Please help understand Electric Potential Loss in a DC circuit What is meant by "Electrons lose potential Energy" when going through a resistor. In a case of falling from a height you would be at a lower altitude and therefore lower potential but how is this analogous to electrons. Do they slow down? Wouldn't this mean that they have just lost kinetic energy? If it is meant that they've lost potential energy due to being closer to the positive terminal of the battery when going through a resistor wouldn't this still hold even when there isn't a resistor? If this is the case electrons should always gradually be losing potential energy when going towards the positive terminal but are said they only lose it going through resistors. How?
 A: *

*Gravitational potential energy is stored when two masses are separated. It is a consequence of the conservative force of gravitational attraction. As they get closer, this energy is released.


*Similarly, electric potential energy is stored when two like charges are separated. It is a consequence of the conservative electric repulsion force. As they get farther away from each other, this energy is released (vice versa for opposite charges that instead attract).
This is not about slowing down. Place a charge next to a point that has a large charge already, and it will want to move as far away as it can. It may reach high speeds while doing this if nothing else slows it down (because the released electric potential energy then is converted into kinetic energy).
When dealing with classical circuits where the free charges are negatively charged electrons, then you might have a battery with a positiv terminal and a negative terminal. Plenty of charges are acculumated at the negative terminal, and none (less) at the positive terminal. We thus call the negative terminal a point of high potential and the positive terminal a point of low potential.  Any electron in the circuit is repelled from the negative terminal, i.e. the point of high potential, and attracted towards the positive terminal, i.e. the point of low potential. In short, electrons "want to" move towards points of low potential.
As it moves along with this electric "push" towards the point of low potential, the stored electric potential energy is released. It might initially be converted into kinetic energy. But the higher speeds is immediately slowed again via resistors in the circuit that absorb this kinetic energy (typically as thermal energy and release it to the air as heat).
Ideally this will not happen without resistors. Ideal wires have no resistance. The electrons would thus accelerate and accelerate and accelerate forever... until they reach the other terminal and stop. In reality, all wires do have some internal resistance. Thus, an ever increasing electron current will eventually cause so much resistance that the wire either slows the electrons down to constant speed or that the wire burns and melts.
A: Electrons lose potential energy when going through wires. Electrons lose potential anytime work is being done on them. They don't slow down, its the opposite much like when we fall and gain kinetic energy  we lose potential energy ( they are DEFINED as the negative of another) wires have resistance (in the definition of resistance there is Length) so when electrons move through a wire  they do lose potential energy
If you would connect a single wire to a battery with no load. When an electron makes a single loop, it would lose all of the voltage.
In order for electrons to move, work must be done on them. By definition making them lose potential energy
Now resistance is a bit slightly trickier because even if the electrons are moving at a constant speed( aka no NET work is being done on them) they still lose potential energy as the work done on the electrons via the E field is balanced by the negative work done by resistive forces
A: Electrons lose potential energy as they move around a circuit from one terminal of a battery to another. If you have n different circuits each powered by a 4.5v battery, then the electrons in each circuit will lose 4.5v of potential overall as they move around it. The overall loss of PE is equal to the voltage difference between the battery terminals.
More generally, electrons will lose PE when they move between one point of higher potential to another point with lower potential.
If you consider the potential along the path of a circuit from one battery terminal to the other, if won't, in general, change at a constant rate. What you will find is that the rate at which the potential changes will go up or down depending on the resistivity of the material the electrons are passing through.
If you have a 4.5v circuit consisting of three equal resistors in series, then there will be a a 1.5v drop over each resistor. If there is one resistor of 8 ohms and five resistors each of 0.2 ohm, then there will be a 4v drop over the first resistor, and an 0.1v drop over each of the  subsequent five resistors. Likewise if there is a 8 megohm resistor and five 0.2megohm resistors, there will again be a 4v drop over the first and 0.1v drop over each of the others.
What you can see from those examples is that the overall drop of PE is 4.5v in each case, but the 4.5v total is the sum of the contributions from each resistor along the way, and the contributions are proportional to the values of the resistances.
In practice, most circuits are made of components joined by wires. Wires are resistors too, but their resistance is very low- usually so low that it can be assumed to be zero. So by convention, people usually use the word resistor to mean a component with a resistance that is much higher than the resistance of the conducting wire, the resistance of the wire being ignored.
Given all that, when text books say that electrons lose PE when they pass through resistors, what they really mean is that the electrons also lose PE when they pass along the wires joining resistors, but the loss of PE along the wires is so small compared with the loss of PE over the resistors it can be ignored.
Of course if you don't have any resistors at all, and the battery is short circuited with thick copper wire, the electrons still lose 4.5 volts of PE over the wire.
A: First of all, why can we define an electric potential along the circuit? Because there is a conservative electric field along the circuit. As the electron moves along the circuit, it is accelerated by this electric field, so the electric field does work on the charge. Consequently, it's electric potential energy decreases. I think you get this part already.
So why doesn't the potential energy of an electron depend solely on its position relative to the terminals of the battery, or on the length of wire it has travelled? The reason is that the electric field along the circuit is not uniform. If the only thing connected to the battery was a wire with uniform resistance, than yes, the electric potential of the electron may decrease uniformly with distance, because the electric field along the wire is approximately constant (even this may not be true, the presence of bends and kinks along the wire may result in a non uniform electric field). But when a resistor is introduced to the circuit, the electric field along the wire will be almost zero, while there will be a large electric field in the resistor. So the electric potential will largely fall over the resistor.
But why should the electric field be 'concentrated' in resistors and not wires? To understand this, we should look at what produces the electric field in the circuit. It turns out that the presence of the battery induces a layer of charge, called surface charges, on the outer surface of the wires/resistors. The electric field strength in the wire is proportional to the gradient of the surface density of these surface charges, that is, how quickly the surface charge density changes. The resistor creates a sort of bottleneck for these surface charges, causing positive surface charge to accumulate on one side, and negative surface charge on the other. This sets up a sharp surface charge gradient over the resistor which generates a significant electric field. Along the wires, the surface charge density is nearly constant, resulting in a negligible electric field.
A: 
What is meant by "Electrons lose potential Energy" when going through
a resistor.

The electrons are given electrical potential energy at the negative terminal of the battery. At the microscopic level they move through the resistor alternatively acquiring  kinetic energy from the force of the electric field and transferring kinetic energy to particles of which the resistor is made due to collisions. This raises the temperature of the resistor causing the dissipation of energy in the form of heat.
The equal gaining of kinetic energy from the field and losing of kinetic energy by collisions results in a constant average kinetic energy and constant drift velocity (constant current) through the resistor. So the electrons on average do not "slow down". They move from having high potential energy at the negative terminal to lower levels of potential energy as they move through a series circuit, while the current remains constant. So the repeated sequence of energy conversion is electrical potential energy, to kinetic energy, to heat.
The analogy with a falling body is what happens after a falling body reaches its terminal velocity due to air resistance. The velocity of the body (analogous to current) remains constant while the body loses gravitational potential energy (analogous to electrical potential energy). Kinetic energy is simultaneously transferred from the falling body to the air molecules by collisions due to air resistance (analogous to the electrical resistance).
Hope this helps.
