Clamp Meters and  Apparent Power - Page 1
Forum section: electronics / beginners

author: killingtime
author_rank: Regular Contributor
timestamp: March 20, 2023, 07:01:04 pm
content: Hi,Would be grateful if someone would sanity check my understanding of AC current meters..If you have an AC device that's heavily reactive and consumes a lot of apparent power, am I right in saying that your run of the mill current meter (clamp or moving iron 2 wire) will register the reactive current flowing into and out of it?So a clamp meter on the primary of an MOT (say) with open secondary would register the 3 amps of reactive current?How does the real power consumed on top of this show on the meter? It's just a larger value right?Can't find anything in meter manuals on this. Lots of articles explaining the difference between apparent and real power, but nothing on meters.And you need a 4 wire setup (voltage and current) with phase measurement for real power measurements - something that clamp meters alone will never achieve.Thanks.

author: RoGeorge
author_rank: Super Contributor
timestamp: March 21, 2023, 10:40:57 am
content: A clamp-meter only measures the current, so it can not calculate the active power correctly for reactive loads.If the load is resistive, for example an incandescent light bulb, then the active power can be calculated close enough assuming the nominal mains voltage is known. Unless the load is a big AC motor, or some heavily switched load without PFC (Power Factor Correction), then measuring only the current (i.e. with a clamp meter) can be close enough to calculate the active power.For high precision measurements, or in any energy meters, all 3 must be known (U, I, and phase between U and I).

author: killingtime
author_rank: Regular Contributor
timestamp: March 21, 2023, 11:14:32 am
content: Hi RoGeorge,I'm aware of the theory.My question was; does a clamp meter (or two wire ammeter) register (measure) purely reactive current.Or to put the question another way, if you were to put a near ideal inductor across an ac supply (theoretically), would a clamp meter register the current in any one wire feeding it?In the above example, no real power is expended in the inductor, the current used to magnetize the inductor is fed back to the supply (ignoring resistive losses), so the net current flow is near zero. I think this is why some older mechanical kwh house meters don't record (or bill you) for reactive power used.

author: radiolistener
author_rank: Super Contributor
timestamp: March 21, 2023, 11:21:05 am
content: Quote from: killingtime on March 20, 2023, 07:01:04 pmreactive load doesn't consume power, it just get it store for a very small period of time and then returns it back to the mains on the AC next cycle. So, when your device has reactive part of impedance, then there is active part of power which is consumed by device and reactive part of power which flows back and forth on every AC cycle from mains to device and back from device to mains. Usually home power counters doesn't count reactive part of power they should take into account just active one. But industrial power counters (for factory and high power machines) can count both - active and reactive part of power separately. This is because reactive part of power add energy loss in wires (due to wire heating), and when load consumes very high power that losses may be significant and needs to be taken into account.Since reactive current flows back and forth through wire in both directions, you cannot see it with a clamp meter, just because clamp meter measure average current over some period of time, so when there are two currents +Ir and -Ir which flows in opposite direction during AC cycle, they will cancel each other on average result. But it depends on AC meter implementation, in some cases reactive current can affect measurement result in unpredictable way, so you can see some random error on some meters when there is reactive current is present. Reactive current can distort AC sine wave, so it also can affect measurement, because AC current measurement depends on a waveform.As said before, in order to measure reactive power, there is need both measurement simultaneously - voltage and current and phase delay between them.

author: armandine2
author_rank: Frequent Contributor
timestamp: March 21, 2023, 11:32:48 am
content: CA F205 Clamp Meternot sure if this is "run of the mill" but it looks like it will give VA var and pf

author: killingtime
author_rank: Regular Contributor
timestamp: March 21, 2023, 12:34:00 pm
content: Quote from: armandine2 on March 21, 2023, 11:32:48 amYes it can, but it needs a voltage measurement to get the relative phase. I was asking after just current clamp meters. Good to know anyway.Quote from: radiolistener on March 21, 2023, 11:21:05 amThis answers my question. I don'twan'tto measure reactive current specifically, I just wanted to know what el-chepo AC current clamp metersactuallymeasured. If it's the slow averaging on the meter side that causes a purely reactive current to read as zero, then using an AC clamp with a much faster response would work - like an O-scope current clamp with a bandwidth of a few 100 KHz. Still just a single clamp over a conductor, but if the bandwidth is there, you'd see the reactive current flowing both ways - for a purely reactive load. In reality there be some resistive component to the load as well, so you'd see the combination of both currents as one waveform.If my understanding of this is correct, that's a limitation of current clamp meters (and moving iron meters) I wasn't aware of.

