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A lot of (good) US stuff is available in Japan through international online sales (Amazon).

However, most equipment purchased in Japan from the US are shipped as-is, without any change, while the voltage is 120 V in the US and only 100 V in Japan.

I'm struggling to assess the impact, if any, of such a difference depending on the type of equipment. The question here is to help me make the right purchase (target being, if possible, no noticeable loss of power).

There are some subtle electrical concepts that I'm not familiar with. To begin with what I (think I) know: \$V = RI\$ and \$P=VI\$ (\$P\$ for Power), that gives the famous $$P=\dfrac{V^2}R$$

My question is, what between A and B below is correct (or something else)?

A) \$R\$ is constant, Power decreases?

In this case, if \$V\$ decreases but \$R\$ is constant, obviously the Power should decrease also, proportionally to the square of the difference, for instance in Japan that would be $$P_{jp} = \dfrac{100^2}{120^2}P_{us} \approx 0.69\text{ }P_{us} $$

So for instance an US 1440 W @ 12A electric kettle would only be 1000 W in Japan (same 12A).

B) \$P\$ is constant, Intensity increases? (\$R\$ varies)

In this case, \$P\$ being constant and \$V\$ being lower in Japan, the device would suck more Amperes to reach the same Power. The ratio is $$I_{jp} = \dfrac{V_{us}}{V_{jp}}I_{us} = 1.2\text{ }I_{us}$$

The US kettle example @ 12A would consume 14.4A in Japan, which is worryingly close to the 15A "limit" of many outlets here.

My question: for instance, are my A or B assumptions below correct depending on these kinds of equipment: (please add any other equipment that make sense)

  • Electric kettle (resistor => A?)
  • Oven / toaster (resistor => A?)
  • Food processor (motor => B?)
  • Tank water pump (motor? => B?)
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    \$\begingroup\$ A) If R is constant, decreasing V will also decrease I. \$\endgroup\$
    – Finbarr
    Commented Mar 22, 2022 at 8:25
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    \$\begingroup\$ B) P will only likely be constant for equipment that runs from power supplies that can adapt to the lower voltage, like laptops, phone chargers, TVs etc. \$\endgroup\$
    – Finbarr
    Commented Mar 22, 2022 at 8:27
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    \$\begingroup\$ First thing would be to check the manufacturer’s specs. For units with switchedmode power supplies, these will normally go down to 90V. These are constant wattage - less volts, more amps. Units with resistive heaters - you’ve worked that out - less Watts. Units with universal motors will spin a little slower. \$\endgroup\$
    – Kartman
    Commented Mar 22, 2022 at 8:30
  • \$\begingroup\$ @Kartman Believe me, I read the specs from A to V :) The problem is most stuff is rated 110~130V. "Units with universal motors will spin a little slower" ok, that makes sense if V is lower, but does it mean the power W gets lower as well? Or, in that motor case, does I increase, leading to more torque (and less speed) for a constant W? \$\endgroup\$
    – Déjà vu
    Commented Mar 22, 2022 at 8:42
  • \$\begingroup\$ If torque stays the same but rpm drops, then power drops. If you’re really concerned,just get yourself an auto transformer to boost the volts to 120. \$\endgroup\$
    – Kartman
    Commented Mar 22, 2022 at 10:39

1 Answer 1

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If the product is not rated for 100Vac e.g. 100~140 or 100~240 Vac, it is not guaranteed,

Resistive loads like kettles and ovens will draw less power not more because R is relatively constant ( except for the PTC effects)

\$P=V^2/R\$

Motors will have a no load RPM proportional to f and torque limited by Vac/Z. where Z(f) is also raised by back EMF proportional to RPM and eddy current losses.

Many appliances may use BLDC motors but label specs must be adhered to, or ignored with great caution for effects on performance or reliability.

You must use a specific example with datasheet if you expect an accurate answer. This is too vague.

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  • \$\begingroup\$ Sorry, not a pro here. What are f and the Z(f) function? Can you please explain the last sentence in a more Layman's terms way? \$\endgroup\$
    – Déjà vu
    Commented Mar 23, 2022 at 1:29
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    \$\begingroup\$ f(requency) and Impedance = Z(f) are terms we use to define load and effects from grid compatibility. in general terms. Current is never constant, but impedance can be at the same frequency and heaters are just resistive and motors will have less power not the same as they are only resistive on start surge \$\endgroup\$
    – D.A.S.
    Commented Mar 23, 2022 at 23:09
  • \$\begingroup\$ Thanks. Have 2 exemples, this grinder that's said to be 100~120V, 70W, and that brewer 120~140V, 1650W. Not sure the specs are detailed enough. \$\endgroup\$
    – Déjà vu
    Commented Mar 24, 2022 at 1:25
  • \$\begingroup\$ Grinder is 100-120VAC 50/60 Hz, 70W 0.8A perfect . . Brewer is not 120-~140 but 220–240 Volts No good \$\endgroup\$
    – D.A.S.
    Commented Mar 24, 2022 at 2:41
  • \$\begingroup\$ Switched from UK to US, but FSR it was still 220V :) It's this one actually which is 110~120 @ 1650W... This one is rated 110, but it's [just] a brewer... I assume the "logical" functions (display, CPU) will keep working @100, and it will warm more slowly, right? \$\endgroup\$
    – Déjà vu
    Commented Mar 24, 2022 at 3:02

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