# How to quantify the difference in power of tablesaw motor rewired to 240 vs 120 volts

I often read comments that the same 1.5 hp motor will be less powerful if rewired from 240 to 120 volts, but the wattage of the machine should be the same at 240 volts as at 120 because the amps just double when the motor is rewired to 120.

Is this power loss real, and if so how would it be quantified, pounds of torque?

• FYI, standard voltages in the US have been 120/240 for more than half a century. I've revised those numbers to avoid confusion with people less in-the-know than us. Feb 10 '20 at 16:22
• By rough and simplified look - considering constant amps indeed going from 240 V to 120 V you reduce power by 50% (P=U*I -> P(240) = 240V * I , P(120) = 240V/2*I -> P(120)=P(240) / 2. Of course electrical motors tend to take higher amps to compensate (according to the load) Feb 11 '20 at 13:04
• What AC frequencies are you changing from and to? 60/60? or 60/50? Feb 11 '20 at 18:37
• The only difference is the slight power loss in the wiring leading to the motor, when the voltage is halved and the current doubled. I would guess that this is on the order of 2%. Feb 12 '20 at 0:50

I don't know who is saying those comments or in what context they are saying, but I don't agree. If the motor is actually designed to be jumpered between 240V and 120V, then it should perform the same in either configuration.

Induction motors of this type have an entirely passive rotor, so no brushes. The only windings are in the non-moving fields, so it is practicable to switch them. How it's done is to wind the fields with 2 wires half the cross-section. For 120V, those wires are connected in parallel. For 240V, in series. The same amount of current flows through each wire in either configuration, so motor performance is the same, and temperature will be the same.

If you are abusing a motor that is not manufactured to be switchable, then all bets are off. Or if you are dealing with a brushed or shaded-pole motor on a smaller appliance, the rules can change, but that is not OP's question.

• "then it should perform the same in either configuration" Only if the specs say so. I've seen plenty of tools that could be used on multiple voltages, but were only optimized for one. Other settings would put it at ~70% of max power.
– Mast
Feb 11 '20 at 8:56
• @Mast SE being a Q&A site, answers are always to answer questions. As such, answers only only exist in the context of questions, and no answer should ever be read as a blanket statement about all things. OP was asking about a table saw, which will surely have a particular type of modular COTS induction motor in it. Those types are as I say. Feb 11 '20 at 12:44
• @MikeBrockington No, it's right. In 240V mode, current goes through winding A, then winding B, dropping 120V per winding. In 120V mode it splits to go through both. Feb 11 '20 at 16:53
• It's worth noting that dual voltage motors typically report different specifications for 50Hz vs 60Hz operation, which is significant. If they're dual-voltage motors (center-tapped windings), this may lead to confusion about the performance differences being related to the different voltages when, in fact, it is more likely the numbers correspond to 120V@60Hz vs 230V@50Hz, and the frequency difference is the major contributor to the performance specs.
– J...
Feb 11 '20 at 18:00
• @J... That's a very good point. If the voltage change is accompanied by a frequency change (e.g. North America vs 5-Continent), then that has a big effect on the motor. Feb 11 '20 at 18:09

You'll need to check the motor manufacturer's specification sheet for your model to be sure but generally there should not be any significant performance difference between running in at 220V vs. 110V. What does happen, however, is that since the current through the motor is going to be 2X at 110V vs. 220V, it will heat up more and that will cause the efficiency to do down due to the hotter wires. This, of course, causes more current to be drawn which generates more heat. Such a motor should be designed so that it delivers its rated output at any supported configuration. Any exceptions will be called out in the documentation.

• It is also possible that there is some voltage loss in the wiring feeding to the outlet where the motor is plugged in. At double the current at 110V the voltage loss will be double and will be a greater percentage of the total supplied at the motor. Feb 10 '20 at 16:15
• nonsense .. the voltage LOSS is calculated by Wiring resistance * Current .. higher V = lower Amp .. thus lower voltage loss as the wiring resistance depends on material, specific resistance (over temperature) and frequency - but frequency and material doesn't vary and lower amps mean usually lower temperature Feb 11 '20 at 13:07
• Excess heat shouldn't happen if the motor is designed properly. We're talking table saw motors, i.e. squirrel cage induction motors jumperable 120/240. Feb 11 '20 at 16:31
• @Harper-ReinstateMonica Not all motors are created equal. Plenty of chinese-made junk out there. It pays to read the fine print on these. Feb 11 '20 at 16:49
• "chinese made junk" ? Is this 1920? Far and away all the best stuff is made in China. Also, basic physics applies to both "junk" and "expensive" items. Feb 11 '20 at 20:40

The motor must support both 120/240 wiring, this will be indicated on the faceplate. If not, it can't be changed. Power output at 120 or 240 should be about the same. The biggest issue is voltage drop which can affect power a bit. Unless you have a compelling reason to convert from 240 to 120 I wouldn't do it. You'll be drawing twice the current for the same amount of "power" (wattage) which means larger wires and breaker.