author: wasedadoc
author_rank: Super Contributor
timestamp: March 21, 2023, 12:44:36 pm
content: Some nonsense is written above. The clamp meter measures current. It cannot know the phase angle between current and voltage so it cannot know what is real current and what is reactive current. Real current and reactive current show on the meter just the same.

author: killingtime
author_rank: Regular Contributor
timestamp: March 21, 2023, 12:52:33 pm
content: Quote from: wasedadoc on March 21, 2023, 12:44:36 pm>>The clamp meter measures currentYes it does, in one direction or the other, shown as a + or - on the display. For a very reactive load (like a capacitor) with very little real resistance, the current is going to flow back and forth within each cycle, averaging the result out if the bandwidth of the meter is many times that of the AC frequency...If I'm wrong, please enlighten me. Happy to say I'm wrong.>> It cannot know the phase angle between current and voltage so it cannot know what is real current and what is reactive current.Agreed.>>Real current and reactive current show on the meter just the same.Not sure what you meant by that.---EDIT---Perhaps you are right wasedadoc. After a little more thought, even for a resistive load on an AC supply, the current reverses (flows back and forth) on each half cycle. If the meter were to average these currents out, the net result would be zero, and that would defeat the purpose of an AC ammeter. Guess I'm going to have to run a test to find out.

author: wasedadoc
author_rank: Super Contributor
timestamp: March 21, 2023, 01:01:45 pm
content: Quote from: killingtime on March 21, 2023, 12:52:33 pmReactive current is only applicable when AC (alternating current) is involved. A clamp meter never shows + or - when set to measure alternating current. Some clamp meters can measure DC current and in that case will indicate + or -.The reasoning that reactive current is going back and forwards and so cancels out is baloney. Real current flows back and forth too. What is the difference?"shows the same" means that whether the current is N Amps reactive or N Amps real, the clamp meter will show N Amps.And if there is both real and reactive current the clamp meter will show the vector sum. Square root of the sum of the squares of the real and reactive currents.

author: killingtime
author_rank: Regular Contributor
timestamp: March 21, 2023, 01:05:16 pm
content: Quote from: wasedadoc on March 21, 2023, 01:01:45 pmAfter a bit more thought I'm inclined to agree with you. It would help if I owned an AC clamp meter. One on order.

author: bdunham7
author_rank: Super Contributor
timestamp: March 21, 2023, 01:52:58 pm
content: Quote from: wasedadoc on March 21, 2023, 01:01:45 pmTo be fair, the terminology itself can be misleading and maybe the term 'recirculating' current captures the reality a bit better. The point of common misunderstanding (as shown in this thread) is that both real and reactive (or recirculating) currents are entirely real. Current meters measure them, as they should, and the wires that carry them heat up in proportion to (I2real+ I2reactive) * R.

author: TimFox
author_rank: Super Contributor
timestamp: March 21, 2023, 03:56:31 pm
content: A clamp or two-terminal AC ammeter will measure the current through the wire.It may sense true-rms current or mean-absolute-value ("average") current, depending on the ammeter's circuitry.It can't tell the difference between "real" and "imaginary" components of the current without a phase reference.For real power measurements, that phase reference is the AC voltage in the circuit, not seen by the ammeter.

author: BeBuLamar
author_rank: Super Contributor
timestamp: March 21, 2023, 10:45:53 pm
content: The amp meter either clamp or in series meter measures current regardless if it's reactive or not. The meter doesn't know if the current is in phase with the voltage or if it's out of phase with the voltage.

