We have a bi-level (tri-level? There's a small entry way and utility room, then an upstairs living area and master bedroom and a downstairs living area with bedrooms) in mid-Michigan.

The downstairs is naturally quite cooler than the upstairs, but during the summer we were having keeping the downstairs warm enough for our young son. We didn't want to raise the temp upstairs anymore but our local HVAC person suggested running the system fan continuously to help evenly distribute the air. And it worked. He did say that it shouldn't increase our costs that much and searching online seemed to come to the same conclusion. Our electric bill did end up higher than previous years but I chalked that up to higher-than-average weather temps, we were keeping things a little cooler than we had in the past due to our child, and summer electricity rates were in effect.

Now that we're getting into Winter, we just received our latest bill and it's again abnormally high. We did cut the fan usage some the billing period prior, so I'm sure it's because we're always running it still.

How much electricity usage should I really expect to see if I'm continuously running the furnace fan? Enough to see a sizeable jump in cost?

And is there a way to track how much power usage my unit is using? Do I need a killawatt or something plugged in? I'm trying to find the model number on it to see if I can find the electric specs and maybe do some calculations there. Our thermostat does tell me how often the heat or AC runs, but the heat is gas so that shouldn't affect it much anyway.

Any other suggestions or things I should try or look for? Thanks,

  • I think that's called a split-entry house. Commented Nov 28, 2012 at 22:35
  • 1
    It is worth mentioning that you should check your filters and replace them regularly. This could cause loss of air flow. I am not sure if it could make the fan work harder if the filter is dirty, but it is worth a look. I do know dirty filters can cause an HVAC system to burn up parts.
    – Tim Quinn
    Commented Nov 29, 2012 at 23:59

8 Answers 8


Where to find the specifications

Listed on the furnace

Somewhere inside the service panel of the furnace, there should be a schematic for the furnace. This may list the electrical specifications of the blower motor. If not, the blower motor itself will have a label on it. It should list the voltage, amperage, horsepower, etc.

Using an Ammeter

An ammeter is used to measure the amount of current flowing through an electrical circuit. You can use a clamp on ammeter to measure the amount of current being used by the blower motor while it's running.

enter image description here

Simply set the meter to measure current (A in the pictured meter), and clamp it on the hot wire feeding the motor.
This method may require working on or near live wires, please use caution and/or consult an expert to take the readings

Calculate Usage

With the above information and Ohm's law, it's easy to determine how much power the motor will use over a given time period.

Ohm's law says that Power (P) = Voltage (V) x Current (I), so we can determine how much power the motor uses at a single instant in time.


P = 115V * 5.2A
P = 598 Watts  

Next we have to determine how many kilowatt hours the motor will use, which we do by dividing by 1000.

P = 598 W / 1000
P = 0.598 kWh

So for every hour the motor runs, it will use 0.598 kilowatts of power.

Total power/day = 0.598kWh * 24h
Total power/day = 14.352kWh
Total power/month = 14.352Kwh * 30
Total power/month = 430.56kWh

Finally, if you multiply this number by the amount you pay per kilowatt hour. You'll see just about how much it costs to run the motor constantly.

Cost = 430.56kWh * $0.07
Cost = $30.14 per month

Results using this process will be an estimate only.
Values used are not actual values, they are example values only.
Calculations may be incorrect, nobody checked my math.

  • 4
    Awesome answer, but I doubt the motor draws 8.2A current consistently. Typically an electric motor like this has a large initial draw to get it started, at which point the smaller draw is needed to keep it spinning. It might draw 8.2 when it first kicks on but then will probably drop off considerably once it is spinning. Commented Nov 28, 2012 at 21:01
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    @maple_shaft I just made up some numbers, and even mentioned that in my "notes" section at the end. "Values used are not actual values, they are example values only.".
    – Tester101
    Commented Nov 28, 2012 at 21:04
  • Awesome answer, thanks! I'll take a look and crunch some numbers...I had a general idea how it would all work but that spells it out nicely. In real world usage does running the fan all the time typically increase costs that much over normal automatic settings that people are hesitant to do it? Commented Nov 29, 2012 at 2:31
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    -1 This will yield a completely incorrect result as it assumes that the motor is a resistive load which it is not. The motor running at constant speed is predominantly a reactive load. You need to distinguish between real, apparent, and reactive power in order to determine the correct result. Using the method in the answer will give you incorrectly high results for fan power consumption because its power factor at running speed is much less than unity.
    – alx9r
    Commented Dec 10, 2012 at 0:05
  • 1
    @Tester101 - What your answer calculates hardly qualifies as an estimate. An estimate would have a usefully small uncertainty associated with it. Your answer does not. The power factor for a fan could easily be 25% or less, which would mean that your "estimate" is wrong by a factor of 4. Perhaps it's as low as 10% which would result in an error of 10x (e.g $4.75 vs $47.53 per month). My point is your answer does not measure power factor and power factor is essential to determining real power use.
    – alx9r
    Commented Dec 10, 2012 at 18:24

I've been using the fan on my furnace much more recently, so I was curious about this as well.

