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What's the equivalent resistance between point A and B?

  1. Jan 26, 2016 #1
    Simple as that, I've uploaded the image. I don't really care about the exact value, I would prefer just to know the method. So how would I find the equivalent resistance between A and B? Note the component in the middle is a constant current source.
     

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  3. Jan 26, 2016 #2

    Averagesupernova

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    Well what do you know about current source? I will give you a hint. An ideal VOLTAGE SOURCE has ZERO internal resistance.
     
  4. Jan 26, 2016 #3
    Ah, so an ideal current source has infinite internal resistance, and hence its just normal series and parallel resistors that are left? But I don't understand, how does current go through something with an infinite resistance?
     
  5. Jan 26, 2016 #4

    cnh1995

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    Ideal current source is modelled as current source with infinite resistance in parallel. This way, no current flows through the parallel resistance and all the current flows through the load. You are asked to find the Thevenin's resistance between A and B. Hence, you need to replace the current source by it's internal impedance, which is infinity in this case. Hence, while calculating RAB, you just open the current source.
     
  6. Jan 26, 2016 #5

    Averagesupernova

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    It doesn't. That's why the circuit you posted is so easily solved. Just ignore the current source.
     
  7. Jan 27, 2016 #6
    Okay, I almost understand. Am I correct if I say, no extra current (say if I connected a battery across A and B) can go through the current source because the current source is designed so that it maintains a constant current. So because no extra current can go through it, it appears as if it has infinite resistance?
     

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  8. Jan 27, 2016 #7

    cnh1995

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    Well, an ideal voltage source in parallel with an ideal current source is an invalid combination in network analysis. Indeed the current source is designed to provide a constant current. If any load is connected to a current source and all the current passes through the load, then there is no "leakage" current and hence, internal (parallel) resistance of an ideal CS is infinite.
    This is a diagram showing source transformation with 'practical' sources.
    upload_2016-1-27_20-27-18.png
    Practical current source with parallel resistance is equivalent to practical voltage source with a series(intetnal) resistance. Note that both the resistances are same. In CS, it appears in parallel(current division takes place) while is VS, it is in series(voltage division takes place). If you use load resistance as 2Ω, you can see, in VS diagram, voltage across load is 5V. It is same as in the CS diagram, where current through the load is 2.5A.
    Edit: First line corrected by Cabraham. It is valid actually.
     
    Last edited: Jan 27, 2016
  9. Jan 27, 2016 #8
    Regarding a voltage source in parallel with a current source, actually it is quite valid, if I may disagree. If the constant voltage source (CVS) is 1.0 volts, and the constant current source (CCS) is 1.0 amps, they can be connected together in parallel. One will transfer power to the other depending on polarity. If the CVS has the plus terminal on top, CCS has arrow upward, then the CCS charges the CVS, with 1.0 watts transferring from CCS to CVS. Reverse the polarity of just 1 or the other, not both, and the CVS transfers 1.0 watt of power into the CCS. Otherwise I agree with the rest of your post, thanks.

    Claude
     
  10. Jan 27, 2016 #9

    cnh1995

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    Right! I was just about to edit:wink:! Actually, two ideal voltage sources in parallel or two ideal current sources in series are the invalid cases. Thanks for the correction!
     
    Last edited: Jan 27, 2016
  11. Jan 27, 2016 #10
    Okay, yeah that makes sense. So was what I said last time about the excess current can't enter the current source (because the current needs to stay constant), hence it has effectively infinite resistance a good/correct way of thinking about it? Because that's the conclusion I have kind of come up with from what you've said. Because then that would explain why the total resistance of the current source as seen by the other components is infinite, when actually only the internal parallel resistance of the current source is infinite and the part where the current actually travels through has zero resistance.
     
  12. Jan 27, 2016 #11
    Thanks for all the replies by the way :)
     
  13. Jan 27, 2016 #12

    cnh1995

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    Sounds right. If the current source is not ideal, the excess current will flow through it's internal resistance. Hence, the voltage source connected across it will "see" the internal resistance instead of infinite resistance.
     
  14. Jan 27, 2016 #13
    Great. Thanks a lot :)
     
  15. Jan 27, 2016 #14

    Averagesupernova

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    Actually, this would be a battery charger. A real world practical example of such a network.
     
  16. Jan 27, 2016 #15

    Averagesupernova

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    I don't see what is invalid about 2 current sources in parallel either. If they were in series I would say that is invalid.
     
  17. Jan 27, 2016 #16

    cnh1995

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    That's what I wrote.
     
  18. Jan 27, 2016 #17

    Averagesupernova

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    I really need to see my optometrist. LOL
     
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