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Internal resistance

  1. May 18, 2008 #1
    The battery of an electric car consists of 30 cells, connected in series, to supply current to the
    motor.

    (a) Assume that the internal resistance of each cell is negligible and that the pd across each
    cell is 6.0V.
    (i) State the pd across the motor.
    (ii) The battery provides 7.2kW to the motor when the car is running. Calculate the
    current in the circuit.
    (iii) The battery can deliver this current for two hours. Calculate how much charge
    the battery delivers in this time
    (iv) Calculate the energy delivered to the motor in the two hour period.

    b)In practice, each cell has a small but finite internal resistance. Explain, without
    calculation, the effect of this resistance on
    - the current in the circuit, and
    - the time for which the battery can deliver the current in part (a)(ii).
    You may assume that the motor behaves as a constant resistance.

    I can do part (a) with ease, and the first part of b also. However, I am unable to do the part of b which asks for the effect of internal resistance on the time for which the battery can deliver the current in part (a)(ii).

    Please can someone help me with this.


    Thanks
     
    Last edited: May 18, 2008
  2. jcsd
  3. May 18, 2008 #2

    Hootenanny

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    Consider the internal resistance as a normal resistor connected in series with the cells. What happens when a current passes through a resistor?
     
  4. May 19, 2008 #3
    Work is done against the internal resistance, resulting in their being a potential difference across the intrenal resistor. This results in the potential difference across the terminals of the cell being less. Consequently, the current flowing will be less. I don't understand how the presence of the internal resistance determines how long the battery can deliver current for. Also, I thought that the current will be lkess than in a(ii), so surely it will be a lower current that flows?

    Thanks
     
    Last edited: May 19, 2008
  5. May 19, 2008 #4
    Is it anything to do with the total energy or charge a cell can deliver?
     
  6. May 19, 2008 #5

    Hootenanny

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    Correct! So if work is done against the resistance, where does this energy go?
     
  7. May 19, 2008 #6
    Does it appear as heat?
     
  8. May 19, 2008 #7

    Hootenanny

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    Indeed it does. So if some amount of energy is radiated away from the circuit as heat, what happens to the total energy of circuit and hence the total energy stored in the cells?
     
  9. May 19, 2008 #8
    The total energy of the circuit decreases, and so less is stored in the cells?
     
  10. May 19, 2008 #9

    Hootenanny

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    Correct :approve:
     
  11. May 19, 2008 #10
    So would this not mean that the current flows for less time?
     
  12. May 19, 2008 #11

    Hootenanny

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    Correct. Since the total resistance of the circuit is greater than without an IR, this means that more energy will be radiated away as heat per unit time and hence, the total energy stored in the cells will decrease at a greater rate.
     
  13. May 19, 2008 #12
    Thanks alot for the help! This is also what I thought. The mark scheme says the opposite though:

    "(total) charge circulated by battery remains the same [or valid energy reasons]
    time for which (reduced) current flows is increased"

    I am unable to understand the reasoning in the markscheme!
     
  14. May 19, 2008 #13
    The examiners report says "Candidates who deduced correctly that the time increased
    usually approached the problem from energy or charge considerations." I dont know if that helps, since I do not understand why the time increases.
     
  15. May 19, 2008 #14
    Can anyone understand the reasoning of the markscheme, it is really playing on my mind!

    Thanks
     
  16. May 20, 2008 #15

    alphysicist

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    Hi nokia8650,

    The wording of the initial question seems a bit strange, but the markscheme seems correct to me. When the current is reduced due to adding resistance, the power output of the (constant voltage) battery will decrease due to:

    [tex]
    P=\frac{V^2}{R}
    [/tex]

    Then, since power is energy change per time, a smaller power will take a longer time to decrease the same amount of energy.
     
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