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Finding internal resistance from 2 circuits

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A battery with an emf of 60 V and an internal resistance (r) is connected to a 10 ohm external resistance. The power lost inside the battery is 50w. The same battery is then connected to a 4.0 ohm resistance. The power loss is 200w. What is the internal resistance of the battery?
    Diagrams:
    http://i.imgur.com/LUF75yR.png

    2. Relevant equations
    P=VI
    P=I2R
    V=IR
    P=V2/R

    3. The attempt at a solution
    P=VI1 4P=VI2

    VI1=VI2/4

    4 x I1=I2

    4(V/(r+10))=(V/(r+4.0)

    4/(r+10)=1/(r+4.0)

    4(r+4.0)=r+10

    4r+16.0=r+10

    r=-2 Ω
     
  2. jcsd
  3. May 25, 2013 #2

    CWatters

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    Science Advisor
    Homework Helper

    Re:

    and

    If the current is x4 different would the voltage drop across the internal resistance be the same?
     
  4. May 25, 2013 #3
    The voltage drop across the internal would be the same as we're using the same source and the only difference is the external resistor
     
  5. May 25, 2013 #4

    gneill

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    Staff: Mentor

    So, for the same resistance (the internal resistance), if the current changes the voltage drop remains the same??? What happened to Ohm's Law?
     
  6. May 25, 2013 #5
    But the thing is that we have r+10 and r+4 equating to our resistance, so couldn't it be the r be the same number though?
     
  7. May 25, 2013 #6

    gneill

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    Staff: Mentor

    Yes, the internal resistance r stays the same. No, the potential drop across r is not constant if the current changes. Ohm's law is, well... it's the law! :smile:

    Try writing an expression for the power developed by r for a given load (say the 10 Ohm load). Knowing that the result should be 50W, solve for r. (you should find a quadratic equation that returns two potential solutions). Do the same for the other load. What do you find?
     
  8. May 26, 2013 #7
    Thank you so much! I enjoy that you are not exactly spoon feeding me the answer but instead going through the logistics of the concepts! I will apply those and come back if a problem arises
     
  9. May 26, 2013 #8
    A new problem has arisen when I use the r value I find to plug into other power loss formulas, ie P=VI and P=V^2/R the answer are different.

    Attempts:

    P=VI I=V/R
    P=60(60/(10+2))

    P=V^2/r
    =60^2/2
     
  10. May 26, 2013 #9

    gneill

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    Staff: Mentor

    The voltage across r is not the same as the voltage across r + 4Ω, or r + 10Ω. That is, the whole emf of 60V does not appear across r.

    Rather than inserting additional steps to find the portion of the 60V that appears across r in each case, employ another expression for the power dissipated by a resistor R with current I flowing through it: P = I2R
     
  11. May 26, 2013 #10
    Makes sense. I guess I wasn't thinking clearly. Thank you so much for the help though!
     
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