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Emf of a battery

  1. Mar 31, 2011 #1
    1. The problem statement, all variables and given/known data
    determine the terminal voltage in each battery.
    Emf1=18V
    r1=1 ohm
    EMF2=12V
    r2=2 ohm
    R=6.6 ohms
    also the batterys currents are opposing each other.
    2. Relevant equations Vab=emf-Ir


    3. The attempt at a solution i used kirchnoffs rule to find the current I , 18-I6.6-12-I2-I=0 and got I= .625 then i thought i would just use Vab=emf-Ir to find the terminal voltage of each battery however im not getting the correct answer... could someone please explain this to me i have a test tomorrow morning and i really need to figure this out! the answers i got were 17.375 for the 18V batt and 10.75 for the 12V batt
     

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    Last edited: Mar 31, 2011
  2. jcsd
  3. Mar 31, 2011 #2

    gneill

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    Staff: Mentor

    Watch out for the current direction and the thus the polarity of the voltage drop it creates on the internal resistors. If current is being forced into a battery's positive terminal it's possible for it to have a higher terminal voltage than its internal EMF.
     
  4. Mar 31, 2011 #3
    so on the larger voltage batt would i use equation Vab=Emf-Ir = 18-(.625*1) and on the batt with voltage being forced across it use Vab=Emf+... idk
     
  5. Mar 31, 2011 #4
    what are the equations you would use in this situation?
     
  6. Mar 31, 2011 #5

    gneill

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    Staff: Mentor

    Yes. It can be helpful to sketch in your current and the polarities of the voltage drops across the resistors. You did fine writing the KVL loop equation.
     

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