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Batteries connected in Empty Circuit

  1. Nov 24, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    Three identical batteries, with finite internal resistance, are connected in series in a closed loop. (Positive Terminal of battery 1 to Negative Terminal of battery 2; Positive Terminal of battery 2 to Negative Terminal of battery 3; Positive Terminal of battery 3 to Negative Terminal of battery 1.) What reading will a voltmeter register if it is connected in parallel across one of the cells, and all the connecting wires have negligible resistance? What will happen to the voltmeter reading if the number of batteries is increased and the connection points of the voltmeter remain the same?

    2. Relevant equations

    Kirchoff's rule:

    [tex] \sum Voltage = 0 [/tex]

    3. The attempt at a solution

    Using Kirchoffs Rule,

    [tex] \epsilon_{battery 1} + \epsilon_{battery 2} + \epsilon_{battery 3} = 0 [/tex]


    [tex] I_1r_1 + I_2r_2 + I_3r_3 = 0 [/tex]


    But since the batteries are all the same, the current flowing through them would surely be equal, thus meaning that the only way to satsify this current would be to have 0 current flow, and hence no voltage? And thus for the second part, adding more batteries wouldn't make any difference?

    Is this the right explanation?

    ???

    TFM
     
  2. jcsd
  3. Nov 24, 2008 #2

    Redbelly98

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    That's correct. Notice that this is one single equation, relating the voltages of the circuit components in a single circuit loop.

    The problem here is that you have neglected the voltage drops due to the internal resistances ...

    ... and here, the voltages of the cells have been left out.

    Can you make ONE single equation from
    [tex] \sum Voltage = 0 [/tex]
    as you go around the loop? If you haven't already, try drawing a circuit diagram including the cells and internal resistances.

    Correct. That helps simplify things, since there is just one current "I", and not different I1, I2, and I3.
     
  4. Nov 24, 2008 #3

    TFM

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    So would the equation be:

    [tex] \left( \epsilon_{B1} - Ir_{B1}\right) + \left( \epsilon_{B2} - Ir_{B2}\right) + \left( \epsilon_{B3} - Ir_{B3}\right) [/tex] = 0

    But isn't everything the same, since they are identical batteries, thus:

    [tex] 3 \left( \epsilon - Ir\right) [/tex] = 0

    ???

    TFM
     
  5. Nov 24, 2008 #4

    Redbelly98

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    That looks right.

    So the measured "terminal voltage" is ____ ?
     
  6. Nov 25, 2008 #5

    TFM

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    Well rearrangeing the formula gives:

    [tex] 3\epsilon = 3Ir [/tex]

    so would the terminal voltage be [tex] \epsilon [/tex], which would be the voltage of the battery?

    TFM
     
  7. Nov 25, 2008 #6

    Redbelly98

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    "Terminal voltage" is the voltage measured at the physical terminals, or
    ε - I r
     
  8. Nov 25, 2008 #7

    TFM

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    I see so the voltage measured is the voltage of the battery minus that lost by internal resistance.

    Adding more batteries won't make any changes, also, since the voltage across thebattery is independent of any other batteries added.

    Does this make sense?

    TFM
     
  9. Nov 25, 2008 #8

    Redbelly98

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    Yes, that sounds right.

    By the way, this strikes me as a rather odd question, since this is not something people would actually do in practice (unless by accident). But I guess as an academic exercise it serves its purpose.
     
  10. Nov 25, 2008 #9

    TFM

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    Excellent.

    Thanks for your assistance, Redbelly94ג:smile:

    TFMגג
     
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