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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 $P=V^2/R$ ?

green_32
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3 Answers3

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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.

David White
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  • Thanks for the answer, but is $R=\rho L/A$ sufficient for comparing the power consumption rates of same appliance but having difference in only one of the parameters($\rho,L,,A$) , will the resistance due to inductance be the same for the appliance for a change in one of these parameters ? – green_32 May 29 '21 at 23:45
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    @green_32, inductance is largely independent of Ohmic resistance, so the particular design choice will determine the answer. – David White May 30 '21 at 15:51
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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.

Whit3rd
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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.

Nakshatra Gangopadhay
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  • If you increase resistance, the device will not draw more power. You can verify this by measuring and comparing the resistance of a 40W incandescent light bulb with a 100 W incandescent light bulb (if you can still find them). – David White May 29 '21 at 23:12
  • thanks for the answer ,isnt the household voltage supply fixed, how did you make the current constant? – green_32 May 29 '21 at 23:53
  • @green_32, an incandescent light bulb operates on resistance only, so it follows Ohm's law. This means that current is not the same for bulbs of different resistance, and in fact, if you put an oscilloscope in the circuit of that bulb, you will find that the current follows a sinusoid because the supply voltage is sinusoidal. This means that the current is NOT constant and the current is NOT the same for bulbs of different resistance. – David White May 30 '21 at 15:56