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Internal resistivity, is this right?

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data
    To a source of current, one by one are added 2 resistances, first R1=1ohm, then R2=4ohm. In both cases, in the same period of time (t=t1=t2), the resistors give the same amount of heat (Q=Q1=Q2). Determine the internal resistance ("r" small r) of the electric source.


    2. Relevant equations
    I=V/R

    3. The attempt at a solution
    I found somewhere a formula which states: r=(V2-V1)/(I1-I2), so including the basic R=V/I into it would give: r=R2-R1, and the result will be 3ohm? looks too simple for an engineering problem.
     
  2. jcsd
  3. Dec 21, 2011 #2

    gneill

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    Staff: Mentor

    I don't see how that formula would apply here. What values would you use for the V's and I's?

    Instead, assume that the current source consists of and ideal voltage source E and a series resistance r. Then work out the power dissipated in the attached resistors for the two cases described. What formulas do you know for the power dissipated in a resistor?
     
  4. Dec 21, 2011 #3
    well, it looks tricky. here is a variant, which was suggested buy another guy on the internet:

     
  5. Dec 21, 2011 #4

    gneill

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    Staff: Mentor

    One problem with the suggested solution -- the total external resistance for the two cases should be 1 Ω and 5 Ω if, as the problem statement says, one resistor is added at a time.
     
  6. Dec 21, 2011 #5
    I think I translated it bad, 'cause I'm not English, but generally it's not added, this is the same power source, only resistors are changed, this is why we have 2 time values which are equal, but are not important, so they didn't gave us them. thanks for the idea :)
     
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