What is the microscopic explanation for why is there more power in a circuit with a smaller resistance?

Question

According to the Drude Model of electron flow in a circuit, the drift velocity is inversely proportional to the resistivity.

$$E=\rho J$$ $$E=\rho Nev_d$$ $$E=\frac{-\rho Ne^2E}{m}\tau $$ $$\rho=\frac{m}{Ne^2\tau}$$

As we can see, a material with a higher resistivity has a smaller relaxation time, and thus a smaller drift velocity, as $\tau \frac{-Ee}{m} = v_d$.

My question is, why is there more power dissipated in a circuit with a smaller resistance? A smaller resistance means there is a higher relaxation time. A higher relaxation time means the electrons collide less frequently with the lattice structure. Shouldn't this mean less energy transferred per unit time since the frequency of collisions must be somehow proportional to the rate of energy transfer?

I feel that I am close to finding an answer, but I cannot fully explain it: In a resistor (compared to a wire), the resistivity is high and the relaxation time is low and so is the drift velocity. Now, because the relaxation time is low, the frequency of collisions must be high. So a lot of collisions occur, but they occur when the electrons didn't really have much time to accelerate. Whereas in a wire, the electrons are moving so much faster, and while they don't collide frequently, a-lot of energy is transferred to the lattice ions as the electrons themselves have so much speed and kinetic energy varies with the square of the speed. In other words, the speed effect outplays the decreased frequency of collisions effect. How can we show that mathematically, if that is the correct line of reasoning?

Another potential explanation is that the relaxation time in the copper wire (relative to a resistor) actually decreases but N, the charge carrier density increases by a greater factor, and thus the total resistivity decreases even though the relaxation time decreased. As the relaxation time decreased we can use this to reason why the power is greater (more frequent collisions).

Answer 1

First, you have to look at your starting assumptions. It is not generally true that there is more power in a circuit with a smaller resistance. So we need to examine the circumstances where it is true and where it is not true to see if we can discover any hints.

It is not true that there is more power in a circuit with a smaller resistance when the two circuits are supplied by identical current sources. In such a case the circuit with the smaller resistance has a lower power. The current is the same in both cases, but the voltage is lower in the low-resistance circuit, and hence macroscopically $P=IV$ leads to lower power.

It is true that there is more power in a circuit with a smaller resistance when the two circuits are supplied by identical voltage sources. In such a case the circuit with the smaller resistance has a higher power. The voltage is the same in both cases but the current is higher in the low resistance circuit, and hence macroscopically $P=IV$ leads to higher power.

So, the hint that we obtain is that we must focus on the current. The condition we discovered is that it is only true that the low resistance circuit provides additional power when comparing two circuits with the same fixed voltage and different current. So your question becomes how does a reduced resistivity microscopically lead to increased current for a circuit supplied by a voltage source?

Since you were focused on relaxation time in your question, let’s hypothetically fix everything else and vary only the relaxation time. A lower resistivity corresponds to a larger relaxation time. Since the circuit is using a voltage source the $E$ is fixed in $\tau \frac{-Ee}{m}=v_d$ and therefore increasing relaxation time leads to an increased drift velocity. Since we are keeping everything else constant in $I=JA=A N e v_d$ the increased drift velocity leads directly to increased current and hence the increased power.

Note that this microscopic explanation is not a general explanation. It only applies to situations where the only difference is the relaxation time, which is never the case in practice. Resistance could also be changed by increasing the wire’s cross sectional area or charge carrier density. Any or all of those would lead to an increased $I$ and hence increased power. In this case the microscopic explanation tends to obfuscate rather than clarify. There is simply no one microscopic explanation which is correct in all cases. The macroscopic explanation is generally correct and hence preferable. Any microscopic feature leading to greater current produces the same result. This is a common criticism I have of the Drude model. It complicates and obfuscates more than it clarifies. In my opinion, this topic is better addressed at the purely macroscopic level.

Answer 2

You wonder: why is there more power dissipated in a circuit with a smaller resistance?

My question: why do you wonder?

To simplify, just look at a simple wired circuit of a power source and a load resistor. Using Ohm's law, you can derive P = V^2 / R, or P = I^2 * R. So it depends on the properies of the power source, if it is a constant voltage source, power increases with a smaller resistance, but if is a constant current source, power would decrease.

More in detail, when modeling the power source as a voltage source with an added internal resistance Ri, then the output power is maximum when the load resistance matches the Ri. The, the output power decreases with smaller and with larger values for the load resistance.