2
\$\begingroup\$

So when electrons leave the negative side of the battery and the voltage get lost across the resistors in the circuit ..how do the electrons lose their energy ? How do the electrons carry that energy that produces the electricity then they LOSE it in the resistor ?

\$\endgroup\$
  • \$\begingroup\$ This question must hold some sort of a record for a large number of answers all with zero votes after several hours. This is not to say it's a bad question, but perhaps it cannot be answered. \$\endgroup\$ – Neil_UK Oct 4 '17 at 11:11
  • \$\begingroup\$ @Neil_UK : added a seventh! \$\endgroup\$ – Brian Drummond Oct 4 '17 at 12:49
  • \$\begingroup\$ Electron scattering is what you are looking for. \$\endgroup\$ – user110971 Oct 4 '17 at 14:09
4
\$\begingroup\$

'How' is a question that's impossible to answer.

You have to start with the observation that when you connect a battery to a resistor, energy is transferred.

Now there are several levels of model we can invoke to summarise how much is transferred. We can also push those models to 'explain' what's going on in other terms that we feel good about, that we think we understand.

The simplest model is circuit theory. The battery delivers a current, from a voltage, and the product is power. It doesn't tell us what voltage is, or current consists of, but it's neat for doing calculations ranging from the very simple to the very complicated. The hydraulic analogy is often invoked to make it more familiar.

Next we could think about mobile electrons. Voltage is a measure of how much energy per unit charge is involved when they move. It doesn't tell us what electrons are, or how energy is transferred. There are a lot of electrons in the conductor, some move, most don't, why the difference? They gain energy by moving 'down' the field from the battery. How does that get transferred to the resistor? Is there friction between the electrons and the atoms in the resistor? Do they hit the atoms? Do they get scattered? All three are defensible explanations, depending on the level of model, and what you want to assume.

At the lowest level of our understanding is quantum mechanics, where electrical effects are mediated by photons (yes, even at DC). Frankly, that does my head in, because I lack the imagination and skill to work with those models. But it underpins the simpler models, and it still doesn't tell you what's 'really' going on.

Just get used to the idea that energy transfer happens, and learn enough equations at the simplest possible level of model to allow you to figure out how much.

\$\endgroup\$
2
\$\begingroup\$

The electrons do not "lose energy". The energy is transferred via the electric field, which can be viewed at the "pushing force". Crudely (an inaccurate analogy) the field pushes the electrons through a resistance and this causes friction through collisions with the atoms.

\$\endgroup\$
2
\$\begingroup\$

Start by observing what happens to an electron released into a vacuum with an electric field across it, say produced by a 12V battery. It accelerates across the electric field, picking up kinetic energy - 12 electron-volts to be specific.

The electric field provides a kind of potential energy, just like gravity does to a Fig Newton or an apple, providing the acceleration. This is why a voltage is called a potential difference, and the field is called a potential gradient. And the acceleration translates potential into kinetic energy.

It doesn't matter whether the vacuum is 1 metre long, with 12V/metre electric field, or 1mm long, with 12000V/metre, the final energy is 12eV. (And you can work out the final velocity using basic Newtonian physics, if you know the mass of an electron (9.1*10^-31kg) and the value of 1 eV in Joules - from the definition of a volt, this is the same as the electron's charge in Coulombs, 1.602*10^-19). The stronger electric field across 1mm implies more acceleration but a shorter journey, which cancel out to the same final velocity.

Now introduce a thousand obstacles - the atoms in a very small resistor. The electric field is the same, so the acceleration is the same, but after 1/1000 of the distance, the electron bumps into an obstacle, losing kinetic energy and momentum, and has to start accelerating again.

That lost kinetic energy is transferred to the obstacle, vibrating it - heating the resistor.

\$\endgroup\$
0
\$\begingroup\$

It is easiest to demonstrate the process with water flow analogy.

