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Homework Help: Resistance of the resistor and voltmeter's readings

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A battery of EMF 12.6 V and internal resistance 0.1 Ω is being charged from a DC source of EMF 24.0 V and internal resistance 1.0 Ω using the circuit shown in the figure below. V1 and V2 are high resistance voltmeters and R is a fixed resistor.

    43ba58dfc733.jpg

    (a) What is the polarity of terminal A of the source?
    (b) I the charging current is 5.0 A, determine the resistance of the resistor R.
    (c) If the resistance of R were changed to 0.9 Ω what would be the reading on each voltmeter?

    Answers: (b) 1.18 Ω, (c) 18.3 V on V1 and 13.2 V on V2.

    2. The attempt at a solution
    Every part I am in doubt whether it is correct or not.

    (a) Since the battery has a polarity of + - Battery + - (of we look from the top to the bottom), I would say that the polarity of the terminal A would be negative (-). Since the current flows from the positive sign of the battery to the negative sign A of the DC source.

    (b) This is only a guess: 24 - 12.6 = 5 (0.1 + 1 + R) → R = 1.18 Ω. Maybe because the voltmeters have large resistance they don't disturb the circuit and so we just calculate the way I did (like it is a simple circuit only with two batteries).

    (c) No idea.
     
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  3. Oct 3, 2016 #2

    gneill

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    When a battery is being charged, the charge that it used up while it was being used in some circuit is being replenished. Which direction should this charging current flow? Would it be the same direction as when the battery is in normal use?
     
  4. Oct 3, 2016 #3
    Conventional current flows through the load from a point of higher voltage to a point of lower voltage. A battery being charged could be considered a load. Note that energy is subtracted from the current.

    Conventional current flows through a source of EMF (e.g. power supply or battery) from a point of lower voltage to a point of higher voltage (energy is added to the current).
     
  5. Oct 4, 2016 #4

    andrevdh

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    (a) When a battery is being charged we have to reverse the chemical processes normally taking place inside of it.
    This is brought about by forcing a current in the opposite direction through it, that is in the opposite direction to what it normally operates in.
    This is brought about by connecting it to another emf source with a voltage higher than it.
    This opposing emf will then push a current in the opposite direction through the battery, thereby reversing the way the chemical processes normally runs and restoring the battery being charged to its formal higher capacity state.
     
  6. Oct 4, 2016 #5
    So A is negative, it goes to a positive battery, which then becomes negative again and the point below A is positive. So A is negative is correct?
     
  7. Oct 4, 2016 #6

    gneill

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    Draw the circuit with the 24 V supply in place and indicate the direction of current flow. What does KVL around the loop say the current should be?
     
  8. Oct 4, 2016 #7
    91a335a88a99.jpg

    I actually think it should be from + 12.6 V to - 24 V because the long line is where current flows from it and not from the short line (that is where electrons flow from). But it was said that current flows from low voltage to high voltage.
     
  9. Oct 4, 2016 #8

    gneill

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    What's the total EMF for that loop, following the direction of the current you've indicated?
     
  10. Oct 4, 2016 #9
    24 V + 12.6 V = 0 V like this?
     
  11. Oct 4, 2016 #10

    gneill

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    You have indicated a current going clockwise. So do a "KVL walk" clockwise around the loop summing up the EMFs (just the sources). Remember that when you traverse a source from its - lead to its + lead it is a gain in potential, and when you traverse it from + to - it is a loss in potential. What is the sum you get? Is it positive or negative?
     
  12. Oct 4, 2016 #11
    So it's -24 V - 12.6 V = 0.
     
  13. Oct 4, 2016 #12

    andrevdh

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    The polarity of your 24 V source is indicated incorrectly in your drawing, but your current's direction is correct, that is the source forces current in the opposite direction to what the battery would normally let it flow, so that the source is charging the battery.
     
  14. Oct 4, 2016 #13
    How did you determine that?
     
  15. Oct 4, 2016 #14

    andrevdh

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    The source needs to oppose the battery in order to reverse the current throught the battery.
    PA040354.JPG
     
    Last edited: Oct 4, 2016
  16. Oct 4, 2016 #15
    In other words since we want to charge the battery, we need to oppose the current flow. And since we know that current flows from + and we have + of the battery as up, therefore we put + at A (the point of the DC source), in that case current will flow from the regular + point from the source and will inversely flow into the battery and charge it. Like this?
     
  17. Oct 4, 2016 #16

    andrevdh

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    Correct - I have added a "drawing" to my previous post.
     
  18. Oct 4, 2016 #17
    And what about (b) and (c)?

    Update: could you tell what is written on your graphs? The image is too small.
     
