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Find the internal series resistance of the battery

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A battery is connected with a resistor R1=4 om and then it is replaced with the resistance 9 om. In both cases the heat released in the same time is the same. Find the inner resistor of the battery.

    2. Relevant equations
    Q=UIt (U-tension; I-intensity, t-time)
    I=e.m.f/R+r

    3. The attempt at a solution
    UI1*t=U*I2*t U and t are constant
    I1=I2
    emf/R1+r=efm/R2+r
    But it results R1=R2
     
  2. jcsd
  3. Jan 7, 2016 #2
    Your assumption that the currents (I1 and I2) are equal is not correct.

    The power delivered to the load is maximal when the load resistor is equal to the internal resistance, so without calculation one can say that it is between 4 ohm and 9 ohm.
     
  4. Jan 7, 2016 #3
    Can you give me a hint to find it correctly? From which formula should I start?
     
  5. Jan 7, 2016 #4

    cnh1995

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    If you sketch the graph of power vs resistance, you'll see it is symmetrical on both the sides of the peak power. Here, since the power is same, where do you think should the resistance be between 4 and 9?
     
  6. Jan 7, 2016 #5

    cnh1995

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    If you want to go mathematically,
    P=I2R will be helpful. Make proper substitution for I in terms of E and r.
     
  7. Jan 7, 2016 #6
    That is not correct: the power is zero at zero resistance (short circuit) and infinite resistance (open circuit).

    @prishila: ok then, one more hint: the current is larger when the load resistance is 4 ohm than when it is 9 ohm.
     
  8. Jan 7, 2016 #7
    r=6 ohm Thank you!
     
  9. Jan 7, 2016 #8

    cnh1995

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    Yep, you're right! I was about to edit..That graph takes infinite time to get back to zero since R is in the denominator. Sorry PietKuip! I need to be careful when guessing the graph without actually plotting it.
     
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