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Determining the total current of many batteries

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data
    The figure represents a part of a circuit, and the potential difference between the f and d is 12Volts, depending on the information shown in the figure, determine:
    1-the reading of the Ammeter
    2-the value of (Vb)₃
    3- the potential difference between (c , b)
    vEhMp.jpg
    2. Relevant equations
    Ohm's law: Vb=I (R+r)
    Kirchhoff's law: ΣI=0
    ΣV=ΣIR

    3. The attempt at a solution
    Seriously, I got confused with all these batteries, and I considered that the total current is 3A, since Vfd=12. bFHZJ.jpg So, I know there may be a lot of mistakes in my solution, and I would appreciate any help or explanation of how those whole batteries would work and produce the total current.
     
    Last edited: Dec 18, 2016
  2. jcsd
  3. Dec 18, 2016 #2

    cnh1995

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    Answers to Q1 and Q2 look good. In Q3, you are asked to find the potential difference between c and b. I think instead you have calculated Vcd (which is also not correct).

    (Also, I've never seen Kirchhoff's circuit laws named as 'first law' and 'second law'. They are called Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL).)
     
  4. Dec 18, 2016 #3

    ehild

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  5. Dec 18, 2016 #4
    I made a mistake, I wrote Vcd, but I meant Vcb, I edited the photo. And in my text book the Kirchhoff's current law can be the first law and the voltage law is named the second law.
    So please, is the answer correct now?
     
  6. Dec 18, 2016 #5

    cnh1995

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    You got 3A flowing through the 3 ohm resistor towards right and the 7V battery has an internal resistance of 1 ohm. This means, between c and b, you have 3A current flowing towards right through a 4 ohm resistor and a 7V source in series with that. That dosen't give 13V between b and c.
     
  7. Dec 18, 2016 #6
    The inner resistance of the battery would be used in Ohm's law, so the PD between the terminals of the battery is (V=Vb-Ir) so it equals 4V and this voltage is in series with the voltage on the 3Ω resistance which I would use Ohm's law to determine it (V=IR) so it equals 9V.
    Then the total voltage would be (9+4=13V), Is that wrong, according to what you see I think that my mistake may be in the way I calculate them, so if you see any mistake please clarify it for me!
    Where : r is the internal resistance, V is the potential difference, I is the current and Vb is the emf.
     
  8. Dec 18, 2016 #7

    cnh1995

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    7V is the emf of the battery and 1 ohm is its internal resistance which is in series with the 3 ohm resistance. So, you have a 4 ohm resistor in series with an emf of 7V.
    Right, but what will be the polarity of that voltage? Is it a potential drop or a potential gain?
     
  9. Dec 18, 2016 #8
    It is a potential drop, so it would be like this:
    V=IR=3*4=12V
    And then should I add them to the emf of the battery, so the total PD would be 19V, or add them to the PD between tne two terminals of the battery and then the total PD would be 16V?
    Sorry if I understand slowly or ask a lot of question:biggrin:!​
     
  10. Dec 18, 2016 #9

    cnh1995

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    No.
    Just redraw the circuit from c to b.
    Draw a 4 ohm resistor (3 ohm + 1 ohm) in series with the battery of 7V emf. Note the polarity of the battery. A current of 3A is flowing through the c-b branch from c to b. So, when going from c to b
    i) What is the potential change across the 4 ohm resistor? Is it a gain or a drop?
    ii) What is the potential change across the battery? Is it a gain or a drop?
    Add the answers of i) and ii) and you'll get Vcb.
    Please don't be!
     
  11. Dec 18, 2016 #10

    gneill

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    I note that the problem statement leaves it a bit vague as to the polarity of the 12 V potential difference between nodes f and d. To me that means that there are two valid solutions to the problem.
     
  12. Dec 18, 2016 #11
    Ok, MRjx2.jpg
    i) V= IR=4*3=12V (a drop)
    ii) V= Vb-Ir=7-3=4 ( a gain because the current of the battery is going out from the positive to the negative) I use the conventional direction.
    So 12+4=16V
    Or what did you mean by the potential change across the battery?
     
  13. Dec 18, 2016 #12
    Could you please explain how would the polarity of 12 V PD between f and d change the case? I didn't understand what do you mean?
    And it is not given in the peoblem statement so I considered the current would flow from f to d.
     
  14. Dec 18, 2016 #13

    cnh1995

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    Right.
    No. The internal resistance is taken care of already in i). So in ii), you must consider the emf of the battery, which is 7V.

    You are right about the potential gains and drops in i) and ii).
    So, what is Vcb now?
     
  15. Dec 18, 2016 #14

    gneill

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    Consider if the current flowed from d to f instead.

    Anyways, you should finish up your analysis of your version of the problem first before considering this second version.
     
  16. Dec 18, 2016 #15
    So it would be 19V?
     
  17. Dec 18, 2016 #16

    cnh1995

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    No. You are adding a drop and a gain.
     
  18. Dec 18, 2016 #17
    So what should I do?, you are going to make me cry.
     
  19. Dec 18, 2016 #18

    cnh1995

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    From c to d, you encounter a 12V "drop" in potential across the resistor and a 7V "gain" across the battery.

    You spent 12 dollars and earned 7 dollars. How many dollars did you effectively earn or spend?
    I don't know if I can make it any simpler.
     
  20. Dec 18, 2016 #19
    I lost 5 dollars so the potential difference would be 5V and that is a drop in the voltage.
    Is this correct?
     
  21. Dec 18, 2016 #20

    cnh1995

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    Right!

    Now you can think about the second version of the problem as gneill mentioned earlier.
     
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