I was writing an answer to a very similar question that was closed and merged in this one, so excuse me if I repeat the other answers. The particularity of this other answer is that it called for an analogy of heat dissipation in the electric circuit with the energy in a spring and the question described the situation in terms of difficulty of passing the resistance.
Here it goes.
This is a simple question, but even though people don't like answering simple questions, it always comes to mind that Richard Feynman would have answered it!
The main issue here will be my pet peeve with a phrase of yours: "to cross the first resistance in the second scenario less energy was required compared to the first".
No, not at all. It's not the energy required to pass through the second resistor, it's the energy lost in this resistor. The difference is enormous from the point of view of thermodynamics. Losing energy is always easy. Since we agreed that energy is lost, let's see what you did. You had a nine-volt battery, with great electrical potential energy. By connecting the poles, with some resistance, you will gradually lose this energy. As for your spring analogy, it's as if you had the spring in the figure below:
The potential energy of this spring is visually clear in the suspended weight. If you think that this weight could be a bucket of sand, you can take sand from it and little by little use (in your favorite way) the potential energy of the sand, throwing it on the ground. The spring relaxes little by little, losing its potential.
This bucket of sand attached to the spring, however, is an ideal that the battery cannot achieve, as there is always a finite resistance between the poles, even if enormous. It's like there's a small hole at the bottom of the bucket. If its resistance is considerable, we can think of a considerable hole in the bucket. In fact, the figure I am suggesting is well represented by the sand clock below, which is a beautiful analogy, as it highlights the relationship between entropy (degradation of energy) with the passage of time.
Its multiple resistances in series could also be seen as a multi-stage sand clock, with the sand falling into a sphere below which, in turn, also has a hole and causes the sand to fall into another container below. Unfortunately, I didn't find a ready-made figure of a "composite sand clock", but I'm sure you get the idea.
Now, with this schematic in mind, we can go back to thinking about your electrical circuit.
Your thought process is delving into the behavior of electrical circuits and the nature of electron flow through resistors. Let's break down your questions and ideas to address the confusion.
Series Circuit and Voltage Drops
When you connect resistors in series, the total voltage supplied by the battery is divided among the resistors based on their resistances. Ohm's Law and Kirchhoff's Voltage Law help explain this:
- Ohm's Law: $ V = IR $
- Kirchhoff's Voltage Law: The sum of the voltage drops across each component in a series circuit equals the total supplied voltage.
If you have a 9V battery and two resistors $ x $ and $ y $ in series, the total resistance $ R_{total} $ is $ x + y $. The current $ I $ through the circuit is:
$$ I = \frac{V_{total}}{R_{total}} = \frac{9V}{x + y} $$
Voltage Drops
The voltage drop $ V_x $ across resistor $ x $ is:
$$ V_x = I \cdot x = \frac{9V}{x + y} \cdot x $$
Similarly, the voltage drop $ V_y $ across resistor $ y $ is:
$$ V_y = I \cdot y = \frac{9V}{x + y} \cdot y $$
Energy Consumption in Resistors
In the scenario where $ y \gg x $:
- The total resistance $ x + y \approx y $.
- The current $ I $ is approximately $ \frac{9V}{y} $.
- The voltage drop across $ x $ is $ V_x \approx \frac{9V}{y} \cdot x $.
Since $ y $ is much larger than $ x $, $ V_x $ becomes very small. Most of the voltage drop (and thus energy dissipation) occurs across the larger resistor $ y $.
Electron Behavior, Energy Dissipation, eletron decision, and Maxwell Demon.
Electrons moving through a resistor experience a potential difference (voltage drop) which causes them to lose energy. This energy loss is what powers devices or is dissipated as heat. The electrons don't "decide" to lose energy; they follow the path dictated by the electric field and the resistive properties of the materials they travel through.
In fact, the term "decide" is curious. For those who understand that the world is explained by the laws of nature, there is no decision, neither by these electrons nor by us humans, everything happens so that entropy is produced and, especially, so that it is produced at the slowest possible pace. , with as little "action" as possible, from the point of view of Lagrangian physics, but that's another story.
Getting back to the resistors:
- In the first scenario (single resistor $ x $): The entire 9V is dropped across $ x $. The electrons lose all their energy (from the battery) across this resistor.
- In the second scenario (resistors $ x $ and $ y $ in series, with $ y \gg x $): The voltage drop across $ x $ is minimal, and the majority of the voltage drop occurs across $ y $. The electrons still lose energy according to the resistive properties they encounter.
Stability and Energy Loss Analogy
Your analogy to a spring is insightful. Systems tend to move towards lower energy states for stability. In electrical circuits, electrons move in response to the electric field established by the potential difference (voltage). The resistors impede this flow, causing a voltage drop (energy loss) which is analogous to how a spring releases stored elastic energy when it is compressed or stretched.
Electrons "prefer" to lose energy in a resistor because:
- The resistive material converts the kinetic energy of the electrons into heat (or other forms of energy).
- This energy loss is a consequence of the physical properties of the resistive material.
