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Terminal Voltage Formula Doesn't Apply?

  1. Feb 27, 2012 #1
    I'm confused as to why I can't simply use the terminal voltage formula here...

    Vt = E - Ir = 1.5 - (.5)(.6) = 1.2 V

    The actual answer is 1.8 V. Is there a difference because it is "find the terminal voltage of the cell while it is being charged?
     

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  3. Feb 27, 2012 #2
    any ideas?
     
    Last edited: Feb 27, 2012
  4. Feb 27, 2012 #3
    Okay, sure I will!
    Any thoughts on the initial question though?
     
  5. Feb 27, 2012 #4

    SammyS

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    That's the answer you would get if the current were flowing in the direction opposite what is given in this problem.


    attachment.php?attachmentid=44478&d=1330401763.png
     
  6. Feb 27, 2012 #5

    gneill

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    Label the potential drops on the circuit according to the current direction. Add them up. What do you get?
     
  7. Feb 27, 2012 #6
    Isn't there only one potential drop? Across the internal resistor?
     
  8. Feb 27, 2012 #7

    SammyS

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    There is the 1.5 Volt cell that's also between the terminals. Right?
     
  9. Feb 27, 2012 #8

    gneill

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    Well, when you "walk" through the circuit in the direction of the current flow, do you have a potential rise or a potential drop as you pass though the battery?
     
  10. Feb 27, 2012 #9
    That counts as a potential drop too? Tell me, what exactly is a potential drop?

    Anyway, okay, if I do that.
    Potential drop across the internal resistor is V=IR = (.6)(.5) = .3
    Adding that to the EMF (another potential drop) is 1.8 V.

    Oh, well there's the answer then. :)
     
  11. Feb 27, 2012 #10
    So you have two potential drops within the battery?
     
  12. Feb 27, 2012 #11

    gneill

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    It depends upon the current direction through the battery and the direction that you "walk" through the circuit. While the polarity of a cell does not change, whether you consider it to be a potential drop or a potential rise depends upon the direction you're going.

    The polarity of the potential change across a resistor on the other hand clearly depends upon the direction of the current flowing though it. So both the current direction and the direction you choose to follow around the circuit matter.

    EDIT: Note that your "Terminal Voltage Formula" assumes that the battery is producing a current that flows out of its positive terminal. In that case the internal resistance causes a voltage drop in the direction of the current flow, and so decreases the voltage that you see at the battery terminals. When the current is forced to flow in the other direction (battery charging), the potential drop across the internal resistance due to the current has the same polarity as that of the cell. So then the battery terminals have a larger total potential across them.
     
  13. Feb 27, 2012 #12
    Hmm... okay... so what would happen if the internal resistor was on the left hand side? Would there still be two potential drops?

    I'm just confused because I haven't had to deal with adding up the potential drop across the internal resistor and EMF value itself before in order to find terminal voltage. I thought the terminal voltage equation would just work really simply.
     
  14. Feb 27, 2012 #13

    gneill

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    The location of the internal resistance doesn't matter. In fact, in a real battery it's distributed in various places in the chemistry of the cell. The simplified model that we work with has an ideal cell in series with a 'lumped' resistance which we call the internal resistance of the battery. Since the cell and resistance are in series, the same current flows through both so the sum of the potential changes is the same regardless of the order in which you add them.
    It's good to know where a formula comes from and what assumptions it relies on so that you can judge its applicability in different situations. In many cases its easier (and safer!) to apply simple, basic circuit analysis rather than memorize a number of particular-case formulas.
     
  15. Feb 28, 2012 #14
    Is this an exception though? Because there have been so many other scenarios where I haven't had to do this. Why now did we have to add up the potential drops? As in, why didn't the formula work?
     
  16. Feb 28, 2012 #15
    Imagine being given two batteries one of 9V and the other of 6V.There are two ways that these can be connected in series.One way gives a total voltage of 15V and the other way(with one of the batteries turned round) gives a total voltage of 3V.Sketch it out with the polarities marked.
    Now look again at the advice given above and think about how the voltages add,with the current flowing one way and subtract with the current flowing the opposite way.
    (if you get confused about current direction remember that there is a charging circuit)
     
  17. Feb 28, 2012 #16

    gneill

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    See the EDIT addition that I made to post #11 in this thread.
     
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