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I know that putting 240V through an appliance rated at only 120V would almost certainly damage it.

But what if I connect an appliance rated for 240V into a 120V AC socket? Apart from degradation in performance, would there be any damage to the appliance itself?

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Not much happens at all :-) about half what you expect, at best. You probably won't damage it, but best not to have to find out. – EJP Jul 13 '14 at 8:24
What's the -1 for? Any comments for it? – shortstheory Jul 13 '14 at 9:04
Maybe because it is off topic here @ EE. DIY would be a better fit. Warning: Do not mess with mains voltage, it really hurts! – Turbo J Jul 13 '14 at 10:44
A fan motor rated for 240V may not start running and the lack of cooling air flow may cause components (including the fan motor itself) to overheat. – jippie Jul 13 '14 at 10:49
+1 to offset the stupid and pointless -1. A reasonable question. – Russell McMahon Jul 13 '14 at 17:07
up vote 10 down vote accepted

Probably nothing serious would happen as has been mentioned already but there is a possibility of a potential serious situation. Consider an appliance only intended for operation on 240V AC but is able to work from (say) 200V to 250V. To do so might mean it uses a switch-mode power supply to regulate the internal DC voltages. Let's say it required 100 watts internally, maybe some form of audio amplifier.

At 250 volts AC it would draw 0.4 amps plus 10 % more for inefficiencies - that's a current of 440 mA. At 200 volts AC it would draw 550mA. At 100 V ac it would try and draw a current of nearly an amp if it were able.

The point is that it will try and draw more current at a lower AC voltage and this could blow an internal fuse or damage the switching transistor - the average current may only be 1 amp but the switching current might be 10 amps. Also, at a lower voltage (with the increase in current) the reservoir capacitor after the bridge rectifier will be struggling to maintain low ripple and between cycles the dc voltage before the switching element may sag to only 50 volts - this means a higher instantaneous current draw on a cyclic basis and possibly more damage to the regulating switching transistor.

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There could be damage. With half the RMS AC voltage, that is, half the force pushing charge through the device, we might expect half the current flow. If the device acts like a simple resistor, that's exactly true. That means 1/4 of the normal amount of power is used by the device.

If the device has capacitive or inductive reactance, and has nonlinear effects, then no. Still, with no specific device as subject discussion, we may as well assume one quarter the power usage.

If that power is primarily working a motor, then the motor will be spinning slower. (duh.) Some motors depend on spinning at a high speed to keep themselves cool. If it's not spinning fast enough, maybe it won't keep itself as cool. But at (probably) 1/4 the power, it's not getting as hot, either. Will friction or load keep the motor from spinning at all?

Whether the cooling effect is diminished in the same proportion as the motor heating, depends on the actual type of appliance, the load the motor is pushing, the presence of voltage regulating circuits, and for all I know, the appliance's astrological birth chart.

That's just considering basic motor physics. The range of parts and physical phenomena in a generic unspecified household appliance is vast, and so it is not possible to rule out some other way that half-voltage input could cause damage.

Short answer: without further info, it's guesswork, but the range of guesses must include the possibility of damage.

There is only one way to find out, assuming you can make the plug fit the socket...

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Thanks! "Plug fitting the socket" shouldn't be too hard with all the international adapters out there :P – shortstheory Jul 13 '14 at 9:06
The rotational speed of an induction motor depends primarily on the frequency of the input voltage. The voltage itself changes how speed and torque relate. A lightly loaded induction motor will likely spin faster at 120V 60Hz than at 240V 50Hz. A heavily loaded motor will likely just stall instead. – ntoskrnl Jul 13 '14 at 10:18
Why are you so specific on motors while OP was quite generic? – Vladimir Cravero Jul 18 '14 at 6:49
@ntoskrnl and a stalled motor will dump all the incoming power as heat in the motor. – Peter Green Dec 17 '15 at 12:19

In a linear situation (an electric blanket, for example), the power will just be reduced to 25%.

Switching power supplies such as PC power supplies (the kind with a slide switch to select the input voltage) will attempt to produce the required output power with the available voltage, and unless some kind of under voltage lockout or thermal protection kicks in could be damaged - the power devices will get much hotter than normal.

