Electronics Math and Power Factor Not Adding Up

[tommygdawg]

Hey all,

I've got an HMI light (one of those ones that runs off a ballast) and I'm trying to do some basic math but it's not coming out right. I've plugged the device into a Kill-a-Watt to measure the amperage and wattage. Here's my issue:

The light is rated for 1200 watts. According to the Kill-a-watt, I'm actually drawing 1040 at approximately 12.6 amps. I took my measured amperage times the voltage and got a wattage of 1,512 watts (12.6*120). Why the discrepancy? I then wanted to understand the power factor of the unit, but the math for that doesn't work out at the 1,040 watts measured. To get my power factor I have to take the rated wattage and divide it by the actual draw (1200/1040). This doesn't work out, it gives me a power factor of 1.15 which is impossible. Now, if I divide by the wattage the math gives me it works (1200/1512 = .79). I should also note that if divide measured wattage by voltage (1,040/120) I get around 8.67 amps. So which is it? Am I running at 8.67 amps or 12.6?

Do these question make sense? I suppose the TL;DR version is: why does my meter read 1,040 watts at 120 volts and 12.5 amps, when the math tells me that I should be reading 1,500 watts at 120 volts and 12.5 amps?

Thank you!

[helius]

You constantly refer to 120 volts but you never said you measured that. Voltage at the outlet can vary +/- 10% or sometimes even more, and exactly 120VAC is unusual.

When a device is "rated" for a certain power output that usually means it is the maximum safe output that the design anticipated. So if there is a dependency of power on the available voltage (could be interpreted as a "resistive" characteristic, or simply as no line regulation), this figure only applies at the highest anticipated supply voltage, which could be ~140VAC.

[Yansi]

Power factor is the ratio of active real power and imaginary apparent power.

imaginary apparent power S is your Volts times Amps you measure with (trueRMS) multimeter. The active power P is what you measure with a special device called wattmeter.

cos phi = P / S

Note: It also does apply that S^2 = P^2 + Q^2

When the power factor "cos phi" reaches 1, it means there is no reactive power Q (Q=0) and S = P.

In reality, the situation is a little more complicated, because there is also a deformation power D, which causes distortion.

For your 1040 watt measured by wattmeter and 1512 voltamperes (beware!! imaginary power is in VA not W!) gives a power factor of 0.69  (1040/1512)

EDIT: Corrected terminology.

[retrolefty]

Power = V x I is true only if the attached load is a pure resistance. This is one case where the Wikipedia explaines it pretty well. Welcome to the world of AC where there is resistance, inductive reactance, and capacitance reactance to account for.

https://en.wikipedia.org/wiki/Power_factor

[T3sl4co1l]

Power factor is the ratio of active power and imaginary power.

Real to apparent*.

The equation is right though

[Yansi]

Sorry, I have some gaps in the exact english ee terminology, sometimes.

I've fixed that. Thanx for correction.

[tommygdawg]

Hey all, I'm still confused but I'll be a bit more specific (and let's just drop power factor for the moment).

I did measure my voltage at the wall and it's right around 120 plus/minus literally 1-3 volts. I don't remember the exact number but it's right around 120. So, for the purposes here, let's use that number.

My main issue is I don't understand where the discrepancy in the numbers is coming from. I'm measuring a pull of 12.5 amps (at the wall) at a voltage of 120 volts. I'm also measuring (at the wall) a wattage of 1040 watts.

According to the math, 120*12.5 = 1500 watts. If I did it in reverse then the math gives me incorrect amperage (1040/120 = 8.67 amps). So my question is, why am I measuring a wattage of 1040 when the math tells me I should be measuring 1500 (or conversely measuring an amperage of 12.5 when the math tells me I should read 8.67)?

Just for the record, I happen to know that these lamps draw quite a high current and there's no way it's *only* drawing 8.67 amps, so it's much more likely that the 12.5 amps figure is more accurate.

[T3sl4co1l]

If you could look at the current and voltage waveforms, you would see them either out of phase, or current looking spikey.  Power != V*I because math.

Tim

[flynwill]

If your kill-a-watt is like mine it will happily report Volts, Amps, Watts, VA, Hz and Power factor.  (As well as tally kWh if you leave it plugged in).

When you measure Volts and Amps separately and multiply them together you get "VA" (for Volt-Amps)  what Yansi referred to as "apparent power".

