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Finding internal resistance & EMF of a battery in a circuit

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    When an external resistor of resistance R 1 = 14 Ω is connected to the terminals of a battery, a current of 6.0 A flows through the resistor. When an external resistor of resistance R2 = 64.4 Ω is connected instead, the current is 2.0 A. Calculate the emf and the internal resistance of the battery.

    2. Relevant equations
    Volts = I * R
    Power= I2* R
    Power= V * I
    (R=resistance)
    (I=Current)
    3. The attempt at a solution
    V = I * R
    V1= 6 * 14
    V1= 84 Volts

    V2= 2*64.4
    V2= 128.8 Volts

    I do not know how to continue
     
    Last edited: Apr 3, 2017
  2. jcsd
  3. Apr 3, 2017 #2

    Drakkith

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    Staff: Mentor

    I think you're missing something:

    How would you find the bolded part?
     
  4. Apr 3, 2017 #3
    No, I am not missing anything. I wrote it directly from a question in a quiz that I got wrong. In reference to the bolded part, I have no clue.
     
  5. Apr 3, 2017 #4

    gneill

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    Staff: Mentor

    It's the implications of the bolded part that you're missing.

    Start by making a sketch of the circuit (say for the first case with the 14 Ω resistor). Write KVL for the loop and show us what you get.
     
  6. Apr 4, 2017 #5

    PeterO

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    Homework Helper

    Nice question. Where is your attempted solution.
    Hint: Treat the circuit as an ideal cell with two resistors in Series - one of them is the Internal resistance, which cannot change, the other is the external (circuit) resistance. You can generate a pair of simultaneous equations to find the Emf of the ideal cell and the internal resistance. You were not asked for the Emf of the cell, so once you have the Internal resistance value you can actually stop.
     
  7. Apr 4, 2017 #6

    Drakkith

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    Staff: Mentor

    When it comes to batteries, do you know what internal resistance is?
     
  8. Apr 4, 2017 #7
    Internal resistance is the resistor inside a battery and next to the emf, which is like the source, and together they give out the terminal voltage right? and I don't understand what KVL means.
    I drew the circuit and if I treat the circuit as an ideal cell with two resistors in series would it mean that
    84 + 6r =emf ?
    and
    128.8 + 2r= emf ?
     
  9. Apr 4, 2017 #8
    OHH
    I got:

    r = 11.2 Ω
    and
    Emf = 151.2 V

    Would that be right??? : )
     
  10. Apr 4, 2017 #9

    gneill

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    Staff: Mentor

    KVL is Kirchhoff's Voltage Law. KCL is Kirchhoff's Current Law. You should look those up as they along with Ohm's Law are the basis of all circuit analysis and you will need them for almost every problem involving circuits.

    What you calculated in your first post was the voltages across the external resistors, which also happens to be the voltages presented at the battery terminals in each case.

    So an expression like "84r" would imply a voltage multiplied by a resistance, which doesn't make sense in terms of Ohm's Law, which would have you multiply current by resistance to make a voltage. Again, look at writing KVL loop equations for each case so that you will have two equations in the two unknowns that you want to find.

    Without doing the math myself I'd say l that looks reasonable for the given data. Are you clear on the steps you took to get there?
     
  11. Apr 4, 2017 #10

    Drakkith

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    Staff: Mentor

    For part 1:
    Since ##I = \frac{V}{R}##, and ##R## is ##11.2 + 14=25.2##, then ##\frac{151.2V}{25.2Ω} = 6A##

    That looks good to me. You can double check part 2 if you'd like.
     
  12. Apr 4, 2017 #11
    Thank Youu!!
     
  13. Apr 4, 2017 #12
    Ok! and Yess I understood the steps. Thank Youu!
     
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