author: MrAl
author_rank: Super Contributor
timestamp: March 22, 2023, 12:09:07 am
content: Quote from: killingtime on March 20, 2023, 07:01:04 pmHi,With a purely real load like a resistor the meter reads the entire real part of the current.With a purely reactive load like an inductor the meter reads the entire reactive part of the current.With a partly real and partly reactive load the meter will read the total current as the 'norm' of the reactive and real parts of the current. This means you take the square root of the sum of the reactive squared and real part squared. That will show on the current meter either an AC clamp-on meter or a regular AC current meter.This is best understood using complex numbers. Complex numbers have a real and what is called an imaginary part. That just means that it takes two numbers to specify one complex number. The imaginary part is shown by a multiplication by the lower case "j" in the English alphabet. For example for X a complex number:X=2+3*jHere X is a complex number, and the real part is 2 and the imaginary part is just 3 (not 3*j).The imaginary part can be negative, and the real part can be negative, so we might see:X=2-3*jwhere the imaginary part is now -3, and we might see:X=-2+3*j or -2-3*jwhere in the first the real part is -2 and the imaginary part is 3, and in the second the real part is -2 and the imaginary part is -3.This relates directly to the current measurement of a load that is both reactive and resistive so it is quite handy to know.To understand this clearly it is probably best to start with a resistor in parallel with an inductor. That way you can measure the current through the resistor by clamping onto only the resistor lead, and you can measure the reactive current by clamping onto only the inductor lead, and you can measure the total current (both reactive and resistive) by measuring the lead that connects them to an AC voltage source such as 100vac at 50Hz.With that in mind, say we have a 100 Ohm resistor in parallel with an inductor of 0.3183 Henries (318.3mH). That's a large inductor but we just want to use that as an example.It should be obvious that they are both in parallel with the line voltage of 100 volts, so the current in the resistor is:iR=100/100=1 ampand the current in the inductor is:iL=100/(w*0.3183)where w is 2*pi*f with f the frequency which is 50Hz.With w=2*pi*50 and that inductor value we get:iL=1 amp (very nearly so actually)So the resistive current (real part) is 1 amp and the reactive current is 1 amp. That is what you would measure with the current meter.The TOTAL current, the current from the line, would be the square root of the sum of the squares of the two:iT=sqrt(1^2+1^2)=sqrt(2)=1.4142 amps. That is the current you would measure in the line itself. The simple current meter can not distinguish between the resistive and reactive parts once the two components are in parallel, and that is what you get with a load that is both resistive and reactive.So you see we have three different current levels to consider.Now the phase difference, if you care to know that or measure that, would simply be the atan() of the reactive part divided by the resistive part.In this case they are both the same so first writing it out:ph=atan(reactivecurrent/resistivecurrent)so we have:ph=atan(1/1)=atan(1)=45 degrees.Note this is the principle angle which does not show if one leads or lags the other. To get that information it is a good idea to use the expressions that have been used since the dawn of time:ELI the ICE man.This helps to remember that in an inductor (L) the voltage (E) leads the current (I),and in a capacitor (C) the current (I) leads the voltage (E).Since upper case i's can look like 1's i'll write the i's in lower case:"ELi the iCE man"and here the lower case 'i' is the current.The most important part of this though is the way the reactive part and resistive parts are summed. Note in the above example we did not just add the two 1+1=2 we had to take the square of each one (1*1 and 1*1) and add those together, then take the square root of that result.This is easy to do on a hand calculator or of course a calculator you have on your computer.If any of this is not clear just let me know. We can go though some more examples.To calculate power, we need to know the phase angle.In a DC circuit the power is:P=I*Ebut in an AC circuit with both resistive and reactive parts, the power is:P=I*E*cos(ph)where ph is the phase angle and I is measured in the line lead which is the total current.Again this is easy to calculate with a hand calculator.In the above example we had E=100 and I=1.4142 and ph=45 so we would get:P=100*1.4142*cos(45)=100 watts.It is interesting to note that this is ONLY the power in the resistor, as there is no power dissipated in the inductor with an ideal inductor. That means we could have, in this example, used just the voltage and the current in the resistor to get the real power:P=100*1=100 watts.Small side note:If this was a series circuit instead of parallel, we would have to calculate the voltage across the resistor to do it this way:P=vR*Ibecause then the resistor is not connected directly across the line.

author: rstofer
author_rank: Super Contributor
timestamp: March 22, 2023, 03:50:06 pm
content: More about ELI the ICE man... The attached plot is for a series RC circuit driven at 1 kHz and, I believe, 1V P-P (but I might be off on that) The R value is 10k and the C value is 0.1 ufd for a time constant of 1 ms the reciprocal of which is 1 kHz.Note that the current waveform (orange) goes through the X axis before the voltage waveform (blue). The current is leading the voltage and the circuit is capacitive - hence ICE (I leads E in a Capacitive circuit). Note that the phase shift is approx 250 us which is about 1/4 of 1 ms which is 90 degrees (1/4 of 360 degrees) and this is what we expect. A 90 degree phase shift.This experiment was done with a Digilent Analog Discovery 2 and the related Waveforms software. The cursors aren't perfectly placed but are close enough for my use.

