Hi, Could anyone tell me the simplest way to measure the above parameters with digital multimeter (if possible).

Side question.
Is the apparent current = rms current(without angle)?

 

Hi, Could anyone tell me the simplest way to measure the above parameters with digital multimeter (if possible).

Side question.
Is the apparent current = rms current(without angle)?

There is no such thing as apparent current.
You are confusing that with real and apparent power.

A DMM can measure the rms AC voltage input to a circuit by measuring AC voltage across the input.
A DMM set to AC current range will measure rms AC current when placed in series with the circuit.
Multiplying the two values will give you apparent power in VA units.

In order to determine the real power you need to determine the phase angle between the AC voltage and current. In other words, you need to determine the power factor.



See this:
https://www.allaboutcircuits.com/te...ent/chpt-11/true-reactive-and-apparent-power/

 

You might want to take a small step back to make sure you understand the purpose of measuring an AC signal by it's RMS value. In simple terms, the RMS values of a voltage and a current, when multiplied together, gives the correct power dissipated in a resistive load. Computationally, the RMS value of a voltage or a current, is computed over one cycle of the AC waveform by squaring it and taking the mean (average) for the squared values of the waveform, then taking the square root of that mean (average). It can be shown that this is the only computed number of a sinewave that gives the correct answer for power in a resistive load.

 

It can be shown that this is the only computed number of a sinewave that gives the correct answer for power in a resistive load.

No. For a sine wave you can measure the peak voltage and divide by sqrt(2).

RMS works for an arbitrary waveform.

Bob

 

No. For a sine wave you can measure the peak voltage and divide by sqrt(2).

RMS works for an arbitrary waveform.

Bob

The first idea would work if you have a peak detecting meter or make a peak-detecting probe and know you have a sine wave.

You can get the RMS value of an arbitrary waveform if you have a true RMS meter. Most digital multimeters either measure peak or average and convert that to RMS.

 

No. For a sine wave you can measure the peak voltage and divide by sqrt(2).

RMS works for an arbitrary waveform.

Bob

That is kind of a tautology since that is just an alternate way of computing exactly the same quantity for a sinewave. Your method provides no insight into why it should be that way and why the square root of 2 has that property.

 

That is kind of a tautology since that is just an alternate way of computing exactly the same quantity for a sinewave. Your method provides no insight into why it should be that way and why the square root of 2 has that property.

What I was trying to clarify is that RMS applies to any waveforrm, while you semed to be saying it was something special for sine waves.

Bob

 

What I was trying to clarify is that RMS applies to any waveforrm, while you semed to be saying it was something special for sine waves.

Bob

RMS voltage and current can be computed for any waveform. On that we agree. I'm not sure about multiplying them together to get power for anything but a sinewave. Is it your claim that this is a valid thing to do? I'm actually agnostic on that issue since it has never come up in my experience.

 

Yes, you can get the RMS value for the voltage across something and get the RMS value for the current through that same something.

Yes, you can then multiply them together to get the apparent power -- but only because that is the definition of apparent power and why the word "apparent" is used to describe it.

But for any non-resistive load, you need additional information in order to get anything more meaningful than that, such as real power or reactive power.

 

Hi, Could anyone tell me the simplest way to measure the above parameters with digital multimeter (if possible).

Side question.
Is the apparent current = rms current(without angle)?

Hello there,

The real and imaginary parts of a current or voltage encodes both the amplitude and the phase shift so in order to measure the real and imaginary parts you have to be able to measure the phase angle and the amplitude. When you measure both of these then you can convert that to the real and imaginary parts. You do need to know what you are measuring though either peak or RMS or whatever.

For example, if you have a resistor of 1 Ohm in series with a cap of 1 Farad driven by a 1v peak sine source with angular frequency w=1 and you measure the peak voltage across the resistor and divide by the resistance or measure the current and the phase angle across the resistor relative to the sine source you will read 1/sqrt(2) peak volts AC and a phase angle of pi/4 (45 degrees).
You then convert this to rectangular form with:
I=A*(cos(a)+j*sin(a))

and in this case we have A=1/sqrt(2) and phase angle pi/4 so we get:
I=(1/sqrt(2))*(1/sqrt(2)+j*1/sqrt(2))

and simplifying:
I=1/2+j/2

which means the real part is 1/2 and the imaginary part is 1/2.