I have a Luxaire G9T100014UPC13C natural gas furnace that is about 9 years old, and a Current Innovations CI300E-WIFI thermostat (which I'm controlling from my iPod).

I'm measuring the overall power use of the furnace -- right where it comes into the cabinet.

My ammeter doesn't have the fastest response, but on startup it is measuring as high as 23A.

When the furnace starts up fully (including gas firing), it is drawing about 5.2 A.

furnace on - 5.2 A draw

When I turn the temperature down, and it goes back to only running the fan, it drops slightly, to 4.4 A.

fan running after furnace turns off - 4.4 A

Now, I am not quite sure how this works out. If I wait until it turns off, and then set the fan to 'On' (instead of 'Auto'), it draws around 9 A.

fan only - 9.0 A

I suppose it's possible that while I have it set to 'Fan On' it runs the fan at a higher speed than it does normally, but it certainly doesn't feel like it. I let it run for 10 minutes before measuring and it was still around 5.2 A, and have repeated this a couple of times a couple hours apart just to be sure (which is why I took pictures the second time). Aside, anyone have any idea why it's so different while only the fan is on vs gas firing?

  • Is it a high efficiency furnace? I believe the fan motor is digitally controlled so that the fan can run slower when you need to draw less heat into the air but I may be wrong. When too much heat is drawn from the exchanger it lowers the temperature of the exhaust gases causing it to condense into corrosive carbonic acid in the flue. This can then drip back down into the furnace causing corrosion. Commented Nov 29, 2012 at 12:07
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    When the furnace is firing, the draft inducer will be running as well. This will account for the extra current draw when the furnace is firing.
    – Tester101
    Commented Nov 29, 2012 at 12:54
  • 4
    If you take the measurements when in 'Cool' mode (if available), you'll likely see a current draw similar to the 'Fan ON' mode. Blower motors often run at slower speeds when in 'Heat' mode, because hot air is lighter and easier to move.
    – Tester101
    Commented Nov 29, 2012 at 12:58
  • @Tester101 I always assumed that the low speed fan for heat mode was because no one wants a house that feels draftier in the winter, while in cooling mode the extra air velocity helps you feel cool
    – mac
    Commented Nov 29, 2012 at 15:10
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    @gregmac As far as I can tell, in all cases in your tests above you are measuring current that is proportionate to apparent power, but not real power. Real, apparent, and reactive power are related but not the same. Your utility's watt-hour meter measures real power. When the fan is running at constant speed, it is most likely predominantly a reactive load with low power factor. So, you will read high current even though it is not using much power.
    – alx9r
    Commented Dec 9, 2012 at 23:54

Nameplate Power Won't Reveal Actual Power Consumption

The horsepower, wattage, current, and/or voltage will not tell you how much power your fan is actually consuming. That is because the amount of power consumed depends on the load that the fan and your duct system imposes upon the motor. In other words, the actual power consumed depends very much on your particular installation.

Power Consumption Follows an Affinity Law

The relationship between fan RPM and power consumption follows an affinity law. Specifically power is proportionate to the cube of the RPM. So cutting your fan speed in half reduces your power consumption by 8x.

This relationship makes clear that reducing fan speed as much as possible is critical to reducing power consumption. However, affinity law alone won't help you determine power consumption in absolute quantities.

Reducing Your Furnace Fan Speed

Many household furnace fan motors are designed to operate at a variety of discrete speeds. This is usually accomplished by applying power to one of a variety of winding configurations built into the motor. Where this is the case, the furnace installation manual usually provides a table of connection wire colors and corresponding speeds for your fan.

The furnace fan speed is meant to be set to an appropriate value for the particular installation. In my experience, however, the speed installers typically set the fan to is much higher than optimal.

Fan Power Consumption Increases With Constricted Air Flow

A fan pushing air through a short length of oversized duct uses much less power than the same fan pushing the same rate of air through small, extremely long, or damped ducts. Reducing constrictions like filters, grills, and closed dampers on the entire air circuit including the return air path will reduce the fan power consumption for a given amount of air flow.