author: TimFox
author_rank: Super Contributor
timestamp: March 22, 2023, 04:14:56 pm
content: Again, the illustration in the response above requires measurement of both the voltage (from the wire to some reference point) and the current (through the wire).This requires two separate devices: an ammeter and a voltmeter.

author: rstofer
author_rank: Super Contributor
timestamp: March 22, 2023, 05:17:37 pm
content: Quote from: TimFox on March 22, 2023, 04:14:56 pmI don't have a clamp-on that simultaneously measures voltage and current. It would be nice but I'm still using the Amprobe meter I bought around 1965. Still works...One neat thing about the AD2 is the fact that both input channels are differential. I can use one channel to measure the voltage across the resistor (hence the current waveform) where neither end is ground and the second channel to measure the voltage across the capacitor.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: March 22, 2023, 07:33:22 pm
content: A clamp on ammeter measures current and current only. There is no differentiation between real or reactive current.

author: rstofer
author_rank: Super Contributor
timestamp: March 22, 2023, 07:39:46 pm
content: Quote from: Ground_Loop on March 22, 2023, 07:33:22 pmMaybe some are better than that:https://www.amazon.com/Extech-380976-K-Single-Phase-Three/dp/B008465KMUQuote

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: March 22, 2023, 07:48:47 pm
content: Quote from: rstofer on March 22, 2023, 07:39:46 pmIt still has to measure both voltage and current. Power cannot be measured. It iscalculatedbased on voltage and current measurements. A simple ammeter only measures current.

author: wasedadoc
author_rank: Super Contributor
timestamp: March 22, 2023, 07:49:53 pm
content: Quote from: rstofer on March 22, 2023, 07:39:46 pmMaybe so but:1. To do those extras you need to contact the leaded probes to the voltage points at the same time as putting the clamp around the current carrying wire.2. The clamp does not measure Direct Current, only AC.3. Small DC currents can only be measured by putting the meter in series in the same way as a conventional multimeter.

author: coppice
author_rank: Super Contributor
timestamp: March 22, 2023, 07:56:30 pm
content: Quote from: rstofer on March 22, 2023, 07:39:46 pmThat's not a current meter. Its a power meter. Connect only the clamp and it measures only the current. Connect only the voltage sense leads and it measures only the voltage. Connect the clamp and the leads and it uses both the instantaneous voltage and the instantaneous current, and can give you all the measurements you listed.

author: rstofer
author_rank: Super Contributor
timestamp: March 22, 2023, 08:29:55 pm
content: No, the referenced meter is NOT a simple clamp-on ammeter but it does answer the questions re: power factor, etc. Adding the voltage probes is all it takes and they have done just that. I'm sure there are many other meters that do the same thing.And it will still operate as a simple clamp-on ammeter if that is all that is required.

author: coppice
author_rank: Super Contributor
timestamp: March 22, 2023, 08:48:31 pm
content: Quote from: rstofer on March 22, 2023, 08:29:55 pmThe utility meter market has driven the cost of making a meter like that Extech one, with accuracies down to 0.1%, very low. There are a few other similar products, but not many. Its a niche market compared to multi-meters or clamp on current meters. Oddly there are multi-meters with a current clamp, which have all the sockets and leads, where the incremental cost of accurate active, reactive, apparent power, and power factor would be small, yet they haven't all migrated in that direction. I guess there is low demand.

author: BeBuLamar
author_rank: Super Contributor
timestamp: March 22, 2023, 10:48:02 pm
content: Quote from: wasedadoc on March 22, 2023, 07:49:53 pmMany clamps can measure DC current. Fewer clamps can measure small DC current but some can measure down to 5mA.

author: coppice
author_rank: Super Contributor
timestamp: March 22, 2023, 11:08:18 pm
content: Quote from: BeBuLamar on March 22, 2023, 10:48:02 pmMany? There are a few meters that use hall sensors or flux gates to measure DC, but they are rather niche. Most use a CT, and a few use a Rogowski coil. They only measure AC.

