Why kettles boil slowly in the US

I saw a tweet recently that intrigued me:








64 thoughts on “Why kettles boil slowly in the US

  1. I’ve been living in France for two years and have gotten addicted to my electric kettle! I would love to take it back to the US with me when I move in a couple of months–can I just use a regular plug adapter and plug it in (even if it means it will boil a bit more slowly)? It doesn’t look like it’s dual voltage, it says 220 – 240 V. Thanks!

  2. If you had a 240 volt outlet installed in your kitchen, you could use your French kettle. Just check that it doesn’t care about 50Hz versus 60Hz. You would need to replace the plug with a NEMA 6-15P plug to match the outlet. (Like a normal US plug, but both blades are horizontal.) Also, if it has a detachable cord, you could probably buy the correct cord with a US plug.

  3. “This is such a problem that many households in the US still use an old-fashioned stove-top kettle.”

    I don’t think slower boiling times is what makes Americans not use electric kettles; I mean, 237 seconds is still considerably faster than boiling water on the stove! Plus look at the case of Japan, which has even lower voltage than the US (100v, the lowest in the world according to this map?), yet electric kettles are in every home, office, and convenience store there. I think it’s the cultural difference of not being a tea-drinking society, rather than an extra minute or two of boiling time, that’s stunted the popularity of the electric kettle in America.

  4. It doesn’t necessarily follow that lower voltage implies lower power; if you halve the resistance of the element in the kettle, you’ll get twice the current and that will compensate for having half the voltage. However, the 15A limit could be a problem.

  5. (Actually, if you’re running your kettle at half the voltage, you’ll want an element with 1 quarter of the resistance. Then you’ll have the same power.)

  6. No, Bruce. You want the power to be high in a kettle, not low. The correct approach here is to use P = IV.

  7. I think both are correct – just stated differently.

    Bruce’s 2nd comment is correct in general. P=V^2/R. Two kettles at equal power require the kettle at half voltage to have one-quarter the element resistance.

    Mr. Reid is correct on P=VI in specific as current is the limitation in this case (as Bruce alluded to as 15A). USA and UK have similar electrical branch current ratings and the kettles operate near the rating so halving the voltage drops the power by half. If you quartered the resistance to maintain equal power the breaker or fuse would trip.

    I have a 13 kW tankless heater under my kitchen sink (USA). But it most certainly doesn’t plug into the wall outlet since it’s on a dedicated 240 V / 60 A circuit.

  8. Yes, Mr. Reid. For a given voltage, the only way to set the current is to set the resistance. V=IR, so I=V/R. As you said, P=IV, so P=V^2/R and R=V^2/P, i.e. you want a lower resistance to compensate for a lower voltage.

  9. The power dissipated in a resistor (which is doing the heating) is I²R. If you reduce the voltage you reduce the current. If you combine a reduced current with a reduced resistance, you are not going to get the same level of power dissipated and the same level of heating.

  10. A kettle that is designed for Europe (say 240V) and is rated at 2000Watts draws about 8.34 Amps, so therefore has an element of about 28.8 ohms.

    Now if you were to plug this kettle into a 120V outlet it would draw around 4.17 Amps with a power of rating of only 500 Watts.

    So it effectively will have only one quarter the heating capability.

    In order to bring the kettle back up to 2000 Watts on 120 Volts, the element would need to be 7.2 ohms and therefore draw 16.67 Amps.

    So if you halve the voltage and want to maintain the same POWER output, you need to reduce the resistance to one quarter of the original value.

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