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?
Best Answer
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.