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Kirchoff's laws, emf and internal resistance

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    two identical cells of emf E and internal resistance r are connected in series.
    a 7 ohm resistor is connected across the combination and draws a current of 0.333A.
    the two cells are now connected in parallel; the 7 ohm resistor now draws a current of 0.375A from the combination.

    calculate the emf and internal resistance of the cells.

    2. Relevant equations

    E - Ir = IR

    Sum of potential differences in a closed loop = 0
    Current entering a junction = current leaving a junction

    3. The attempt at a solution

    i managed to get one equation from the series circuit:
    2(E - 0.333r) = 0.333 x 7

    and an equation for the parallel circuit:
    E - 0.1875r = 0.375 x 7

    but this didn't give the correct answers from the book, which says that E = 1.5 and r = 1
     
  2. jcsd
  3. Sep 15, 2009 #2

    berkeman

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    Welcome to the PF. Good job showing your work -- that makes it a lot easier.

    I don't see how the solution of E = 1.5V can be correct. The 0.333A through the 7 Ohm resistor gives a voltage of 2.33V, so connecting two 1.5V batteries in parallel can't get you up there... Any chance of a typo or misunderstanding in the problem? Your equations look correct to me.
     
  4. Sep 15, 2009 #3
    im a bit suspicious of the answers in the book, theyve been wrong before!
    but i dont want to assume that its always the book thats wrong, so maybe ive missed something?

    the equations i got give E = 4.5V and r = 10 ohms, which i think seems too high.
     
  5. Sep 15, 2009 #4

    berkeman

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    That's the same numbers that I got with your two equations (which I think are correct equations for the described problem). Either set of numbers (book's or ours) work for the first equation, but only ours fit the 2nd equation. What would be the current for the 2nd case (parallel batteries) to give the book's answers?
     
  6. Sep 15, 2009 #5

    Redbelly98

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    I agree with the two equations you set up. In fact, the result of the parallel circuit implies an EMF of more than 0.375*7=2.625V, so the 1.5V answer has to be wrong.

    EDIT: berkeman beat me.
     
  7. Sep 15, 2009 #6
    you can get the answers in the book by taking the currents in the parallel equation as being the same, both 0.1875- which gives r = 1, but i dont see why this would be the case at all. surely taking one complete loop and applying kirchhoff's rule, the current through each of the cells would be half that through the resistor?
     
  8. Sep 15, 2009 #7

    Redbelly98

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    Yes, each cell generates 1/2 of the total resistor current.

    Berkeman was saying that the parallel current of 0.375A (0.1875A through each cell) is inconsistant with the book answers of 1.5V, 1Ω per cell. So, what would the total current in the parallel circuit have to be if each cell has 1.5V and r=1Ω, and they are connected in parallel to a 7Ω resistor?
     
  9. Sep 15, 2009 #8
    i worked it out to be 0.2A.

    ive just posted another question on this, where my answer disagrees with the answer in the book. i think the solution should be to get a new book!
     
  10. Sep 15, 2009 #9

    Redbelly98

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    Looks right. So the mistake in the book is either with the final answer, or that they gave the wrong current for the parallel circuit case.

    Getting a different book is an option if you are doing self-study.
     
  11. Sep 15, 2009 #10

    berkeman

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    :rofl: Classic!
     
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