  19. Oct 4, 2016 #18

    andrevdh

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    Things get complicated :)
    You have to add little r's (internal resistors) for both the source and the battery to your drawing.
    Didn't the problem give the readings on the voltmeters? I get the impression that they might have for (b)?
     
  20. Oct 4, 2016 #19

    andrevdh

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    Ok, gneill suggested an easy way to solve for (b).
     
  21. Oct 4, 2016 #20
    If the charging source has its current going from + to the + of the battery, then the battery is charging. And when + goes to - then the battery is not being charged, right? It is being consumed / used?
     
  22. Oct 4, 2016 #21

    SammyS

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    Yes, that's essentially correct.
     
  23. Oct 4, 2016 #22
    If the battery is powering a load, that is true. However, if the battery is the load then current flows in the opposite direction. Considering the graphic symbol for a voltaic cell, the long, thin line represents the positive terminal, and the short, thick line represents the negative terminal.
     
  24. Oct 5, 2016 #23

    andrevdh

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    An emf source can be considered to be a (positive - conventional current) charge pumping station - that is it circulates the charge around the circuit
    PA050355.JPG
    These two pumping stations are assisting each other, thereby adding their pumping ability or emfs
    PA050356.JPG
    These two are opposing each other - pumping against each other
    PA050357.JPG
    we find that whoever pumps hardest determines the direction of the current in the circuit, if E1 > E2 then the (+) charge will circulate clockwise in the circuit ....
    also if E1 = E2 no current will be present in the circuit.

    In the case of charging the battery we find that the battery we can restore the chemical state when the charging source forces the charge to flow in the opposite direction throught the battery. This reverses the chemical reactions in the battery, something that does not work for all batteries of cause.
     
  25. Oct 5, 2016 #24
    Alright, I think I got it. Since we know that the DC source is charging the battery, therefore the A point on the graph should be positive. In case we were not told that the DC source is charging the battery or that we were told that they are assisting each other, in that case we would say that the A point is negative after looking at the battery.

    For (b):
    As @gneill explained the KVL, this is what I got:
    029f6ac37555.jpg

    And so the walk is: 24 - (1 * 5) - 5 R - (0.1* 5) - 12.6 = 0
    5 R = 5.9
    R = 1.18 Ω... This should be correct + I do understand what I did now.

    Though a couple of questions:
    1. Does it matter where we put the internal resistance? Before or after the source / battery? In my case I did put both of them above the source and battery. As I understood it, it doesn't matter where I put them, only that they should be near the source / battery.
    2. Does it matter where I start the walk? I started from the bottom left corner. Was I required to walk from the upper left corner? Is there a rule? Or if we just start from a corner and then come back to that same corner -- that's what is required?
    3. In the original picture we have the battery that had two long lines and two short lines and a dotted line in between. Is it OK if when re-drawing I just represent them as I did on my graph? I'm a bit confused on this part, since I didn't see anywhere a proper explanation what is the difference between just two lines (+ -) and the one that we have in the original graph. As I see it -- the battery is composed of two batteries (+ - and + -). This part is particularly important to understand since one can assume that each + - and + - is equal to 12.6 V or that we need to divide the number by two and draw + 6.3 V - --- + 6.3 V -.

    Update:
    I easily managed to do (c) : )!

    Since we have a new resistance of the R resistor, we need to find the new current (ccrrent is the same through the circuit since voltmeters have large resistance and don't consume current (at least we assume so)):
    24 - 1 I - 0.9 I - 0.1 I - 12.6 = 0
    2 I = 11.4
    I = 5.7 A.

    Then we do the small circuit DC + V1: 24 - 1 * 5.7 - V1 = 0, so V1 = 18.3 V. And then we make a larger circuit DC + V2: 24 - 1 * 5.7 - 0.9 * 5.7 - V2 = 0, so V2 = 13.17 V.
     
    Last edited: Oct 5, 2016
  26. Oct 5, 2016 #25

    andrevdh

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    1. Correct, since they are resistors the current will pass through them anyway.

    2. Correct again, since one traverses the whole of the loop it does not matter where you start, the terms will still be the same, just in a different order :)

    3. Correct again, I think the intention is to show that it consists of various separate cells connected together. In the case of a 12V lead-acid battery there will actually be 6 cells I think since each one generates slightly above 2V. Many batteries consists of several cells connected in series to give a higher voltage. Open some of them if you have the inclination.

    (c) the emf of each battery is just reduced by the voltage drop over their internal resistors - this is what the voltmeters indicates - the reduced emf.
    The way you wrote it worries me a (little :) ) bit - not seriously though. To get the voltage V1 you just to start at the one connection and walk around to the other connection I think. That should give you the V1 voltage so that is just amounts to
    , same with V2. The values will come out + or - depending on which way you walk around, usually it should be from the black lead of the voltmeter to its red lead.
    Well done!
     
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