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EMF/internal resistance question

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data
    http://session.masteringphysics.com/problemAsset/1042319/4/YF-25-68.jpg" [Broken]

    Above is a picture of the circuit!
    The question is: "What is the terminal voltage of the 4.00-V battery?"

    2. Relevant equations
    I=V/R
    V_ab = E - I*r


    3. The attempt at a solution
    So first, I calculated the current in the loop...
    the voltage is 8v-4v = 4v
    and the total resistance is...6+.5+9+.5+8 = 24 ohms
    so I = V/R =4/24 = 1/6

    from this...
    V_ab = 4v - (1/6)*(.5) = 3.917v

    however, the program I am using (masteringphysics) responded "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures" when I entered this in...
    I have no idea where I'm going wrong because in a later question, I used the same method and got the right answer...help!
    Thanks!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 12, 2010 #2
    you may have used the wrong number of significant figures
     
  4. Feb 12, 2010 #3
    no...I double checked that
     
  5. Feb 12, 2010 #4
    how many significant figures are there in 4.00 V
    hom many in 3.917 V
     
  6. Feb 12, 2010 #5
    I entered the value twice, once as 3.917 and the second time as 3.92...both times it had the same error message.
     
  7. Feb 12, 2010 #6
    Sorry, I didn't see this at once. What is the direction of the current? And what is the direction
    of the potential difference across the upper 0.50 ohm resistance?
     
  8. Feb 12, 2010 #7
    the direction of the current is clockwise...but i'm not sure what you mean by the direction of the potential difference...the voltage is higher on the right side (in the picture) and lower on the left side...
     
  9. Feb 12, 2010 #8
    If the current goes clockwise, it will make the left side of the top 0.50 ohm resistance more positive to the right side. Therefore the potential difference has to be added to the 4.00V of the battery.

    If the 8V voltage source is replaced by a wire the current would go in the other direction and the potential difference across the internal resistance of the top battery would have to be subtracted from the EMF of the battery, but it has to be added here.
     
  10. Feb 12, 2010 #9
    Thanks a lot! that worked and it does make sense...I was just plugging numbers into the formulas that I have and not really thinking about the problem lol.
    Thanks again!
     
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