Does high resistance imply low power consumption in household appliances? As we know that, the voltage in households are fixed, so for  a particular appliance, will increasing the resistance decrease the power consumption  rate according to $P=V^2/R$ ?
 A: Each alternating current electrical device is designed for a certain power requirement.  Because the device is running off of alternating current, there is more involved than Ohm's Law.  Resistance for alternating current comes from "classical" resistance, and from inductance, whereby the alternating magnetic field of an inductor resists changes in electrical current.  This means that the engineers who design these devices use both resistance and inductance to achieve the power requirement that they want.
A: An appliance is not a resistor.   There are three important ideal linear components in electric circuits, which are the linear-response resistor, the current-lag response inductor, and the current-lead response capacitor.   To dissipate power in a circuit with only those components, the resistor is the essential
element.
The dissipation of power by an appliance, though, can include transformation
of that power into light, or radio, or chemical reactions, or lifting weights,
which are NOT resistor-like power dissipation (with the exception of incandescent  lamps).   So, while the power company will charge you for the
power used, just as they would for a resistor, the actual character of the power usage cannot be inferred from any resistor component, or measure of simple resistance.
Indeed, an electrical generator has some internal (wire windings) resistance, but
it has an actual NEGATIVE 'power consumption rate' by electrical
measurement.  The  internal resistance is not  negative, however.
A: Yes, that should be the case if the power supply was constant. But remember, your appliance needs to work ! So, if you increase the resistance, the appliance will simply 'draw' more power, and the consumption would go up.
$ P = V^2/R $
However,
$P = I^2R$
According to this formula, power should increase.
The point is,you need to look at the system more carefully. Measure, voltage, current and the resistance.
