Consider a simple series circuit with 2 light bulbs. We know that there will be a voltage drop after the current passes through the first light bulb, but the current remains the same. If we were to observe the electrons traveling the circuit before and after they reach the first light bulb, how would they differ if not by the rate they are traveling? It is obvious they have more energy before reaching the first light bulb than they do after, but what does this energy difference look like on a subatomic scale?
[Physics] How does a voltage increase/decrease affect the electric potential energy of a charge
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Related Solutions
Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric field and the nature of the material they move in.
When you hook up a circuit of some kind to a voltage source (battery, generator, wall-wart, whatever), the electric field already present between the terminal of your power supply causes some electrons in the wires to move around. In the course of doing that the electric field gets re-arranged to point along the wires and through the components and so on. There is rather a lot of shuffling that goes on immediately after the power supply is hooked up and I am going to mostly ignore it to focus on what happens when a (short-term) steady-state is established.
Circuit basics
Observe the circuit in its operating state: the electric field points along the wires and though components in various way. In some places that field $E$ is weak and in some places it is strong, and in some places the electrons flow fast and in others slow, but there are two rules that must be obeyed:1
The current (number of electrons passing a point) is the same throughout the circuit. This follows because I have restricted out consideration to time when things aren't changing, and if more were passing point A than point B (a little further down the circuit) electrons would be piling up in the space in between them.
The total voltage change around the circuit has to be zero. This is because voltage is a function and can have only one value at any point in space, so if I follow any path that comes back to itself the changes has to equal zero when it returns.2
These rules are written in terms of voltage and current, but previously I was talking about electric fields, so what's the relationship between them?
The current comes into play in the form of Ohm's Law: $V = I R$.
The potential change in a section of circuit with length $d$ and constant electric field $E$ is $\Delta V = E d$, so we can write the voltage rule as $ 0 = V_{ps} - \sum V_i = V_{ps} - \sum_i E_i d_i$ where $V_{ps}$ represents the voltage gain of the power supply.3 Rearranges this gives us $$ V_{ps} = \sum_i V_i = \sum_i E_i d_i \,.$$
One last thing before we're ready to answer the question: the electric field in the wires is usually assumed to be very small compared to the electric field in other things like resistors. Therefore, we can ignore the $Ed$ contributions from the wires in working the maths. This isn't true for very long wires or for very fine wires under low voltages, but we're going with it anyway.
How do the electrons "know"
Consider a very simple circuit with a switch in it. A resistor (numbered 1) is connected directly to the battery and to the input of the switch. From the switch the current gets back to the battery either directly or through a second resistor (numbered 2).
The circuit starts with the switch set so that only one resistor is involved. When we hook it up, the fields rearrange themselves such that we have very weak fields in the wires and a very strong field in the resistor: $V_1 = E_1 d_1 = V_{ps}$. To make the current rule work, we have a lot of electrons moving slowly in the wires and a few electrons moving very quickly in the resistor (think of car flowing).
$t = 0$ The original state of the circuit has a field in resistor 1 $E_1 = V_{ps}/d_1$ and very weak fields everywhere in the wires. There is no build up of electrons anywhere and the current flow is steady throughout.
$t = 0 + \epsilon$ The switch has changed state, but the electric fields have not re-arranged yet, so there is zero field in resistor 2. Electrons continue to move through resistor 1 at the same rate as before, when they get through it there is no field to move them through resistor 2. They begin to pile up between resistors 1 and 2. As they do that they begin to reduce the field in resistor 1 and increase it in resistor 2.
$t = 0 + (2\epsilon)$ Now there is a little field in resistor 2 and a little less in resistor 1. Current has started to flow through resistor 2 but there is still less than is flowing through resistor 1. More charge is building up between them and that is driving the field in 2 up more and the field in one down more.
$t = 0 + (\text{several }\epsilon)$ The field in resistor 2 has risen and the field in resistor 1 has dropped until they are almost matched. Current flow through the 2 resistors is almost the same with only a small amount more coming through resistor 1. The charge between them has almost, but not quite stopped changing and that means the fields in them are also almost fixed.
$t = 0 + (\text{many }\epsilon)$ The field in resistor 2 has risen high enough that it's current matches that in resistor 1. This represents the new current of the circuit as a whole and is lower than the original current.
What we learn from this consideration is that any time the flow of electrons is faster through one part of the circuit than another, electrons will accumulate in such a way as to re-distribute the electric field in the circuit so that the flow is more even than before, and that this process happens continually until the flow becomes uniform throughout the circuit. The strength of the electric field is also related to the voltage change over each component and will be adjusted until the total is equal to the supplied voltage.
1 Rules written in a form that applies only to series circuits. For a more complete version, look up Kirchoff's Laws.
2 This is true when you neglect magnetic induction.
3 I am assuming that there is only one power supply. The full treatment of Kirchoff's laws can relax this restriction.
I was just wondering what happens in a circuit in terms of different types of energy transformations.
You can think of this just like gravity. See below.
If you apply a voltage to a circuit then electrons start moving (very slowly).
Voltage is a name for electrical potential difference.
- There is a gravitational potential energy difference between the floor and a higher shelf, because a book wants to fall down, if it could.
- There is an electrical potential energy difference between two points in a circuit, if the charge wants to move to another point.
But the electrons are also now moving once the voltage is applied as there is a current so would they have kinetic energy as well?
Yes.
- A book falling from a high shelf is trading the gravitational potential energy into kinetic energy.
- A charge moving because a voltage is suddenly applied, is trading the electrical potential energy into kinetic energy.
Some information I have found online says that electrons collide with atoms in a bulb and this is why the filament heats up. But this would imply that kinetic energy is being transformed into heat [...]
Yes.
- Drop a steel ball into a fluid, and the fluid resistance because of viscosity (because of "bumbing" into many water particles) with pull out some of the kinetic energy, slowing it down. This kinetic energy loss is indeed converted into heat.
- A charge "bumbing" into atoms similarly are slowed down, and the loss in kinetic energy is converted into heat.
[...] and this can't be correct because surely any change in kinetic energy would alter the current flowing?
It does alter the current flowing. Without the resistance, the current would be much, much higher.
- The steel ball falling through the water is slowed down and soon reaches a constant speed (terminal speed).
- The charge moving through a resisting material is slowed down and soon reaches a new equilibrium speed, where the "push" that makes it move (the voltage) balances out the resistance that holds it back.
The end-result is indeed a slower flow, i.e. a lower current, than without this resistor present (in that particular part of the circuit).
Also, in a series circuit, if you measure the potential difference between any 2 points after all of the loads then you always get zero. I was just wondering why it is zero because don't the electrons keep moving even after passing through all the load in order to get back to the power source so surely they cannot do this without some form of energy?
Remember Newton's 1st law. If charges exit the last resistor, they are no longer slowed further down, no. But the speed they came out with will stay. If nothing prevents them along the wire alone, they will not speed up, no, but also not slow down. So they continue.
And should one be stopped by whatever reason, the next charge will come and push it forward (like-charges repel). This is the case when the wire is not a completely perfect conductor but has a little resistance.
Best Answer
There is absolutely no difference between the electrons. Every electron is identical. The difference is in the entire system of charges.
Imagine a clock that you can run by allowing a weight to fall. What is the difference between the weight when it is raised to the top and when it has fallen to the bottom? The weight itself is identical. The difference is that at the top, the weight is higher in a gravitational field. This position requires energy to reach compared to the floor.
Similarly, the circuit has an electric field and charges have different amounts of energy due to their position in the field.