voltage without connecting a dedicated load.
I can't really follow your logic here, first you make the assumption that the source is a simple AC or DC outlet or a battery, and since the source does not behave like a simple AC or DC outlet or a battery, there must be something wrong with my information. Or maybe you should entertain the possibility that you assumption and conlclusion is somehow logically flawed.

Originally Posted by KlausST
In your measurement results I can not see a true R. I see a complex load. Thus the results will depend on waveform. You are free to ignore this.
I'm personally not ignoring anything, but I will extend you the same courtesy; you are free to ignore the measurement results presented, dismiss them without valid reason, or refrain from assisting to correct/validate them.

Originally Posted by KlausST

Mathematics:

P = V x I
and R = V / I (rearranged: V = R x I, I = V / R)

Now take: P = V x I .... and replace "I" with "V/R" according above formula "I = V/R".
The result is P = V x (V / R) and solve this: P = V x V / R = V^2 / R.

This means P = V x I and P = V^2 / R
In your case this is not true.
There are just three possible reasons:
* The measurement of I is wrong
* The measurement of V is wrong
* or "R" is not a true ohmic "R". Maybe it is non linear, maybe it is complex...

I donīt know which reason generates the problem. I can only guess.
Your conclusion would be absolutely logically correct if there actually were only three possibilities. But I have no desire to debate the additional possibilities with anyone on the internet, I prefer to look at the measurement results only. So if you feel that the I measurement is wrong, then please suggest what I can do to correct it, likewise if you feel that the V measurement is wrong, then please suggest what I can do to correct it - I believe that would be the right thing to do.

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Originally Posted by Akanimo
What is the essnce of C1?
C1 + C3 forms a capacitative 100:1 voltage divider

likewise

R2 + R3 forms a resistive 100:1 voltage divider

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

There are just three possible reasons:
* The measurement of I is wrong
* The measurement of V is wrong
* or "R" is not a true ohmic "R". Maybe it is non linear, maybe it is complex...
Looking at the waveforms, I'd guess that at least two of the three points apply. Respectively it's impossible to determine what's actually happening when only looking at the waveforms. It's necessary to validate the measurements and the resistor impedance independently.

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

Originally Posted by FvM
Looking at the waveforms, I'd guess that at least two of the three points apply.
I would say at least one, but I am willing to entertain the possibility of two.

Originally Posted by FvM
It's necessary to validate the measurements and the resistor impedance independently.
I vote in favor of that suggestion, I already spent \$1.000 on the CT, if the CT is yielding incorrect results I can at least claim my money back. If anyone knows anyone reputable that is willing to validate the results on a pro-bono basis, or at least at a civilized price point, please let me know.

In any case, be it right or wrong, I decided to go with the P = U x I time being, for the simple reason that "R" does not exhibit any signs of joule heating even after a few hundred of "pulses".

Thank you.

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

You can also model what you are doing with SPICE.

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

A point that makes me completely distrust the waveforms in post #1 is that they start "out of nothing".

So we e.g. can't determine the As integral applied to the Pearson probe. According to datasheet, it's typically saturated at 0.0006 As, which is apparently exceeded in the experiment. Depending on the preceding current waveform, the falling current measurement may be simply transducer saturation. If (unexpectedly) no saturation happens, there will be still a current droop due to 140 Hz lower cut off frequency.

Pearson probes are fine instruments, but you need to understand their limitations.

Similar problems might occur with other measurements too.

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

Originally Posted by FvM
A point that makes me completely distrust the waveforms in post #1 is that they start "out of nothing".
In stead of expressing your personal distrust maybe you should start by posing me a question. Obviously the traces don't start "out of nothing" I just chose to start at that particular point when load is applied, prior to that point both traces are 0V complete flatliners. As I tried to explain KlausST, which waveform do you expect to see when your probe is connected to e.g. a open switch or near infinity resistance(?)

Originally Posted by FvM
So we e.g. can't determine the As integral applied to the Pearson probe.
How would you determine the applied integral when the integration is done inside the probe(?)

