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.

Hi,

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*j

Here 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*j

where the imaginary part is now -3, and we might see:

X=-2+3*j or -2-3*j

where 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 amp

and 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*E

but 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*I

because then the resistor is not connected directly across the line.