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I've this doubt regarding resistor: suppose I make a simple circuit with a power generator of 3 Volts and 1 Ampere and a resistor of $10 k\Omega$ I know from Ohm law that the current flow $i$ is:

$$I=\frac{V}{R}=\frac{3 V }{10000 \Omega}=0.3mA$$

So if I place my multimeter in series after the resistor I can measure exactly that current , and if I place the probe in parallel I can measure $3V$ so in this case I think that the resistor regulate the charge flow to 0.3 mA but in fact is the whole circuit to dissipate this current after the resistor regulated it, because I can see it passing through the multimeter ( and indeed 0.7 Amp give by the generator are not dissipated).

Shorting out the circuit with no resistance attached I can't measure an increase of temperature (I think because I'm not working with high power) but if a use a smaller resistor (e.g. $2 \Omega$) I can fell it getting hot so I infer that a smaller resistor dissipate more energy (in this case I've $1 Amp$ passing through it) but I think that this is no more a good charge regulator because it dissipate a lot of energy through Joule effect. So I don't know if exists materials that can be a good "charge regulator" (e.g. a good resistor) and material that can be good "energy dissipator" (e.g. good heater). I've tried to make my homemade analogous resistor cutting $3,5 cm$ of a kanthal wire with a resistance of $57 \Omega / m$ but I can't measure correctly if this dissipate better the energy. Can anyone tell me what really do a resistor?

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    $\begingroup$ Just touch an underrated resistor and you'll know :-) $\endgroup$
    – Carl Witthoft
    Commented Mar 16, 2014 at 15:31
  • $\begingroup$ Indeed, from personal experience I know that one can get burned this way. The mechanism that causes a resistor to heat up is the inelastic collisions that the conduction electrons that move through the material undergo with the constituants of the material. Thus energy is transferred to the material resulting in its heating up. $\endgroup$
    – Urgje
    Commented Mar 16, 2014 at 15:51
  • $\begingroup$ @Urgje thanks what I don't understand if an ideal resistor can act like a pure current flow modulator without dissipating energy with heat... $\endgroup$
    – G M
    Commented Mar 16, 2014 at 16:03
  • $\begingroup$ The very nature of a resistor causes it to dissipate energy in the form of heat when attached to a power source. But if you connect a device to a power source through a resistor you can regulate the current through the device this way. However, semi-conductor based current regulators do a much better job. $\endgroup$
    – Urgje
    Commented Mar 16, 2014 at 16:15
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    $\begingroup$ Yes, that sounds strange but it is clear that the smaller the resistance the larger current flows. This means that more inelastic electron collisions take place, i.e., more energy is dissipated. $\endgroup$
    – Urgje
    Commented Mar 17, 2014 at 8:52

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The heat generated is the wattage dissipated, namely $W = VI$, so if the resistance is lower, the current will be higher, and if the voltage remains the same, you get more heat.

If you short out the battery (provide a very low-value resistor) then the other resistance in the circuit will take the heat, namely the battery itself, and the voltage across the short circuit will be zero. This is a good way to burn out a battery.

EDIT: OK, you're saying it is counter-intuitive that more resistance means less heat. Let me try to explain it. First, let's assume the voltage source has very low internal resistance compared to the resistor you are experimenting with, like, say, a 12-volt car battery.

Now, you put your resistor R of, say, 1 ohm across the poles of the car battery. So 12 amps will flow, so 12*12 watts of heat come off. 144 watts, that's a lot of heat. (In fact, you should probably be using big light bulbs instead of little resistors.)

Now, you take two resistors R and tie them in parallel. How much resistance do they have? R/2, because 12 amps will flow through each one, for 24 amps total. (R = V/I = 12/24 = 1/2 ohm) Each resistor sees 12 volts and 12 amps for 144 watts, but what does the pair of them see? 144 watts in each one, so 288 watts together. So by cutting the resistance in half, you doubled the heat (assuming the source has low internal resistance).

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  • $\begingroup$ Thanks a lot for the answer, what I don't understand is if the wattage dissipated can be grater in a determinate type of resistor with the same resistance. Can a resistor act only modulating the charge flow without dissipating energy? $\endgroup$
    – G M
    Commented Mar 16, 2014 at 16:09
  • $\begingroup$ @GM: No, because in any moment in which there is a voltage across the resistor and a current flowing through it, energy is lost. A resistor will lose it through heat. Something like a motor will lose it through mechanical work. A capacitor or inductor will lose it by building up energy in its field. For a resistor, it will generate heat - there's no other way for it to behave. $\endgroup$
    – Mike Dunlavey
    Commented Mar 16, 2014 at 20:48
  • $\begingroup$ Ok, but when I add the resistor with a greater resistance there is less dissipation but the difference from input and output current (i) is greater so the effectiveness of the resistor seems not linked with the dissipation itself. In other words the dissipation of energy seems an effect but not the way the the resistor uses to regulate the charge flow. Thanks for the help for the moment I upvote you I will accept a more detailed answer! $\endgroup$
    – G M
    Commented Mar 16, 2014 at 21:16

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