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Bit of trouble understanding basic electronics

  1. Oct 24, 2005 #1
    I'm having a little difficulty understanding a few key electronic concepts and am hoping a some of you may be able to shed light.

    1) In a power source (i.e. battery) I understand that E = V / D. Despite the size and mass of an electron being incredibly small; when enter the field- do they accelerate (as they gain kinetic energy)? [This would make sense because work is being done on them.]

    2) As they exit the battery, do they then travel at a constant velocity, assuming there is negligent resistance in circuit wires (i.e. obeying Newtons first Law of Inertia)?

    The thing I'm finding most difficult to comprehend is what exactly happens in a voltage drop. My book states that, for series circuits, the current is constant but the voltage is divided proportionally to all resistance.

    3) My understanding for the above is that the kinetic energy each electron possesses (in this case- its motion provided by a power source) is given by E = ½mv². As these electrons collide with atoms in a resistor they transfer some of their Kinetic Energy as Heat. Mathematically, this loss of Kinetic energy must mean their velocity is also be decreasing (in relation to the E = ½mv² formula); which contradicts the books statement that the current is constant throughout a series circuit.

    Obviously there is a problem with my working somewhere... could someone please explain where I'm going wrong?

    Much appreciated
     
  2. jcsd
  3. Oct 24, 2005 #2

    berkeman

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    Welcome to PF, Metro. The source of the electrical energy (a battery or a charged capacitor or a photocell or whatever) creates an electrical potential at its output terminals. When you connect a resistor between those terminals, you get a voltage across the resistor material, which is essentially an electrical field throughout the volume of the resistor. Electrons are driven by the electric field, and flow through the resistor. The higher the resistance, the shorter the mean free path of the electrons between their collisions. So the higher the resistance, the more collisions per second, so the flux of the electrons (the current) is lower. It's the flux, or average velocity of the electrons that's related to the current, not the instantaneous individual velocity (which does increase in the direction opposite to the electric field vector between collisions). Just think of the analogy of a Japanese Pachinco (sp?) machine -- more pins in the way means a lower overall current. And yes, if you measure the voltage along the body of the resistive material, you will get a linear decrease in voltage from the top + side to the bottom - side.
     
  4. Oct 25, 2005 #3

    Cliff_J

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    Metro - my advice would be to step back and look at the flow of electricity as something more than the flow of an individual electron. Its more like enough stuff cancels out on the equation that you can simply look at the net effect and basically ignore the minutia.

    So in a simple series circuit, current flow is constant along any point in that circuit. This is very important. If there are 3 million electrons per second 'entering' one side, at any point along the way there will be effectively 3 million electrons per second flowing. So therefore there will also be as many 'leaving' the circuit and returning to the power supply.

    The net potential along the way will change depending on the resistance, and again its not the behavior of an electron at question but the overall average behavior.

    So there is no contradiction, because current is representing the collective behavior and not that of an individual electron. And if an electron does loose some of its kinetic energy as heat but still continues on its journey, it will still enter and exit the material, just leaving with slightly less energy than entering. But it is still traveling across it, or more importantly it is "in essence" traveling across it because it may impart its potential to other electrons and they will repeat the process along the conductor.

    So an individual electron may only move one atom but its potential is transferred like how each 'ring' in a slinky spring may only move a fraction of an inch but the wave may move a few feet across the length of the slinky. Or how each air molecule will displace only a tiny amount but a sound wave can travel a very lond distance. Or the energy released in an earthquake is large enough to travel around the world and seismologists can measure it yet each molecule of rock is moved very little.
     
  5. Oct 25, 2005 #4

    russ_watters

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    I'll be a little more blunt: kinetic energy plays no role here whatsoever, so it is useless to think about the velocity of electrons.

    Yes, electrons move and yes, voltage acts like a "force", but that does not make electricity a kinetic energy. Remember what the definition of an "ampere" is: it's the moving of a specific quantity of electrons (a coulomb) through a circuit every second. So voltage doesn't need to change the speed of the electrons flowing through a circuit to change the energy (power, actually), just the quantity.
     
  6. Oct 25, 2005 #5
    Wow, thanks for all the replies! I think I'm slowing catching on...

    I think I understand what you mean by the voltage being 'applied'. To make sure, I'm wondering what happens when an electron leaves the high potential side of a battery and enters the external copper wire?

    Does an electron that leaves the battery collide with (and orbit) its neighbouring copper atom (or electron?) in the process dislodging the neighbours valence electron, and transfering it to the next. Then is this process is repeated over and over?

    If this idea is true- the general idea would be that the transfer of energy is extremely quick- but the movement of each individual electron is rather slow. And if I'm right, wouldn't it be better to think of current in terms of the amount of charges flowing through any given point every second... not the speed of an individual electrons on their journey through the circuit?

    Thanks a lot for all your help- I love this stuff; I'm just having real difficulty understanding it...
     
  7. Oct 25, 2005 #6

    russ_watters

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    Wow - you're pretty damn close. Yes, electrical energy moves in a wave, similar to a pressure wave, and individual electrons don't move very far or very fast for electrical energy to be transferred*. The one major clarification I'd make is, remember what is so special about metals: metallic bonding". AKA, a "sea of electrons". Electrons in a metal are not fixed to individual atoms, and as a result, they can move, with very little resistance, through the metal.

    *edit: in fact, with alternating current, an individual electron from your local power plant never does make it to your house!
     
  8. Oct 26, 2005 #7
    A wave- like a sound wave travelling through the air? Aha! Beautiful!



    So... say, in copper- although the valence electrons can easily jump into the conduction band, they only ever really move small amounts in open space... because they soon collide with the nearby atoms (and their electrons) transferring their energy in the collision (just like air particles in a sound wave)?
     
    Last edited: Oct 26, 2005
  9. Oct 26, 2005 #8

    GENIERE

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    It is more correct to think of a wire as something that guides the transfer of energy from point A to point B. The acceleration of the electrons due to the applied voltage generates an EM field, a field that propagates external to the wire. The electrons do not transport energy; it is borne by the EM field that is generated by the electrons. The EM fields propagate at the speed of light in the medium surrounding the wire but the group velocity of all generated fields is somewhat slower.
     
  10. Oct 26, 2005 #9

    berkeman

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    Remember that the electron has a negative charge, and flows in the opposite direction of the "current". So electrons flow out of the - terminal of the power supply, and are attracted to the + terminal as they flow through the external circuit.
     
  11. Nov 27, 2005 #10
    To answer the question about a voltage "Drop," try this link about Kirchoff's Law.

    http://www.allaboutcircuits.com/vol_1/chpt_6/2.html

    Kirchoff Law States that, The Sum of all Voltages in a loop will equal 0."

    Lets say you have a circuit with 3 1KOhm resistors in series and a battery at 9V. The total resistance would be 3KOhms and would have 9V Across the 3 resistors. Each resistor would "drop" a voltage proportional to its value of resistance to the total resistance. 1K / 3K = .3333, then .3333 * 9V = ~3V.

    Again the total resistance is 3K, the voltage is 9V, the current through the circuit is 3mA, (Ohms Law, I=V/R) Each resistor is 1K and again using Ohm's Law (V=I*R) the voltage across each resistor is 3V. So you start at the Positive Terminal of the battery, and go with the electron flow of Current. +9V-3V-3V-3V=0. If you start at the negitive terminal of the battery its, -9V+3V+3V+3V=0

    ::::+9V-::::-3V+::::-3V+::::-3V+::::
     
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