Connect with us

Basic Physics Question

Discussion in 'General Electronics Discussion' started by chopnhack, Jan 25, 2016.

Scroll to continue with content
  1. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    While learning about current, the text states that charge/current is not consumed. KCL shows this as does the law of conservation of energy.

    What I don't understand intuitively is how charge/current is not consumed if they produce work?

    In more concrete examples we see energy given off with a resultant change in the materials, i.e. - Fire, gives off heat, light, etc. while consuming the energy source (wood was the source of potential energy until liberated by the fire).

    Can someone explain this to me with respect to how current in electricity is not consumed, yet still does work?

    Thanks!
     
  2. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Current is the measure of how many electrons are flowing through a circuit.
    Water travelling down a pipe does not consume water, but the pump consumes energy in moving the water. Electrons are no different.

    The source consumes energy pushing/pulling electrons in the circuit to do work.

    This is how I understand it at least.
     
    Tha fios agaibh likes this.
  3. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    In this simple example, the battery is the source or 'pump' to produce our potential and also the source of our current. The light bulb is our resistor. When the switch is closed, the difference in potential between terminals is realized, and a flow is established. Electrons move from the negative terminal to the positive (current). As they pass through the light bulb, work is done. Electricity is 'used' and converted into light and heat.
    How?
    If 1 coulomb is passing through the bulb per second, 6.25x10^18 electrons are moving between terminals of the battery and through the bulb, how is the energy transferred to the bulb and yet all the electrons are measured as returning to the other side of the battery?


    [​IMG]
    taken from : https://www.teachengineering.org/vi.../cub_electricity/cub_electricity_lesson05.xml
     
  4. Bluejets

    Bluejets

    4,277
    909
    Oct 5, 2014
    Energy is not created or destroyed but changed from one form to another.
    At least as far as we know it in this universe.
     
  5. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    Yep, covered in first sentence. Didn't want anyone to think this post was going into a free energy machine discussion ;-)
     
    Gryd3 likes this.
  6. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    The net charge does not increase or decrease in a closed system according to the Law of Charge Conservation. KCL states that the rate of charge flow into a point is equal to the rate of charge flow from a point. KCL is a consequence of the Charge Conservation Law. The Law of Energy Conservation does not apply here.

    A better question is why should it? It takes force to move a charge against an electric field. The farther the charge is moved, the more energy is needed (energy=force times distance). At the end of the movement path, the charge is still there. Why would you think it is consumed? Same with a wire. Do you think that fewer charges exit from the end of the wire than entered it?

    Fire is a consequence of a chemical reaction. Why are you confusing it with electrical circuits?

    I just did above.

    Ratch
     
    Arouse1973 likes this.
  7. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    Your question leads me to believe that you do not really, really know what voltage is. Voltage is the energy density of a quantity of charge. Like charges don't get along too well together, so it takes a certain amount of energy to gather like charges into a certain amount of volume. It takes more energy to add more charges into the same volume. It takes even more energy to reduce the volume while maintaining the same charge. In the MKS system, voltage is joules/coulomb. That means one volt of energy density of the charge is one joule/one coulomb.

    The voltage applied across the above light bulb causes a electric field across the bulb terminals. The charge carriers (electrons) will be repelled from one terminal and attracted to the opposite terminal. During their trip through the bulb, they will encounter deflections and diversions caused by the atoms of the filament. This is known as resistance, and it is a quantum process, not a physically mechanical one. This causes heat to form, which lights the bulb and reduces the energy density of the charges. When the charges exit the bulb, their voltage is less, meaning their energy density is less, but no charges were "consumed".

    Ratch
     
    Last edited: Jan 26, 2016
    Arouse1973 and chopnhack like this.
  8. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    Does this imply that the voltage drop that occurs across the filament accounts for the energy released from the bulb?

    Can you explain the quantum process further? (In simplistic terms.)
     
  9. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    Partly. Energy = Voltage x Charge. So that means that if more charges lose some of their energy density, more energy is lost as heat.

    That is not pertinent to your understanding of this explanation. There is plenty of material that explains it if you are interested.

    Ratch
     
    Last edited: Jan 26, 2016
  10. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    This is where I am getting hung up.... We know there is a voltage drop across the filament, but there is no change in charge, thus all of the change in voltage must account for the energy portion of the equation. So some energy is lost as heat, through the undefined quantum process (that yes, I would like to know more about as I requested above as it seems pertinent!).

    Why do you say partly? It seems that if the energy density is lost through the filament as heat and then the filament produces light, we can simply say that reaction 1 is electrical energy passing through the filament producing heat. The light is reaction 2 when the metal becomes super heated.

    Can you elaborate?
     
  11. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    Yes, the energy which the charges expend flowing through the resistance is lost as heat. Less energy means a lower energy density of the charge or lower voltage where they exit the bulb. That is usually called a voltage drop.

    Correct, why should there be? There will be twice as much energy lost if two coulombs go through the bulb compared to only one coulomb.

    False conclusion. You can put a million volts across a device, but if no charge flows through the device, it will not dissipate any energy.

    So you are going to study it then, right?

    Because charge flow is also a factor in energy dissipation.

    Yes, loss of energy causes loss of energy density. Analogously, if you remove some salt from seawater, the salinity is less. Who cares what the heat does? Whether it lights a bulb or warms a heating pad, who cares?

    Ratch
     
    Last edited: Jan 26, 2016
    davenn likes this.
  12. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    To summarize tonight's lesson as I understand them at this point: Charges are not lost or consumed as they flow through the circuit. But the energy done by these charges is imparted by the change in potential.
     
    Arouse1973 likes this.
  13. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
     
    Arouse1973 likes this.
  14. BobK

    BobK

    7,682
    1,686
    Jan 5, 2010
    This is why I like my voltage as height analog.

    Think of water falling through a waterwheel. No water is lost, but energy is lost as the height of the water is changed.

    Mass in a gravitational field is equivalent to charge in an electric field. Potential in the first case replaces height in the second case.

    Bob
     
    chopnhack, Gryd3 and davenn like this.
  15. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    Thank you both Ratch and BobK - I think I have a much firmer understanding of voltage and current!
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-