If the load were purely resistive (eg traditional incandescent light bulb) the Voltage and Current waveforms would be exactly in phase and the VA and Watts readings would match.

However with a reactive load, like your ballasted lamp, the waveforms are out of phase.  Which means during the cycle of the AC waveform there are times when power is flowing towards the load and other times when the power is flowing back into wall (having been momentarily stored in the magnetic fields of the ballast).   So to measure power in this situation you must measure the instantaneous current and voltage at many points through the AC cycle, do the Watts=Volts*Amps calculation at each point (keep track of whether the power is positive or negative) and add it all up.   That's exactly what your Kill-a-watt is doing which is what makes it such a nice little meter for $40.

If you select "VA" on the kill-a-watt it should display the same 1500 you get when you multiply the Volts and Amps.  Select power factor (PF) it should report Watts/VA.

The VA number is important.  Those 8.67 Amps are real amps and are heating up the wiring just like 8.67 amps of purely resistive load would.  That's why power transformers are usually rated by "VA" and not by "Watts".

[coppice]

If the load were purely resistive (eg traditional incandescent light bulb) the Voltage and Current waveforms would be exactly in phase and the VA and Watts readings would match.

That's not quite true. A water heater is a resistive load. An incandescent light bulb isn't. The temperature, and therefore resistance, of a lamp filament varies so much through the mains cycles that most lamps produce about 10% to 20% THD in their current waveform, even when the voltage waveform is a pure sine wave. The harmonics in the current affect the RMS current measurement, and hence the apparent power measurement. They do not affect the active power measurement, because this is only sensitive to things that correlate between the voltage and current waveforms. The result is that active and apparent power readings do not match, even though the fundamentals of the voltage and current waveforms may be exactly in phase.

[flynwill]

That's not quite true. A water heater is a resistive load. An incandescent light bulb isn't. The temperature, and therefore resistance, of a lamp filament varies so much through the mains cycles that most lamps produce about 10% to 20% THD in their current waveform, even when the voltage waveform is a pure sine wave. The harmonics in the current affect the RMS current measurement, and hence the apparent power measurement. They do not affect the active power measurement, because this is only sensitive to things that correlate between the voltage and current waveforms. The result is that active and apparent power readings do not match, even though the fundamentals of the voltage and current waveforms may be exactly in phase.

Hmm well there may well be incandescent bulbs out there where the filament fluctuates in temperature as you say, but I would assert that most don't (or the effect is very small).  Since I had the bits easily to hand I plugged a 150 W incandescent bulb into my kill-a-watt, and the PF display bounced back and forth between 0.99 and 1.00.  (Of course I  haven't tried to do the math, perhaps you could have 10-20% THD in the current waveform and
still have a power factor 0.99).

But yes there are plenty of loads (like just about any DC power supply) where the current waveform will be in phase with the voltage but  very distorted leading to differences between VA and Watts just as you say.

In any case this is way beyond the OP's concerns.

[coppice]

Hmm well there may well be incandescent bulbs out there where the filament fluctuates in temperature as you say, but I would assert that most don't (or the effect is very small).

I assume this is a faith based belief, because if you try actual current waveform THD measurement on normal commercial incandescent lamps for room lighting you will find considerable THD. I don't know how the kill-a-watt measures power factor. Do you? There are several approaches. The influence of ADC noise on small current measurements means most meters derive power factor from active and reactive measurements. However, reactive measurement is also problematic when harmonics are present, unless you have the cash needed to build a full scale power quality analyser.

[bostonman]

I'm digging up an old thread. Some time ago, maybe a year ago a similar discussion was discussed, but I can't locate it.

Instead of reading the entire thread over, basically the OP didn't get the correct wattage when multiplying V and I. The discussion of apparent and reactive power was the cause.

My wall voltage is 120V AC RMS. When I measure current through a reactive load (i.e. not purely resistive), is the current I'm measuring correct and just the total power is affected by the PF?

A few times over the last few months I've measured current through a device to figure out the hourly cost by simply multiplying V and I and hadn't considered the PF because I thought the current measured was being factored into it.

Although I have one of those wall meters, I'm curious whether having a current meter (DVM) in series with the device is giving the correct current and how I would know what the power factor is if I didn't have a wall meter so I can calculate the correct hourly cost.

[SteveThackery]

When I measure current through a reactive load (i.e. not purely resistive), is the current I'm measuring correct and just the total power is affected by the PF?

