I was talking to someone about trying to dissipate the most heat from a metal crucible (essentially just a resistor $R$). He argued that you wanted the resistor to have a high resistance because $P=I^2R$, so $P\propto R$.

But I thought back to messing around with resistors and batteries, where I remember that if I attached a much lower resistance to a battery, it would begin to heat up much faster than a bigger resistor. This also makes sense in the context of the power law, because $P=IV$ and $I=V/R$, so $P=V^2/R$. So $P\propto 1/R$ and a smaller resistor dissipates more heat.

The conclusion I came to is that, in the first scenario, the machine in question supplies current and that current is driven across the resistor, so it's appropriate to use $I^2 R$. However, in the second scenario, $V$ is set and $I$ is determined by $R$ from there.

Is that correct?

  • $\begingroup$ Yes. But generally increasing R will decrease I by quite a lot, so you'll probably end up having to put more batteries just to make sure your I remains the same (for first case). $\endgroup$
    – t.c
    Commented Oct 14, 2014 at 17:38
  • $\begingroup$ You will draw the most power (and hence dissipate the most heat) when the resistor value matches the internal resistance of the source, as DanielSank states. This is a truth that any sound engineer knows, because it determines how loudspeakers should be matched to amplifiers. $\endgroup$
    – Hot Licks
    Commented Oct 15, 2014 at 3:26
  • $\begingroup$ @HotLicks: Interestingly, when dealing with high frequency systems the condition for no reflection at a load is that the impedance of the load should match the characteristic impedance of the incoming transmission line (this is just like matching dielectric constants in e.g. immersion microscopy). $\endgroup$
    – DanielSank
    Commented Oct 15, 2014 at 4:48
  • $\begingroup$ @DanielSank - Yes, the matching impedance rule pops up in multiple places. It's more surprising when it isn't true (though I can't think of a place where it isn't offhand). $\endgroup$
    – Hot Licks
    Commented Oct 15, 2014 at 11:55
  • $\begingroup$ There's actually a lot of good information on this at the following wikipedia entry: en.wikipedia.org/wiki/Electrical_resistance_and_conductance $\endgroup$
    – raddevus
    Commented Oct 16, 2014 at 11:42

3 Answers 3


It depends on the internal resistance of the source.

Fist consider a "voltage supply". What does "voltage supply" even mean? A voltage supply is supposed to output a fixed voltage no matter what we connect it to. Is this even possible? Suppose we connect the two terminals of the voltage supply together through a piece of wire, i.e. a really low resistance load resistor $R_L$. The current output should be $V/R_L$. If this is a 9 Volt battery and my resistor is 0.1 $\Omega$ then I'd have a current of 90 Amps. A 9 Volt battery most certainly cannot output 90 Amps.

The way to model the limitation in the battery's maximum output current is to imagine that it is an ideal voltage supply in series with an internal resistance $R_i$. Now when we connect it to a load resistor $R_L$, the total current is $I=V/(R_i + R_L)$. If $R_L\rightarrow 0$ then the current goes to $I_{\text{max}}=V/R_i$, a finite value. In other words, the internal resistance sets a maximum output current.

Since the total current is $I=V/(R_i + R_L)$, the total power dissipated in the load resistor is

$$ P_L = I^2 R_L = V^2 \frac{R_L}{\left(R_i + R_L\right)^2}. $$

In order to find the value $R_L^*$ for which the power is maximized, differentiate with respect to $R_L$ and set that derivative equal to 0:

$$ \begin{align} \frac{dP_L}{dR_L} &= V^2 \frac{(R_i + R_L)^2 - 2 R_L(R_i + R_L)}{(R_i + R_L)^4}\\ 0 &= V^2 \frac{(R_i + R_L^*)^2 - 2 R_L^*(R_i + R_L^*)}{(R_i + R_L^*)^4} \\ 2 R_L^* &= R_i + R_L^* \\ R_L^* &= R_i \, . \end{align} $$

This is the result to remember: the power dissipated in the load is maximized when the load resistance is matched to the source's own internal resistance.

Now, any circuit you would reasonably call a "voltage source" must have a low internal resistance compared to typical load resistances. If it didn't then the voltage accross the load would depend on the load resistance, which would mean your source isn't doing a good job of being a fixed voltage source. So, because "voltage sources" have low output resistance, and because we showed that the power is maximized when the load resistance matches the source resistance, you will observe that the load gets hotter if it's low resistance. This is why you found that with batteries, which are designed to be voltage sources, the lower resistance loads got hotter.

Current sources are the other way around. They're designed for high internal resistance so you get a hotter load for a higher load resistance.

But in general, you get more power in the load if it's matched to the source.

  • $\begingroup$ As a consequence, the voltage source gets undesirably hot as well (it dissipates the same power at the optimum). $\endgroup$ Commented Oct 15, 2014 at 10:25

It depends on if your power supply is constant voltage or constant current. Usually, it's the former, so it means that the $P = V^2/R$ is the more appropriate one to use. Therefore, a smaller resistor will dissipate more power in this situation.

In some situations (electromagnets is one that comes to mind), the load is driven with constant current, so larger resistors dissipate more power.

  • 1
    $\begingroup$ I think you mean 'smaller resistor'. $\endgroup$ Commented Oct 14, 2014 at 17:38
  • $\begingroup$ Sorry, I was editing at the same time ;) $\endgroup$
    – Gremlin
    Commented Oct 14, 2014 at 17:38
  • $\begingroup$ This is correct. See my answer for more details on this idea. $\endgroup$
    – DanielSank
    Commented Oct 14, 2014 at 18:48

You are right, as is Eoin's answer. I'm only answering to show one way to think about it that is useful to me when people bring up this common misconception.

Imagine you have a pair of terminals with some voltage between them (like a common 1.5V battery). Nothing's connecting those two terminals together. Except for air, that is. Is there any current? No, so air's like an infinite resistance. Is the air heating up? Clearly not.

So, there: infinite resistance $\rightarrow$ zero power.

Sure, I'm talking constant voltage here, which seems to be the norm in casual discussions.

  • $\begingroup$ (Yeah, try this experiment with constant curretn - but don't try it at home) $\endgroup$ Commented Oct 15, 2014 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.