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Energy flow in an electrical circuit

  1. Mar 29, 2014 #1
    Hi, could someone explain how energy and voltage operate in an electrical circuit. I'm really confused because I was reading a book on misconceptions in electric circuits which said that energy is not associated with the charges in a circuit. It goes on to say that if it did then you would have to wait for electrons from the battery to arrive at the the load. It says that a common analogy of voltage and energy is using freight carts. If you had a 6V battery then the battery gives the carts 6J of energy and the carts take them and drop them off at the load and then they go back to the battery to get re-filled. It says that this analogy is completely wrong and misleading. How should you view potential difference then? If the potential difference across a bulb is 6V then doesn't this mean that each coulomb of charge loses 6J of energy going across the bulb? But then how does this apply to a.c. because charge makes no net movement in any direction, they just vibrate back and forth due to the electric field? So what then does 6 joules per coulomb actually mean?
     
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  3. Mar 29, 2014 #2

    Nugatory

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    Staff: Mentor

    One electron pushes on the next pushes on the next pushes on the next.... and so forth, and then eventually one of the electrons delivers energy as it is forced through the potential difference at the other end. Alternating current delivers energy because it doesn't matter which direction we're pushing the charge.

    You might try imagining a long rigid rod with you holding on to one end. You can do work at the other end by pushing the rod (DC one direction) or by pulling it (DC the other direction), or pushing and pulling back and and forth (AC); and also that nothing is making any sort of round trip.
     
  4. Mar 29, 2014 #3
    Thanks for your answer that was really helpful. How would you interpret a p.d. of 6 joules per coulomb across a bulb which is driven by ac current? In a dc circuit I would say that each coulomb of charge passing through the bulb loses 6J of energy but since ac charges do not move, how do you interpret this in terms of joules per coulomb?
     
  5. Mar 29, 2014 #4

    nsaspook

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    The returning carts in a loop analogy is misleading because when we calculate the power flow from battery to load we don't see the kinetic (6J in your example) energy in the movement of charge (electrons) in a loop being dumped and coming back empty, we see a one way directional energy movement from the battery to load on both wires that matches the flow of power.

    http://sites.huji.ac.il/science/stc/staff_h/Igal/Research Articles/Pointing-AJP.pdf
     
    Last edited: Mar 29, 2014
  6. Mar 29, 2014 #5

    Nugatory

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    With AC, the charges do move - first in one direction and then in the other. You will get just as tired running back and forth for an hour as you would running in one direction for an hour, and the charge can do just as much work moving back and forth as it would flowing in direction.

    You can connect an oscilloscope (which is basically a device for displaying voltage changes that happen too quickly for the needle of a DC voltmeter to respond) across the light bulb. The oscilloscope will show that if the average potential difference is 6 volts (one volt is equal to one joule per coulomb) then the instantaneous potential varies from about -8.5 volts to +8.5 volts so first the charge moves one direction then the other, but it's doing work and lighting the bulb in both directions.

    (Why 8.5? Well it just so happens that that's the peak value that makes the average come out to 6; google for "RMS peak" or "root mean square" for more detail).
     
  7. Mar 30, 2014 #6
    Thanks Nugatory. This is a great explanation and has really helped me!!!
     
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