Imagine that the battery is a water tower filled with water and water droplets represent electrons. Water pressure at the bottom represents voltage and water flow represents current. Ground represents positive terminal, the one where water (electrons ) will flow when allowed. In this system water it self does not have any energy, it is because it is held up high it can do some work when falling down.

Same with electrons: battery is just a chemical device that collects and stores some electrons at the negative terminal. If a load is connected between negative and positive terminal, electrons will start to flow through it and heat it in the process - do work. Same way water held up high can do work (and does in hydroelectric plants) when falling down. Energy is stored in the system, not water or electrons themselves.

Hope this helps.

\$\endgroup\$
0
\$\begingroup\$

Its called a voltage, the battery has zillions of ions that want to move an electrode to combine with electrons. All of the ions combined create an electrical potential called a voltage across the terminals of the battery. That makes the electrons on one side of the battery want to 'flow' to the other.

If you put a resistor in series with the battery, it gives them a pathway to flow to the other side of the battery. As they travel through the resistor they bump into other atoms and lose their electrical potential and create heat.

\$\endgroup\$
0
\$\begingroup\$

Energy cannot be stored in conductors unless they are long or coiled into inductors. The Energy is E=1/2LI^2

Conductive Atoms stay in tact and their electrons never lose energy, but the energy to excite them can be lost. The flow of electrons is very slow (drift velocity) but the waves of collisions is fast (c, speed of light) dependent on the surrounding insulation which slows down the speed of light inside the (dielectric e, which creates a capacitance, C).

The atoms themselves are lossless. (Conservation of Energy)

But the atoms on conductors have electrons that are easily excited (lol) and escape orbit and jump to the nearest atom with a spare space in their shell in the outer orbit.

The conductor property also has resistance which is the effect of the collisions that can dissipate heat in the exchange from bouncing around with a flow of charges i.e. I^2R=Pd losses. (Ohm's Law)


Electrical Energy is only stored in reactive elements, L or C where Pc = 1/2CV^2 but is generated by moving coils or magnets or batteries and lost with anything that has resistance with current flowing thru it.

Batteries are chemical electrolytics like e-caps , except roughly 1000x (1k) higher capacitance per unit volume with a chemical makeup to have a galvanic cell reaction voltage.

All insulators are dielectrics, even a perfect vacuum. The dielectrics can store charges and will have some contaminants or particles that can transfer charges that lead to breakdown voltage... even if it is a really small fraction like a 100 particles per zillion (lol) in a high vacuum chamber <4 torr. At this point the breakdown voltage /mm rises has risen sharply. (Paschen's Law)

\$\endgroup\$
  • \$\begingroup\$ In inductors energy is stored in a magnetic field, not conductor it self. Same way it is stored in an electric field in a capacitor. \$\endgroup\$ – Jurkstas Oct 4 '17 at 7:12
  • \$\begingroup\$ Yes we ought to say that, but we measure it by I and V, since you cannot have one without the other. ( H fields and charge flow & E field and charges ) That's what reactive parts (L&C) do is create the fields which in turn create magnetic or electric forces which in turn stores power and over time stores energy in the interval standardized by 1 second. in Joules or watt-seconds. So it becomes synonymous or equivalent energy depending on choice of units. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 4 '17 at 8:42
  • \$\begingroup\$ *But you are correct, the Magnetic field is outside the conductor and Electric field in caps is between the conductors. and you cannot store the energy without both zones, the conductors and the fields around them or between them respectively. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 4 '17 at 8:42
  • \$\begingroup\$ I had no intention to nitpick. It is just that the first simplifier statement does not fit your further detailed explanation. It does fit the question's level though. While rest of the answer seems to be overcomplicated and steers to broader context a bit. But I'm new here, let me know if my comment is inaccurate. \$\endgroup\$ – Jurkstas Oct 4 '17 at 12:57
  • \$\begingroup\$ your perception is good. I tend, to make my answers suit a broader audience and sometimes too much detail, and push some to learn more at the expense of confusing some. It was a hasty reply , so oh well WYSIWYG. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 4 '17 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.