From P = V^2/R it’s clear that when the battery is fed through a voltage doubler before reaching a resistor, it outputs four times as much power as when it’s not passed through the voltage doubler. This is also demonstrated by the fact that with the voltage doubler, the resistor spins twice as fast, but the battery spins four times as fast. Clearly then, the effective resistance at the input of the voltage doubler must be one fourth of the resistance of the resistor connected to the output. This feels really unintuitive though. Reasoning in terms of mechanical advantage, it really feels like the resistance should be halved, not quartered. Anyone have an intuitive explanation for this?

Hey @AverageHuman, your intuition is correct, the effective resistance is halved. This is where the junctions get a bit funny and not fully analogous with real electronics. You have to think of each output of the junction as independent circuits, each driven by the same voltage. This is achieved by the junction acting as a 2:1 mechanical advantage when configured in the voltage doubler setup. In the voltage doubler, you are driving a single resistor with both circuits, so the effective resistance is 1/2 (or the voltage is doubled, same difference in the eyes of current). Take a look at this:

You can make an equivalent circuit to the voltage doubler using two resistors of 1/2 the value of the resistor in the single resistor voltage doubler; everything rotates at the same speed and you see double the voltage across the resistor(s). If you look at a single resistor that is 1/2 the value of the resistor in the voltage doubler, you see that the resistor turns at the same rate as all the others, but the battery turning at half the speed. With the voltage doubler, you are effectively making two independent circuits with 1/2 the resistance, but because you are only using a single battery, it has to put out 2x the power.

Here is a mechanical explanation of the quadrupling of the power output:

since there is a 2:1 gearing the force applied on the resistor is twice as high thus it seems possible that the resistor spins twice as fast (compared to no gearing).

Now since there is a 2:1 gearing between the resistor and the battery the battery has to run twice as many revolutions as the resistor and therefore the battery runs a factor 2*2 faster.

Hope it makes sense.