So you can measure the real and imaginary parts but you do this by measuring the amplitude and phase angle. You need a meter that can measure phase though, unless you are willing to calculate that somehow knowing the circuit components.

The secondary issue that came up was the use of RMS values to calculate average power.
The test for whether or not RMS multiplication works on different kinds of waveforms is:
RMS1*RMS2=P1

where RMS1 and RMS2 are the two RMS values in question and P1 is the integral of i*v over the period. Since this varies for different waveforms you can test with those waveforms to be sure. If that equality is satisfied then it works. If it is not satisfied then it does not work with those waveforms.

For example if we have voltage wave A*sin(w*t) and current B*sin(w*t) then the two RMS values are:
Rms1=A/sqrt(2)
Rms2=B/sqrt(2)

and multiplied this is:
P1=Rms1*Rms2=A*B/2

and the average power calculated directly is:
P2=(1/T)*integral(A*sin(w*t)(B*sin(w*t)) dt, over one period,
P2=A*B/2

so here P1=P2 so we got the same results either way.

 

To get the true power for an arbitrary current and voltage waveform you can multiply the instantaneous voltage by the simultaneous instantaneous current and average the values over a waveform cycle.
The number of samples needed for this calculation is at least twice the highest frequency of any harmonics in the waveform.

 

To get the true power for an arbitrary current and voltage waveform you can multiply the instantaneous voltage by the simultaneous instantaneous current and average the values over a waveform cycle.
The number of samples needed for this calculation is at least twice the highest frequency of any harmonics in the waveform.

Hi,

Yes that's the definition of average power.
Pavg=(1/T)*integral(v(t)*i(t),t)

integrating over the period.
I had shown one example but did not get to do any more yet.

 

To get the true power for an arbitrary current and voltage waveform you can multiply the instantaneous voltage by the simultaneous instantaneous current and average the values over a waveform cycle.
The number of samples needed for this calculation is at least twice the highest frequency of any harmonics in the waveform.

In general, to get the average power this way, the number of samples needs to be MUCH higher that twice the highest harmonic. Just imaging trying to determine the real power of a single sinusoid using only the voltage/current values at three evenly spaced points in time.

 

In general, to get the average power this way, the number of samples needs to be MUCH higher that twice the highest harmonic. Just imaging trying to determine the real power of a single sinusoid using only the voltage/current values at three evenly spaced points in time.

Yes, if you are only doing the measurement for one cycle.
So you'd have to average the readings over many cycles for such a low sample rate (assuming the sample rate is not a integral multiple of any of the signal frequency components).

 

Hi,

I think a waveform that is non sinusoidal requires a lot of samples, but a pure sine does not require that many samples because the signal is recoverable from a min number of samples because it's a sine wave which of course has special properties that an arbitrary waveform does not possess in general.

I'd have to review some sampling theory to get more detailed on this, but we can do that too i guess.
All we have to do is generate a sine wave of some frequency and then see what it takes to recover the amplitude.

 

Thanks all the good people for explanation and the time they took

 

I think a waveform that is non sinusoidal requires a lot of samples, but a pure sine does not require that many samples because the signal is recoverable from a min number of samples because it's a sine wave which of course has special properties that an arbitrary waveform does not possess in general.

Yes, a sine wave can be reconstructed through sin(x)/x interpolation if it's sampled at more than 2 samples per period.

Except that a signal that's a true pure sine wave doesn't exist in nature. It may look like a sine wave but it will have other components to make it non-sinusoidal. And for the other components that have a higher frequency than your base sine you'll need to either remove them (filter) or take them into account when sampling or otherwise you end up with aliasing artifacts that make your sampled data invalid.

As for measuring apparent power, real power and reactive power, a DMM is the wrong device for that unless your UUT has a Power Factor of 1 (i.e. no phase shift between voltage and current). If not then you should use a power analyzer or good scope, preferrably with power analysis capabilities.