author: wasedadoc
author_rank: Super Contributor
timestamp: March 22, 2023, 11:09:14 pm
content: Quote from: BeBuLamar on March 22, 2023, 10:48:02 pmMy comments were directed solely at the specific Extech meter, not clamp meters in general. That fact was obvious from the first of my comments which was deliberately omitted when quoting parts of my post. I do possess a clamp meter which can measure both Alternating and Direct Current using its jaws.

author: MrAl
author_rank: Super Contributor
timestamp: March 23, 2023, 01:33:19 am
content: Quote from: rstofer on March 22, 2023, 03:50:06 pmHello there,Where are you measuring that current?I ask because 90 degrees is when we have an ideal capacitor. When we have a resistor in series with a capacitor the typical phase shift is less than 90 degrees. It could be close to 90 degrees, but only for very small resistor values.Just to note, the phase shift current to voltage is:ph=atan(1/(2*pi*f*R*C))

author: MrAl
author_rank: Super Contributor
timestamp: March 23, 2023, 01:36:56 am
content: Quote from: coppice on March 22, 2023, 11:08:18 pmHi there,I think they are much more common these days. For example, i got a DC/AC clamp on meter several years ago price was around $60 USD, but a few weeks ago i got one for $40 USD. Several sold on the web like on Amazon.They are a bit more difficult to use on DC however. The measurement has to be preceded by a 'zeroing' of the meter, and then the meter and wire should not be moved much in order to read the measurement of the DC current.Some of them will respond to the Earth's magnetic field so the meter can not be twisted after being zeroed out or it will go off zero before you can take the actual measurement. This happens mostly for low current measurements.

author: james_s
author_rank: Super Contributor
timestamp: March 23, 2023, 02:59:51 am
content: Current is current, it doesn't matter if it's reactive or not, it's the same thing and a current meter will read the same whether the load is resistive or reactive. The difference between true power and apparent power is the phase difference between the voltage and current waveforms, which is called the power factor, that affects the *power* calculation, not the current.

author: EPAIII
author_rank: Super Contributor
timestamp: March 28, 2023, 04:50:19 am
content: Yes, some time ago I was pleasantly surprised to find out that modern clamp-on meters can measure both AC and DC. They use Hall Effect sensors. And this is not just expensive ones, I have one that cost only $40, USD that has this capability. Not highly accurate, but it does measure DC current as well as AC.It does have a digital VOM built in and I need to read the instructions again to see if it can measure power.Isn't modern tech wonderful?Quote from: BeBuLamar on March 22, 2023, 10:48:02 pm

author: coppice
author_rank: Super Contributor
timestamp: April 06, 2023, 10:42:01 am
content: Quote from: Ali uxama on April 06, 2023, 08:46:03 amThere is no such thing as reactive current. There is only current. The concepts of active, reactive and apparent arepowerissues, not current or voltage issues. Until you start integrating voltage and current over time, to measure power, active, reactive and apparent have no meaning at all. Things consume active power. They do not consume reactive power. Apparent power is a mish mash (the third side of a triangle for pure sine wave voltage and current) of active and reactive power. The active component is consumed, and the reactive component is not. If its consumed it can end up as heat somewhere down the line. Reactive power can't heat anything. Most confusion in understanding these concepts comes from this kind of sloppy use of words. Use words more precisely and it all falls into place.

author: paulca
author_rank: Super Contributor
timestamp: April 06, 2023, 10:54:59 am
content: I think that's why they call it "apparent", apparently.Anyway. I have had a coil clamp meter on my mains electrical tails for 3 years. The "total power returned" is -1.2kWh. That's after a net consumed of something like 18,000kWh can't remember.It's interesting and maybe relevant because that figure should be 0. I do not have an generation equipment. The meter account for all 3, as it has a current AND voltage tap. So it's not reactive power. At some point my house pushed current outward that it did not get back. Where did it come from? Where did it go? (Cotton eye joe!).The reality is, it's probably just aggregate software rounding errors.Just in case nobody has reminded folks yet... a current coil/clamp around a main cable that is unloaded/unburdened is extremely dangerous.

author: coppice
author_rank: Super Contributor
timestamp: April 06, 2023, 11:35:30 am
content: Quote from: paulca on April 06, 2023, 10:54:59 amMost clamp on CTs have a built in burden resistor, so its harmless to leave them unconnected.