Originally Posted by FvM
According to datasheet, it's typically saturated at 0.0006 As, which is apparently exceeded in the experiment. Depending on the preceding current waveform, the falling current measurement may be simply transducer saturation. If (unexpectedly) no saturation happens, there will be still a current droop due to 140 Hz lower cut off frequency.

Pearson probes are fine instruments, but you need to understand their limitations.

Similar problems might occur with other measurements too.

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Hi friends.

I bear some good news and some even more good news to those reading this thread.

Negative Resistance Obviously no one offered to validate my measurements, so I had to do the next best thing, request a friend to repeat the experiment - and the results are the same. It turns out what I stumbled upon is a physical resistor exhibiting true negative resistance, or a complex energy source as others might call it.

Conservation of Energy The energy is conserved because whatever comes out of the Resistor must necessarily enter the source's power supply, and when it does (it does so instantaneously) - the power supply performs work on it at a cost equal to the amount released by the Resistor. In other words "easy comes - easy goes" so no need to spooked.

So the only novel thing here is that I can demonstrate true negative resistance on command, since I can't really see any applications for it - it's pretty useless.

Thank you.

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

Hi,

In stead of expressing your personal distrust maybe you should start by posing me a question.
no one offered to validate my measurements
You want others to do something....Is this really the ob of forum members to work for you - for free?

We can only validate information when you give informations that can be validated. Photos,

****
So the only novel thing here is that I can demonstrate true negative resistance on command, since I can't really see any applications for it - it's pretty useless.
You want us to belive in all you say and in all your measurements.
**** OK, Iīll try. I donīt question any of your statements.

Just to clarify: We donīt talk about "differential" negative resistance. We talk about real negative resistance.
Then according your given informations the "device" will only produce energy. Your measurements do not show a single piece of time where it consumes energy. (while a true resistor only consumes energy).

Then the only conclusion can be (and don't treat this as joke or offense): You found the perpetuum mobile. Even better: you solved the energy problems of our world.
--> Your invention is far away from being useless. Itīs the most important invention ever.

*****End of Trial

Back in reality: I simply donīt believe it.

But I donīt have the knowledge you have, I donīt have the device in my hand to validate the measurements .... and I donīt want to be accused not to spend enough time for you..... Thus Iīll leave this thread.

Good luck

Klaus

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

Originally Posted by KlausST
Hi,
Then the only conclusion can be (and don't treat this as joke or offense): You found the perpetuum mobile. Even better: you solved the energy problems of our world.
--> Your invention is far away from being useless. Itīs the most important invention ever.
I don't appreciate your irony. Where do you find excess energy? The negative current goes back to the source's power supply, the power supply performs work (Work as in W = F * d * cosθ) on it at a energy cost equal to the amount released by the Resistor.

Energy in = Energy out

Which I confirmed by measuring the energy consumed by the power supply.

So any perpetuum mobile will ONLY exist in your mind.

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

Obviously the traces don't start "out of nothing" I just chose to start at that particular point when load is applied, prior to that point both traces are 0V complete flatliners.
"Prior to that point" isn't quite exact, you have actually cut the rising edge and we can't know how long is lasts. The voltage and current transfer functions may involve either non-linearity (e.g. core saturation) or a impulse response of some time duration, probably both. In this case, the hidden rising edge echoes in the visible response.

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.

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

Originally Posted by FvM
"Prior to that point" isn't quite exact, you have actually cut the rising edge and we can't know how long is lasts. The voltage and current transfer functions may involve either non-linearity (e.g. core saturation) or a impulse response of some time duration, probably both. In this case, the hidden rising edge echoes in the visible response.

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.
I had to make a new measurement to record the full pulse and a bit more. The only change I have made is the nominal value of "R" I lowered it from 1.1M to 883K which resulted in higher voltage and longer pulse duration.

For arguments sake, even if I didn't know how to properly use the CT, as it has been suggested, the current would still go negative at some point while the voltage is positive. Now that the pulse is longer, I can record up to one and a half cycle of the current race, after that the scope can't register any current. I assume the slight sinusodial dampening is due to parasitic inductance in the whole circuit.

I have attached the data file if anyone would like to make their own calculations. Just remember to multiply the voltage trace with a factor of 100 (due to the voltage divider), and the current trace is 1A=1V.

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