A few times over the last few months I've measured current through a device to figure out the hourly cost by simply multiplying V and I and hadn't considered the PF because I thought the current measured was being factored into it.

OK, the current you are measuring is "correct" (within the accuracy of your multimeter), but the calculation you are using:

power (in watts) = volts x amps

...might not be.

That equation is only valid when the voltage and current are in phase (or when working with DC). It is up to you to establish whether that is true.  The current you measure with your multimeter does not "take into account" the power factor because your multimeter does not measure the phase shift between the voltage and the current.

The key takeaways:

1/ watts = volts x amps is only true when the current and voltage are in phase (or when working with DC)

2/ your multimeter does not measure the phase angle, so it cannot take that into account.

3/ resistive loads do not cause a phase shift between voltage and current, so you can use the formula: watts = volts x amps

4/ capacitive or inductive loads can cause a phase shift between voltage and current, so you cannot use watts = volts x amps (when there is a phase shift)

Does that help at all?

[radiolistener]

As already mentioned, multiplying current by voltage gives you the apparent power in volt-amperes (VA), not the real power in Watts consumed by your lamp. This value represents the total power flowing through the circuit — part of it does useful work, while the rest circulates back and forth between the lamp’s ballast and the mains due to reactive components. The real power (what your Kill-A-Watt reports in watts) is lower because it accounts for the phase shift between voltage and current.

Some of the reactive power returns to the grid and is effectively used by other loads connected to the same line. But it still flows through your wiring and reactive elements, causing a small amount of extra heating losses (wire heating), so the total energy drawn from the mains is slightly higher than the lamp’s actual power consumption.

How does this happen? The mains voltage is alternating, and your lamp’s ballast contains reactive components — typically inductors or capacitors. When the voltage rises, the reactive components in your lamp’s ballast (inductors or capacitors) store energy in their magnetic or electric fields, drawing power from the mains. When the voltage falls, they release this stored energy back into the power grid. In effect, part of the energy taken from the grid earlier in the cycle is returned later, slightly shifting the phase of current relative to voltage. As a result, over a full AC cycle, some power is only exchanged with the grid rather than consumed, while the rest is converted into useful work or heat by the lamp.

[Terry Bites]

I very much doubt that the electonic balast draws a sinusoidal current.
If it did, then it would likely be PF corrected and if it is then everything should tallly.
The wattmeter thingy you have is unlikely to be sophisocated enough to make an accurate measurment of the current waveform.
At best it will be average responding not RMS.
The RMS current value needs to be adjusted for "crest factor", even really good DVMs cant do that. Cheap watttos, well... you get what you pay for.
Check your waveforms on a scope if you're really worried.
If you still cant sleep read the attached.

[bostonman]

The reactive power is the part I knew, but it's more about the physical measuring. I've assumed that when I measured current with say my Fluke (obviously connected in series), the measured current was the actual current after the reactive losses.

A practical example is that recently I measured the current for a stereo amplifier. The 120VAC RMS is always just that (ignoring line fluctuations) and I measured 0.9xxA RMS (let's call it 1A RMS to make the math easier).

When I saw the 1A, I thought: due to the reactive components in the unit and inefficiencies, it's normal current consumption may be (using a random number) 0.7A, but the losses and the "real" current is 1A. Therefore the amps I'm reading is "real" and power being used is 120W or 120/1000kWh (which is the number I'd use to multiply all the associated costs on my electric bill to find the money spent per hour to run the amp).

[CaptDon]

Wrong, the 12 amps he measured is heating the wires not the calculated 8 amps. That is why power companies add a surcharge for poor power factor. The wiring, switchgear and transformers must support the actual measured amperage draw and yet with a very poor power factor the energy consumed and paid for will be lower than the apparent power. It would really upset the power company if the Apparent power were 100KW but you were only paying for 30KW of real power. To the original poster, Until you get a grasp on the phase relationship between the amperage and voltage your simple math will never work. Try to get a handle on this example. A theoretical 'perfect' inductor with no resistance could have a power factor of ZERO if the phase relationship is a true 90 degrees. It could draw 10 amps at 120VAC and consume NO POWER what-so-ever!!!

[radiolistener]

The reactive power is the part I knew, but it's more about the physical measuring. I've assumed that when I measured current with say my Fluke (obviously connected in series), the measured current was the actual current after the reactive losses.

