I recently moved to Australia from the United States and brought a few stepdown transformers to make use of some appliances I didn't want to repurchase down under.

I have a 2000W transformer that I use for my rice cooker and vacuum and was wondering today if the electricity usage function is

kWh = (Hours used) * (Wattage of transformer)


kWh = (Hours used) * (Sum of wattages of appliances drawing power)

Or possibly neither? At prices up to $0.33 a kWh it's an important thing to figure out!

As well, most of my devices say they operate at 120V but my transformer outputs at 110V, will this cause problems?

  • \$\begingroup\$ No. It does consume magnitizing current when unloaded, which depending on your grid owner, cost or does not cost money. You will also have hysteresis losses in the core which costs money. Is it hot? \$\endgroup\$ – winny Apr 14 '18 at 13:29
  • \$\begingroup\$ Sum of wattages of all appliances, plus a few percent internal losses, plus a flat rate (probably in the 50-100W region) for its own magnetising current. \$\endgroup\$ – Brian Drummond Apr 14 '18 at 15:38

The power drawn by a transformer from the wall is equal to the power drawn by its load, plus the losses of the transformer itself.

With a well designed transformer, these losses will be a small fraction of rated throughput. Some losses (core hysteresis and core eddy currents) are constant regardless of transformer load, that is, they're dependent on the transformer input voltage. Copper losses are dependent on the transformer output current.

Generally, transformers are designed to be fairly efficient off-load, you would not expect losses to exceed 1% of rated load. Losses at full load may be several percent.


Assuming the transformer is ideal (it's not, but it's pretty darn close to ideal), then the power drawn through the transformer is equal to the power the transformer is drawing from the wall.

Therefore, if nothing is connected to the transformer, the transformer draws almost nothing from the wall.

It's the second one. To be pedantic, it's actually kWh = (hours used) * (sum of power draw) / (efficiency) where efficiency is a number between 0 and 1.


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