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RE. Passive Convention?

  1. Aug 23, 2006 #1
    I'm having a lot of difficulty understanding a really simple concept in my first year physics text. See the diagram below...

    http://www.avhq.com.au/~metrosexuality/ele_pass_conv.JPG

    "First, consider Figure 1.5-1a. When the current enters the circuit element at the + terminal of the voltage and exits at the −terminal, the voltage and current are said to “adhere to the passive convention.” In the passive convention, the voltage pushes a positive charge in the direction indicated by the current. Accordingly, the power calculated by multiplying the element voltage by the element current p = vi is the power absorbed by the element. (This power is also called “the power received by the element” and “the power dissipated by the element.”)"

    I have read over that paragraph hundreds of times; and still don't understand it... If a positive charge is entering the + terminal of a voltage element, wouldn't the voltage be pushing the charge against the flow of current?

    Likewise-

    Next, consider Figure 1.5-1b. Here the passive convention has not been used. Instead, the current enters the circuit element at the −terminal of the voltage and exits at the +terminal. In this case, the voltage pushes a positive charge in the direction opposite to the direction indicated by the current. Accordingly, when the element voltage and current do not adhere to the passive convention, the power calculated by multiplying the element voltage by the element current is the power supplied by the element.

    If a positive charge is entering the - of a voltage element, wouldn't work need to be done IN the direction indicated by the current to keep the current flowing?

    It just doesn't make sense...

    Can someone please help clarify this? It's driving me crazy...

    Thank you in advance :)

    Adam
     
    Last edited: Aug 23, 2006
  2. jcsd
  3. Aug 23, 2006 #2

    berkeman

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    Staff: Mentor

    The text's wording is a bit inane, but the first case is just like a resistor, and the second case is like a battery.

    Just start out picturing electrons flying around in free space, under the influence of an electric field. The "current" flows in the opposite direction of the electrons.
     
  4. Aug 23, 2006 #3

    SGT

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    No, the´positive charge is attracted by the - terminal, so this is the normal behaviour of a passive element: positive charges enter by the positive terminal and exit by the negative one.
    I don´t like the expression "power dissipated by the element". This only happens if the element is a passive resistor. A passive capacitor stores power in it´s electric field and a passive inductor stores it in it´s magnetic field.
    An active element, like a power source or a negative resistor delivers power to the rest of the circuit, so the current flows from the negative to the positive terminal.
     
  5. Aug 23, 2006 #4
    Consider:

    ++++++++++ -> + [ELEMENT] - ++++++++ ->

    The way I see it- is that the force experienced by + charges as they approaches the + element is one of repulsion... which constitutes voltage pushing the charge in the opposite direction to the current? Same thing when it leaves the - terminal; it would be attracted back to it, again voltage pushing it in opposite direction to current flow...
     
    Last edited: Aug 23, 2006
  6. Aug 23, 2006 #5

    berkeman

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    I don't like talking in terms of positive charge -- that's non-physical for this situation. Electrons are flowing in the circuit, and they carry negative charge. We just use the convention that positive "current" flows in the opposite direction of the electrons.

    And the voltage drop across an element is the loss of potential energy by the charges as they transfer energy to phonons in the resistive material or metal (or as they generate photons in LEDs, for example). The fact that the voltage is lower at the bottom of a resistor doesn't mean that is a charged node with a negative charge that is going to repel or attract charge. The EMF around a circuit is the force that pushes the electrons along.
     
  7. Aug 24, 2006 #6
    Thank you :)

    So the fact that the current is entering a + or - of a voltage element has nothing to do with the repulsion or attraction of actual charges?

    I'll try say that in a slighty different way...

    The + and - terminals of such a voltage element are used only as convention... They don't necessarily mean Positively Charged Side/Negatively Charged Side etc?

    The orientation of the + or - terminals allow us to determine the voltage drop, or gain- and hence changes in electrical potential?

    That sound about right?
     
  8. Aug 24, 2006 #7

    berkeman

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    Mostly right. The only thing I'd change in your statements in the quote is that the EMF is what supplies the force to push the electrons around the circuit. The sum of the voltage drops and increases around the loop is the line integral of E dot dl, around the loop, right? And the forces that push the electrons around the loop come from that EMF. The source of the EMF can be supplied by a battery, power supply, changing magnetic flux, or whatever.
     
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