My new mini refrigerator is plugged into a 50 ft extension cord that is plugged into my house. Does it cost more to run because of the cord?

  • 2
    Yes - you're losing some power in the cord. But it's probably negligible depending on how much power your refrigerator uses, and how big the wires are in the cord.
    – Johnny
    Commented May 4, 2016 at 2:22
  • 3
    @DanielGriscom The voltage is only the same at the other end with no load, every (real world) wire has resistance, so when it carries current it will have a voltage drop. But, like I said it may be negligible. For example, with a very lightweight 18 gauge cord and a refrigerator that draws 5 amps when in use, the voltage drop would be around 3V, so it'd be losing around 15W in the cord. (or around 2.5% of the power used by the 'fridge). For a more significant load, say a 12A hair dryer, the voltage drop would be around 7V.
    – Johnny
    Commented May 4, 2016 at 2:46
  • 3
    @Daniel, if some power is lost in the cord, the refrigerator will run longer or cycle more often to achieve the same longer term cooling effect. Commented May 4, 2016 at 13:53
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    @Daniel: moving heat from a 5C space to a 20C space takes n Joules per hour. Heating a 50m cord takes m Joules per hours more than a 2m cord does. m + n > n. QED. Commented May 4, 2016 at 14:05
  • 1
    Mandy, I would just put my beer in it and not worry about it. The amount of difference probably won't buy you a beer per year. (Hey I'm a poet). 😉
    – ArchonOSX
    Commented May 4, 2016 at 20:11

4 Answers 4


Yes, it does cost you money using the extension cord.

Grabbing a random number out of the air, let's say the fridge draws 2 amperes. If you have a 50 ft. 16 AWG cord, that's 1.996 watts. If the cord was 50 ft. 14 AWG, that'd be 1.256 watts.

Using the energy star standard of $0.12 per KWh. The 16 AWG cord costs you $0.00023952 per hour, while the 14 AWG cord would cost $0.00015072 per hour.

There's also the problem that the manufacturer's documentation likely tells you not to use an extension cord.

- I have no idea how much current a mini fridge draws, so I simply grabbed a number out of the air. If somebody wants to comment with a realistic number, I'll rerun the calculations.

Using @CarlWitthoft's calculation of 0.7 amperes, and resistance values from NEC chapter 9 table 8. The cost can be figured as follows.

18 AWG cord

VD = Length x 2 x current x ohms per foot
VD = 50' x 2 x 0.7 A x 0.00795 ohms
VD = 100 x 0.7 A x 0.00795 ohms
VD = 70 x 0.00795 ohms
VD = 0.5565 volts

Watts = Volts x Amperes
W = 0.5565 V x 0.7 A
W = 0.38955 watts
kW = 0.00038955

Cost = kW x $/kW
Cost = 0.00038955 x $0.12
Cost = $0.000046746

16 AWG cord

VD = 70 x 0.00499 ohms
VD = 0.3493 volts

W = 0.3493 V x 0.7 A
kW = 0.00024451

Cost = 0.0002445 kW x $0.12
Cost = $0.0000293412

14 AWG cord

VD = 70 x 0.00314 ohms
VD = 0.2198 volts

W = 0.2198 V x 0.7 A
kW = 0.00015386
Cost = 0.00015386 kW x $0.12
Cost = $0.0000184632

So for every hour the fridge is actually running, you'll pay the above amount extra because of the cord.

If you're using the 18 AWG cord, and the fridge runs constantly for a year. The power lost to the cord would only cost about $0.40.

$0.000046746 x 24 hours = $0.001121904
$0.001121904 x 365 days = $0.40949496

So basically, reading this ridiculous answer has likely cost you more than the cord will.

  • "Per hour" - of the compressor running, so an accurate calculation requires knowledge of the duty cycle of the compressor in your conditions.
    – TomG
    Commented May 4, 2016 at 11:55
  • Right, because you only have voltage drop during the duty cycle. 10A@10% will be 25 times the loss than 2A@50%. Since loss in watts is proportional to current squared. Commented May 5, 2016 at 0:12

Pulling some stats from advertised products:

Gauge Conductors: 16/3 SJO Cord Length: 50 ft. Color: Black Watts: 1625 watts OSHA Listed: Yes Plug Type: Grounded Number of Outlets: 1 Maximum Amperage: 13 amps

AWG rating: 4.016 ohms/1000 ft , or 0.4016 ohms for your 50-footer, round trip. Using I^2R, which isn't exactly correct for AC due to phase shifting, you'd get 0.4016 watts per amp drawn.

Looking at ratings page found somewhere:

Compact refrigerators, typically ranging from 1.7 to 4.4 cubic feet, receive an Energy Star rating when they consume no more than 239.42 kilowatt-hours per year for manual defrost versions, or up to 318.4 kWH/year for units with partial automatic defrost.

So let's pick a conservative 365 kW-h/yr because that's a nice clean 1 kW-h/day :-), and at a conservative (IMHO) 50% duty cycle that's 1kWh/(24/2) = 83.3 watts when running. 83 watts --> roughly 0.7 amps @ 120 V.

All of which means I'd cut Tester101's conclusions significantly both because I think he overestimated the extension cord's power loss and because I grabbed a different estimate of the power drawn by the fridge. YMMV


No. It's a poor way to hook up a permanent load, but it does not waste anything like "as much power as the fridge uses" - it might cost you in life of the fridge, due to low voltage when the compressor is starting making it fail sooner than it should - but that would depend in part on the gauge (size) of the wire in the cord.

  • If it is a heavy duty cord it may not cause a low voltage problem but most are not.+ for the answer.
    – Ed Beal
    Commented May 4, 2016 at 18:17

The very best way to answer your question is to purchase an energy measuring device and compare power consumption at the house end with the refrigerator connected directly at the house and then connect the refrigerator at the end of the long cord. Keep the measuring device in the same location for both of the comparison measurements.

One device you could consider is the Killawatt meter that can be found online or at many hardware stores. It can show voltage, current and kilowatts consumed.

  • You'll need either two of these devices or two very long test periods to get a reliable result. We're looking for maybe a 1% difference. Commented May 4, 2016 at 19:02
  • At least 24 hours per test. If the difference is hard to detect, that is not a detection failure, it is the answer. Commented May 5, 2016 at 0:07

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