The most susceptible to damage are appliances like small refrigerators that require enough motor torque to get past the compressor torque humps. With low voltage in (such as a brown-out), the compressor can stall (reducing the motor back-EMF to zero) and thus draw a much higher current than usual, all of it converted to heat. As a bonus, any cooling fans will not operate at full efficiency, if at all.

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If you have a 240 VAC motor which is running at 90% efficiency and is putting out one horsepower into a load, then since one horsepower is 746 watts, it'll be consuming about 75 watts in order to deliver that 746 watts to the load.

That's about 820 watts, total, and since the input to the motor is 240VAC, it'll be drawing about 3.4 amps once it gets up to speed.

When it's just starting, however, it can easily draw ten times that current and dissipate it in the stator's winding resistance, so that power would be 240V * 34A = 8160 watts, and the stator's winding resistance would be 240V / 34A ~ 7 ohms.

Now, if you were to connect 120V to the motor and the static load on the shaft was high enough to keep the rotor from turning, then that 120V would see only the stator's 7 ohm winding resistance, and it would cause the statpr's winding to dissipate: P = E² / R = 120V² / 7R = 2057 watts!

Then, since the motor was designed, ostensibly, to rise a fixed temperature above ambient with 75 watts, steady state, being dissipated in the stator's winding, 2057 watts in it would certainly cause some damage after even a short time...

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I have a radial arm saw that was given to me by a friend. It started slow and did not have much power to cut ( it bogged down easily and the motor smelled hot) I was going to take it to have the motor windings re-wound and upon disassembly I discovered that it was a dual voltage motor(120-240V) that had been plugged into a 120v outlet but was tapped for 240V operation. After I moved the input wiring to the 120V tap and reassembled the motor it started much faster and ran with the expected cutting ability of the 3/4 hp it was rated.I no longer smelled "Hot". I know thios is not a technical explanation but a practical example for comparison. The 120V supplied to the 240V motor "Worked " sort-of, but, it did not allow the motor to perform at its peak efficiency or expected torque.

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Many electronic devices nowadays are rated for any voltage between 100 and 240V, for precisely this reason.

However, anecdotally, there are situations where too low a voltage can cause damage. A few years ago, I owned a particularly cheap phone, which came with a particularly cheap charger, which was only rated for 230V. When running on 115V, I found that the charger would not charge the phone, and indeed appeared to discharge it.

Given the low quality of both the phone and the charger, I suspect the charger simply transformed and rectified the voltage, which would mean the voltage dropped from 4.2V to 2.1V - too low to charge a Li-ion battery.

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I'm in the process of field testing a 277v single phase motor I got for nothing at 120v. I am using it as a whole house fan. It came with a squirrel cage fan and housing. I had planned to hook it up on a double pole switch feeding a three pole to rig it for either 240 or 120, but after hearing it run furiously at 240v, I decided to dial it back to just 120v.

The current draw from this motor at 120v is ~2.3A and at 240v it is ~ 4.5A. I have heard a lot of otherwise smart sounding people on here state that as voltage goes down, the amperage goes up. But I really think that a lot of people are forgetting that this is only the case when the output power remains constant. I'm sure that with the motor running slower, it will have more heating from trying to attain its design speed and always failing, but I think it'll hold in there.

So far, the motor has no problem getting to speed and doesn't seem to get too hot.

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I thought these articles would explain it better: http://www.marcspages.co.uk/pq/2410.htm http://www.hardwaresecrets.com/anatomy-of-switching-power-supplies /2/SMPS110/220V

A 230W power supply will require 2Arms at 115V, and 1Arms at 230V. The storage capacitor will be delivering such a current almost continually (except during the portions of the cycle when it is being "topped up" via the rectifiers).

From personally conducted tests, SMPSs die at about 60VDC or 45VAC - some are lower, some higher, this is an average (and also very dependent on hard the power supply is working at the time).

In some domestic equipment, which has a narrower voltage operating range, the higher current because of a lower operating voltage may start to strain the switching transistor either through the heat the insufficiently cooled transistor needs to dissapate, or simply the current approaches the maximum for which the transistor was designed. Either (or both) of these factors can drastically reduce the working life of the part from many years to merely a few weeks, or months at best.

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You answered the opposite question from the one the OP asked. – Peter Green Dec 17 '15 at 12:23

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