Intuition insists that there is no difference, but in practice, operating a field-reconfigurable dual-voltage motor at 120V/115V instead of 240V/230V will tend to reduce the available torque... but the discrepancy between theory (but it's the same!) and reality (no, not quite!) is caused by reasons that are entirely external to the motor.

The (correctly-wired) motor sees exactly the same conditions regardless of the series/parallel wiring of the motor leads -- as long as conditions are static and the load on the motor (the wood against the blade, its hardness, thickness, and the force pressing it into the blade) is not varying. Of course, with a table saw, operating conditions are quite variable.

A table saw motor draws more power when there's wood against the blade than when the blade is just spinning freely with no work to do. More work to do means increasing the current, which the motor does automatically as it "tries" to maintain is designed rotational speed under load.

Consider...

• Voltage drop in actual volts (not percentage) on a circuit of a given wire size and length is directly proportional to current draw.

• Current drawn by a motor increases with workload.

• Current is doubled when voltage is halved, so the voltage drop on the circuit is larger with the higher current required by the lower voltage. (The wire size is increased, of course, and this is another factor in voltage drop, but the reduced voltage drop from the larger wires is not sufficient to counteract the increased drop from the doubled current, unless wires much larger than code requires are used at the lower voltage.)

• The current increase under load is also doubled when the voltage is halved.

• The power supplier's transformer may be able to better handle the demand when the entire secondary is involved in supplying the peak current, rather than only half of it.

...so the motor typically experiences a more aggressive voltage drop under load when operated at the lower voltage.

And that's where things get ugly, because of another fact about AC motors: the rated torque is available for delivery to the workload only when the motor is operated at its nameplate voltage, and falls off following the law of inverse squares as supply voltage decreases. (Bad things also happen with overvoltage, but that's out of scope of this answer.)

As available supply voltage is reduced by increased voltage drop, torque decreases by a factor of the square the reduction from nameplate voltage... so a motor running at 10% undervoltage (90% of rated) develops torque of only 0.9 x 0.9 = 81% of rated capacity... and, as the physical load on the motor surges (e.g. when you slide a piece of wood into the blade) the current increases and the voltage drop increases more dramatically when configured for 120V than when configured for 240V.

• I don't remember what the source was but I had the understanding that providing a tablesaw with 240V makes it safer because it's less likely to bind up. I think this answer explains why. Feb 12 '20 at 15:10

Typically if the motor is capable of supporting multiple voltages then it has multiple windings that are either wired in series for the higher voltage and parallel for the lower voltage. You can usually see on the nameplate the amperage be marked with two numbers separated by a slash like 16A/8A, the lower number being the higher voltage.

Either way you still get 120v across each winding, if wired for 240v it just makes the circuit supporting it less stressed. If you try to force it it to perform more than the rated work and lug the motor down you may find that using the higher voltage configuration recovers better.

I didn't have enough rep to put a comment. I did the math once for my 1.5HP table saw and the wiring that went to it. I changed it from 14-2@120VAC to 10-2@240VAC (I had a bunch of #10-2 wire laying around), assumed about a 50 ft run. Accounting for just the voltage drop difference on the way from the breaker to the table saw, it came out that you might be able to get an extra 16W of power at the motor if it was drawing the full amperage (I^2*R). That turns out to be ~1.4% extra power at full load. I doubt you would ever notice that small of a difference.

You can find the resistance per foot of various wire guages online, such as: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.html

14 AWG is 2.525 ohms/1000ft 10 AWG is 0.999 ohms/1000ft

``````14AWG: 2.525ohms/1000ft * 50ft = 0.12625 ohms
10AWG: 0.999ohms/1000ft * 50ft = 0.04995 ohms
``````

I^2 * R Losses

``````14 AWG @ 12A (120VAC) = 18W
10 AWG @ 6A  (240VAC) = 1.8W
``````

As others have mentioned, you may lose more due to heating of the coils, but it is probably isn't enough to notice.

• Hello, and welcome to Home Improvement. As you noted, this is really a comment, not an answer. With a bit more rep, you will be able to post comments; in the meantime, please take our tour so you'll know how best to contribute here. Feb 11 '20 at 2:01
• LOL why on earth would you do it that way? You certainly know that increasing voltage calls for smaller wire not larger. The wire bump to 10 AWG for 240V was gratuitous and nonsensical. In effect your answer measures 2 things at once, telling us little about either one. Feb 11 '20 at 16:25
• I had to run a new line anyway. As the saying goes if its worth doing, its worth overdoing :) I also had jointer that was 240V@30A that I needed to run a line for, so I just ran the same wire type for for both since I had plenty of it. Still shows the point that 1.5HP motor isn't going to see much of a difference even when way overdoing it. Feb 12 '20 at 17:38