Measuring Actual Power Consumption Requires a Particular Type of Clamp Meter

Furnace fans typically get their power from a controller or relay within the furnace. It is at this location that electrical measurements can be made to determine fan power consumption. In order to measure the actual power consumed by a furnace fan, you need to do be able to do the following:

  • on a common timebase,
    • measure the waveform of voltage applied to the motor, and
    • measure the waveform of current through the motor
  • calculate the product of the above two waveforms
  • find the time average of the product waveform over several cycles

The above is effectively what a real power meter does. The result will be in watts (W) which is what your utility power meter integrates over time to determine a watt-hour (Wh) reading which they use to calculate your bill.

The above is pretty complicated to do. There are fairly common, specialized test instruments designed exactly for this purpose. The following are two examples:

There are some conceivable other ways of measuring real fan power consumption, but they are more complicated, error prone, or dangerous.

It is not possible to measure AC power consumption of a motor with a multimeter. Nor is it possible with an ammeter-only. This is because AC power consumption involves a phenomenon called power factor which cannot be determined using those tools. In order to measure power consumption, a single piece of test equipment must measure both voltage and current over time (think thousands of samples per power cycle) and process that data to determine power consumption.

  • This is the answer I came here for. Thanks. This makes me want to get a Sense (or other at-the-panel measuring system) so that it's a bit easier to test and view live results of changing things. Otherwise it's just heuristics from looking at what I did a month ago and seeing if there was a change on my bill. Kill-a-watt is great for one-off devices at least.
    – kamranicus
    Commented May 19, 2021 at 2:34

My power meter shows instantaneous real power consumption and when I run my furnace fan, it consumes around 500 watts.

If you leave it on 24x7, this will be quite noticeable on your monthly electric bill.

There is work being done on measuring and improving the power consumption of the motors used to move the air but that only helps if you buy a new system.


All the calculations show that 500+W furnace motors are not cheap to run 24 hours a day. I think the following product or something like it are much cheaper to operate and do well balancing upstairs and downstairs temperatures. Place the fan in a stair well so it blows strait up



Power consumption by the main blower and induction blower, if present, can be expected to vary between about 500 watts and 1,000 watts for the most common, single speed arrangements. Good design practice suggests the furnace should run %25-50 of the time under cold conditions, with %33 being commonly quoted as ideal under most conditions. The same guidelines hold true of refrigerator design.

500-1000 watts is no small draw by any measure. It's about 3-8X the draw of the average refrigerator, and as the design duty cycles are similar, the electrical costs of running a furnace can be expected to be 3X-8X as great. It's about 1/3-2/3 the draw of notoriously high power appliances like hair dryers, vacuum cleaners, microwaves and toaster ovens: but those appliances are generally operated for short periods of time.

In short the electricity cost to run such a furnace can be very significant.

Depending on their design, dual stage furnaces can save considerably on electricity costs. For a given amount of heat delivered, a furnace will operate for a longer period of time under first stage or low heat operation than it would under second-stage operation. This might lead one to conclude that their is little advantage, perhaps even a significant penalty, in terms of overall electrical consumption of a dual-stage furnace.

But whereas air demand is linearly proportional to heat output, power consumption varies non-linearly with air demand and delivery from the blower motor. The efficiency of a real-world system rises sharply as demand for air decreases. At higher flow rates, proportionally more is lost to turbulence and back-pressure across non-ideal ductwork, filters, and delivery vents.

More sophisticated schemes offer additional opportunities for electrical savings. In a fully modulated furnace, the above principal can be taken to even more efficient extremes. Furthermore, in order to achieve the wide range of speeds required, electronically commutated blower motors are a virtual necessity. Like the brushless DC motors that are ubiquitous in computer fans and appliances such as cordless power tools, cordless vacuum cleaners, and electric vehicles, these motors are more efficient than the inductive motors they commonly replace.

  • Smells like AI-generated, and doesn't directly address the (very old) Question, which already has a (very old) accepted answer. Please take the tour to find out how to meaningfully contribute here; AI answers to old Questions aren't helpful. Commented Apr 4 at 15:14

Well- These just Est Averages- Good Enuf for Gov't work !

