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Internal Resistance

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data
    When a 10 ohm load is placed across the terminals of a battery, the terminal voltage is 11.0V. When a 100 ohm load is used instead, the terminal voltage is 11.9V. What is the internal resistance of the battery?


    2. Relevant equations

    V=IR

    3. The attempt at a solution

    I do not know how to approach this question but I do know that I have to use V = IR. I also know that the terminal voltage occurs after the voltage drop from the internal resistance. Is this right? A hint on how to approach this question would be nice! Thank you
     
  2. jcsd
  3. Feb 25, 2014 #2

    NascentOxygen

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    Draw the circuit you are dealing with, representing the battery itself as an ideal voltage source in series with a resistor, the resistor value at this stage being unknown, so call it Ri.
     
  4. Feb 26, 2014 #3
    I did do that. I feel like this question is impossible?
    So, terminal voltage = IR

    when using 10ohm resistor
    terminal voltage = 11 = I(10)

    when using 100ohm resistor
    terminal voltage = 11.9 = I(100)

    current should be the same in both cases, but when solving for current it isnt the same?
     
  5. Feb 26, 2014 #4

    phinds

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    Why would you think the current would be the same in both cases?

    When Nascent said draw the circuit, what he meant was DRAW THE CIRCUIT. And put it here where we can see it. And show your work relative to the drawing.
     
  6. Feb 26, 2014 #5
    also, the answer to this question is 0.917
     
  7. Feb 26, 2014 #6
    If the same battery is used, shouldn't the current in both cases be the same?
    Also here is a drawing of the circuit

    Untitled.png
     
  8. Feb 26, 2014 #7
    after finding the currents for both cases, use the relationship you listed, V=IR. The internal resistance should be equal to [itex]\frac{ΔV}{ΔI}[/itex] in this case. which will get you the answer you had.
     
  9. Feb 26, 2014 #8

    phinds

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    I continue to be dumbfounded at how you can imagine that would be the case. Do you believe that a battery feeding a million ohm resistor draws as much current as the same battery feeding a 1 ohm resistor?
     
  10. Feb 26, 2014 #9

    NascentOxygen

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    If it is the same battery, then its model will be the same: same ideal voltage, same internal resistance.

    Sorry, my mistake. I should have said, draw the circuits, because as you'll appreciate there are two circuits to deal with here. The first "when a 10 Ω load ..." and another "when a 100 Ω load ...". Two different circuits.

    So draw those two circuits, marking on each all the information you know.
     
  11. Feb 26, 2014 #10

    NascentOxygen

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    It must have units.
     
  12. Feb 26, 2014 #11
    Okay i got the answer, thank you. the internal resistance is 0.917 ohms,
    but can you please derive how you got [itex]\frac{ΔV}{ΔI}[/itex] ?

    thank you
     
  13. Feb 25, 2016 #12
    If anyone still wants to know how here:
    V = terminal voltage, E = EMF of battery, I = current from tests (will be different depending of resistances), r = internal resistance.
    V = E - Ir....... This eqn states that the voltage exiting and entering the terminals is equal to the total emf the battery would have without the internal resistance, minus the little bit of voltage drop in the battery due to internal resistance.
    Now subtract these equations derived from both scenarios (E is a constant so it cancels out)
    (V1 = E - I1r) - (V2 = E - I21r)
    Now:
    V1 - V2 = -I1r + I2r
    Factor out the r:
    V1 - V2 = (-I1 + I2)r
    divide:
    (V1 - V2)/(-I1 + I2) = r
    or:
    ∆V/∆I = r
    To Find I by the way, use V = IR, make it I = V/R, use the values given in the question, and plug em in.
     
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