A practical example is that recently I measured the current for a stereo amplifier. The 120VAC RMS is always just that (ignoring line fluctuations) and I measured 0.9xxA RMS (let's call it 1A RMS to make the math easier).

When I saw the 1A, I thought: due to the reactive components in the unit and inefficiencies, it's normal current consumption may be (using a random number) 0.7A, but the losses and the "real" current is 1A. Therefore the amps I'm reading is "real" and power being used is 120W or 120/1000kWh (which is the number I'd use to multiply all the associated costs on my electric bill to find the money spent per hour to run the amp).

What you’re measuring with your Fluke is indeed the RMS current — that is the actual current flowing through the wire. However, apparent current does not equal real power consumption when reactive components are present.

The real power consumption (in Watts, what you pay for) is:

P = Vrms ​× Irms ​× cos(ϕ)

where ϕ is the phase shift between voltage and current caused by inductors or capacitors in the device. So even though your meter reads 1 A RMS, not all of that current is doing useful work, part of it is reactive, circulating energy back and forth between the device and the mains. That’s why V × I overestimates the real power consumption.

The reactive portion of the current does not get consumed by your device — it is returned to the grid and can be used by other loads. However, it still flows through your wiring and reactive elements, causing a small amount of heating losses in the conductors. As a result, the actual energy drawn from the mains is slightly higher than the device’s real power — typically only 1–2 W — while the rest of the reactive power is fed back into the grid.

[radiolistener]

Wrong, the 12 amps he measured is heating the wires not the calculated 8 amps. That is why power companies add a surcharge for poor power factor. The wiring, switchgear and transformers must support the actual measured amperage draw and yet with a very poor power factor the energy consumed and paid for will be lower than the apparent power. It would really upset the power company if the Apparent power were 100KW but you were only paying for 30KW of real power.

It’s important to clarify that the grid does not “supply 100 kW while you only consume 30 kW” in the sense of wasted energy. In reality, reactive power drawn by a consumer is returned to the grid and can be used by other nearby loads. The situation becomes significant only for large industrial consumers with megawatt-level loads and no nearby consumers to absorb the reactive energy — in that case, the reactive current flows over long transmission lines back and forth, causing noticeable losses due to transmission line resistance. To mitigate this, utilities install reactive power compensation (capacitors or synchronous condensers), and industrial users may pay a small surcharge for high reactive currents to cover infrastructure costs.


This situation is somewhat analogous to a high-Q antenna. Suppose a transmitter delivers 100 W to the antenna, with 70.7 V at the transmitter output and 1.41 A of current. Inside the antenna, voltage may reach 10'000 V and currents up to 100 A, but this does not mean the antenna consumes 10'000 × 100 = 1'000'000 W. The actual power consumed remains 100 W, the large voltages and currents inside the antenna are due to reactive energy circulating within the resonant structure, accumulated during previous carrier oscillation cycles.

Unlike a high-Q antenna, the mains grid has a much lower effective Q due to its low frequency and distributed nature. Nevertheless, the principle is similar: the power grid can be seen as a distributed resonant network that temporarily stores and releases reactive energy, with currents circulating through inductive and capacitive elements without directly contributing to real power consumption. The power grid can briefly store energy from the generators in its distributed reactive components. In the event of a sudden generators outage, this stored reactive energy can continue supply the network for a very short period of time.

[MrAl]

Hey all,

I've got an HMI light (one of those ones that runs off a ballast) and I'm trying to do some basic math but it's not coming out right. I've plugged the device into a Kill-a-Watt to measure the amperage and wattage. Here's my issue:

The light is rated for 1200 watts. According to the Kill-a-watt, I'm actually drawing 1040 at approximately 12.6 amps. I took my measured amperage times the voltage and got a wattage of 1,512 watts (12.6*120). Why the discrepancy? I then wanted to understand the power factor of the unit, but the math for that doesn't work out at the 1,040 watts measured. To get my power factor I have to take the rated wattage and divide it by the actual draw (1200/1040). This doesn't work out, it gives me a power factor of 1.15 which is impossible. Now, if I divide by the wattage the math gives me it works (1200/1512 = .79). I should also note that if divide measured wattage by voltage (1,040/120) I get around 8.67 amps. So which is it? Am I running at 8.67 amps or 12.6?

Do these question make sense? I suppose the TL;DR version is: why does my meter read 1,040 watts at 120 volts and 12.5 amps, when the math tells me that I should be reading 1,500 watts at 120 volts and 12.5 amps?