-1,000 Sq ft ranch home, that The Wife and I remodled as our Retirement cottage Home on a Lake in Wi. , on Winter wknds, while still working our last 2 yrs..in the City of Chicago area during the week. -But 3.5" Foam Insulated Walls,( R 24) , Double Insulated Windows and Doors! and 8" -R 56 Foam Insulated Attic! + Electronic Air Cleaner as well as Last yr, we R&R all Lighting to use LED lights now .( Our State of Wi. Provided Energy Refunds of upto $100 per yr to R&R Our Lighting/Lite Bulbs) & a 100% Tax deduction from State Taxes on any excess! and for our New Heating system , Insulating our Home, Walls, Windows and Doors..all 100% Tax Deductable ( ave -25% savings) -*I also Installed a Small 250 whr Solar panel! and a *NG On Demand HWH ( Hot water Heater) with a $200 State Refund adn 150% tax Deduction on the bal . -the Hse Used to use the ave of 547 kw a Mo. ( $80 mo) in winter & ave $70 mo in *NG heating.. *NG = Nat Gas -It now uses less than the ave of 25% of that for both Electric and NG..

1-Our Electricity In Wi. cost the ave of 0.146 cents per KWhr! not 0.07 cents kwhr

2-In summer cost 0.17 cents kwhr! 5 mos of the yr. and 0.13 kwhr the other 7 mos a yr .Thus the ave of 0.146 cents per KWHr for the yrly ave.

3- During our Heating season, On Average- Our Blower motor on our Furnance runs only for 4 min every ave of 20 min. x3 per hour = 12 min an hour.. Ave out to @ x 24 hrs = 288 min per 24 hr day by 60 min = 4.8 hrs every 24 hrs . 3 of the 6 mos a heating season it only runs the ave of 2x per hour and during the coldest months or 3 mos a yr, it runs the full 3x per hour.

4- But for Simplicity and at the MAX usage, using the 3x per hour / 5 hrs per 24 hr day ! 5. Using 300 watts per hour for our Furnance Blower motor for our 1,000 Sq ft Home x 5 hrs a day = 1,500 watt hr useage = 1.5 KWhr a day x 0.146 cents kwhr ? = to about 0.75 cents a day use x 30 days mo = 22.5 kwhrs x 0.146 = $3.30 a month

If you run your Blower Fan 24/7? You should be running one that runs at least 75% Slower speed ( CFM)

If uses like ours of 300 watts Hr Blower? x 25% use speed = 75 whr Or needing only 25% of the 100% CFM needed to keep the air circulated.. If your home needs More? You need a New and larger Blower Motor to push More CFM ! so it can run at 25% Speed vs 100% Speed. 25% of 500 = 125 whr useage.. x 24 hrs? be about 3kwhr a 24 hr day use x our CPKWhr of 0.146 = 0.44 cents a day cost or about $13.00 US a Month cost.

Using a HYBRID 1 KW Electric & Pellet wood Heating Furnance ( we get the Pellets for free from our Township. 6 x 60 lbs bags = 360 lbs a season, 1-60 lb bag in a Feeder , using the ave of 2 lbs per day ) with a 24 hr blower fan system that feeds the (Sealed ) Air Ducts , that uses 25 watts an hour = 600 watts per 24 hr day ! Just to keep the Heated air circulating thru-out the house and Attached Single Car Garage! at ave 68 degrees Temps, 24 hr day!

*I got the Small Solar panel system Used from a Business that R&R theirs and they got bigger one's and cost me Nothing ! Did cost me $200 to hire 2 guys to Install it on my South side Garage Roof. Took them about 1.5 hrs to do it plus the wife made them 2 Sandwhiches and 2 Beers ! and a Piece of Homemade Apple Pie /Ala-mode during the Spring ..

You can buy Used Solar Panels if you search around and save -50% or more ! Alot of Businesses buy smaller ones to test them and tax deduct the costs and give them away as well ! and, Last but not least! We bought Sotck in the 2 utility co.'s here ( WEC and LNT) when we moved up herto retire in 2003 and their Dividends alone more than pay for our Utility bills and then some ! and housing cost only 40% as much as it did in Illinois( Chicagoland area)

Eversince Wi. Became a Republican State, they want to Attract (and Keep) Middle and Upper Income Residents to stay in Wi. ! Getting alot of us Illinoians to move up here to retire! and its working! over 55% of Homes are owned by Former Illinoisians Like us ! and over 85% of the Summer Homes area as well !

Hope some of this helps, someone else!


A single phase motor has a power factor. The formula for power of a single phase motor is voltage time current times power Cosine of the power factor. The power factor for a motor is determined by the manufacturer but I do not know what a typical power factor is for a single phase blower motor. If you can find this out and measure the voltage and the current its easy to figure power consumption. Just multiply the power times the current times the cos PF and that gives wattage. Then convert to kilowatt hours. Ie.......a 100 watt light bulb on for ten hours uses a kilowatt of electrical power. Then multiply the kW hours by your utility rate.

  • This isn't what the OP is looking for, and it adds another unknown (PF) to the mix.
    – Doresoom
    Commented Dec 9, 2014 at 19:36

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