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What is the controversy- current and resistance

  1. Dec 22, 2007 #1
    http://img339.imageshack.us/my.php?image=controversypo0.jpg

    In the above link i have shown a circuit, where in the DC supplies are ideal (i.e. r = 0).
    What current will pass through resistance R ,... V1/R or V2/R.
     
  2. jcsd
  3. Dec 22, 2007 #2

    Hootenanny

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    Why do you want to know?
     
  4. Dec 22, 2007 #3
    You cant divide by 0
     
  5. Dec 22, 2007 #4

    Hootenanny

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    Although true, totally irrelevant.
     
  6. Dec 22, 2007 #5

    mda

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    I am guessing (since the image won't load) that you've got two different supplies in parallel with a load resistor?

    Essentially you cannot have r=0 exactly... you'd have to take a limiting case which is highly dependent on the details of the power supplies. Examples include known r, limited current, perhaps the supplies cannot sink current etc...

    Once you have the non-ideal characteristics of the supplies specified it is reasonably easy to determine the result.
     
  7. Dec 22, 2007 #6
    As I understand this, R is not 0; r(internal) is zero and you can do the ideal case without too much worry.

    So, what do you know about currents at a junction?
     
  8. Dec 22, 2007 #7
    i want to know the current in R.
    r is the internal resistance of the batteries, which i have assumed to be zero.
     
  9. Dec 23, 2007 #8
    It is just two cells (power supplies) in parallel to a resistor of value R

    The talk about little r confuses things, so just say in the ideal case. Since this is intro physics and a very basic circuit.

    Two cells in series (say 2x1.5V, 500mAh types) gives you what current and what voltage if acting as a battery ?
    Two cells in parallel (say 2x1.5V, 500mAh types) gives you what current and what voltage if acting as a battery ?

    If you cannot answer this basic question, then I suggest picking your the first topic that pops up in google for say 'cells in parallel' and have a read

    http://www.batteryuniversity.com/partone-24.htm
     
  10. Dec 23, 2007 #9
    ah.. thx for your answer. .. i m in the process of learning basics only.
    In the parallel combination there, both the voltages are same, i.e. 1.2V, so the potential difference is 1.2 V. But if suppose one battery is 1.2 V and other is 1.5 V, then what is the potential difference we should say finally.
    The way i have understood is that (1.2 V and 1.5 V ); 1.2 V is a lower voltage, so we will take it will just give low run time, and the final voltage across will be 1.5 V only.

    So, for the question that i asked....the current i = max (V1, V2)/R.
    If there is something wrong in my understanding, can u please correct me if possible.
    thx again
     
  11. Dec 23, 2007 #10
    If you really want to learn, go and do it...

    Get yourself some batteries (not mians power supplies), a couple of resistor and voltmeter and ammeter.
    Have a play and see what you get.

    That should answer your question...
     
  12. Dec 23, 2007 #11
    i wish i had that facility with me to experiment. Can you just tell me if i m thinking in the right direction?
     
    Last edited: Dec 23, 2007
  13. Dec 23, 2007 #12

    Hootenanny

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    Sounds good to me :approve:
     
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