Thank you!

When you calculate power factor you end up with a value less than or equal to 1.

PF=cos(Angle)
where Angle is the angle between current and voltage.  The Angle can be in rads or degrees you have to make sure your cosine function gets the right one.

The only problem with that is that it is mostly for pure sine waves.  I don't know exactly what your circuit is, and if it has a current wave that is not sinusoidal, then you have to use a more complicated formula.  That is:

PF=Pavg/(Vrms*Irms)
where
Pavg is the average power (use the integral formulation for that),
Vrms is the RMS value of the voltage in the time domain,
Irms is the RMS value of the current in the time domain.

This illustrates that it may be harder to calculate the power factor depending on what the load is like.

[golden_labels]

Whether it’s a linear load or not also affects the measurement. Is your multimeter “True RMS”? If not, it will underestimate or overestimate RMS measurements for anything other than sinusoid waveforms. May be a few percent, but may be as much as over 40% error in the worst case.

[SteveThackery]

The reactive power is the part I knew, but it's more about the physical measuring. I've assumed that when I measured current with say my Fluke (obviously connected in series), the measured current was the actual current after the reactive losses.

A practical example is that recently I measured the current for a stereo amplifier. The 120VAC RMS is always just that (ignoring line fluctuations) and I measured 0.9xxA RMS (let's call it 1A RMS to make the math easier).

When I saw the 1A, I thought: due to the reactive components in the unit and inefficiencies, it's normal current consumption may be (using a random number) 0.7A, but the losses and the "real" current is 1A. Therefore the amps I'm reading is "real" and power being used is 120W or 120/1000kWh (which is the number I'd use to multiply all the associated costs on my electric bill to find the money spent per hour to run the amp).

I don't think you are reading the replies and properly thinking them through. The current you are measuring IS THE REAL CURRENT!  Forget this nonsense about "normal" current and "real" current. Forget this nonsense about current "after reactive losses". Read this next sentence and don't move on until you have really understood it:

There is only one value for the current; it is the current you measure with your multimeter.

There is in no sense "normal" current and "real" current, and "reactive losses" is a meaningless term.  THE CURRENT YOU MEASURE IS THE CURRENT.

You still don't seem to have grasped the fundamental principle: a reactive load causes a phase shift between the current and the voltage. That is all it does, and that phase shift is what causes the normal wattage calculation to go wrong.

Now read this and, again, don't move on until you have truly got your head around it:

When there is a phase shift between the voltage and the current, you cannot use the formula watts = volts x amps.

When there is a phase shift between the voltage and the current, the wattage consumed is less than the product of the volts and the amps. How much less depends on the degree of phase shift.

Remove from your mind any concept of "reactive losses". Insert into your mind the concept of a phase shift between the voltage and the current. That is a true picture of what is happening.

The power factor (PF) is the cosine of the angle of the phase shift, and varies between 0 (90 degrees phase shift) and 1 (zero phase shift). To calculate the actual power consumption of a reactive load, use this:

watts = volts x amps x PF

[bostonman]

Unless I'm wording/phrasing this wrong, I have known and do understand say a resistor is purely resistive and there for V * I whereas a motor is V * I * PF.

To reiterate and think out loud for a moment to clarify my previous (or maybe still) confusion: What kept puzzling me is which numbers are real. The 120VAC RMS is real and constant, that can't change (ignoring fluctuations). When I measure current with my Fluke in series with the unit and see 1A, that's also real and can't change (again, ignoring fluctuations as the device heats)?

Now if it were a resistive load, that would be 120W. If it's a motor and say the phase is 45 degrees, that would be 70.7W (COS of 45 times 120 * 1), correct?

The confusion came into play, how do I know which meter is giving me a real number based on apparent and reactive (hopefully it's okay to use those terms at this stage). I'm only assuming my plug in kilowatt meter will give me the correct wattage that includes the PF, but is the only way to know by placing both voltage and current on a scope and figuring out the phase angle?

[mawyatt]

Your RMS Voltage and Current Meter readings are Scalars, they need to contain Phase information, thus become Vectors. You need the Relative Phase between the Voltage and Current to calculate the Real and Imaginary Reactive Powers.

A modern DSO is a good tool to allow the viewing of the Voltage and Current waveforms and their relative phase, some even include math capability allowing even more detailed analysis.

Best