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Circuit theory -- Current source, voltage source, resistor in series

  1. Sep 16, 2014 #1
    1. The problem statement, all variables and given/known data
    the voltage Vc in the figure is always equal to
    ret_zpsb7be4b6c.jpg

    2 Relevant equations



    3. The attempt at a solution

    From kirchoffs law,
    -Vc+4+5=0
    Vc=9 V
    will Vc always equal to 9 V or it will vary?
     
  2. jcsd
  3. Sep 16, 2014 #2

    rude man

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    Right.
     
  4. Sep 16, 2014 #3

    phinds

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    Since the point of this forum is to help people learn how to get their own answers, I would not normally give a direct answer to such a simple question, BUT ... I really think this one is very tricky for a beginner to wrap their heads around because the circuit is so unrealistic to the real world, so I'll make an exception in this case and explain it to you rather than just offering hints.

    Vc can be anything you would like for it to be.

    The ideal current source effectively cuts off the circuit from the rest of the world (but requiring that the rest of the world be able to handle a 2amp suck from it. The voltage at Vc is whatever the rest of the world sets it to be (but keeping in mind that the rest of the world has to be happy with a 2amp suck). The voltage at the point between the 2amp source and the 2ohm resistor IS always at 9V, but the current source doesn't care how much voltage is across it and because of the way that particular circuit is constructed, the 5volt source is irrelevant to the rest of the world and has no impact on Vc.

    So let's just say arbitrarily that we want Vc to be 100 volts. Simple ... just put a 200 volt source in series with a 50 ohm resistor and the 50 ohm resistor hooked to the current source and you drop 100 volts over that 50 ohm resistor, with 2 amps flowing through it, thus keeping the current source happy at 2 amps and Vc at 100 volts. Similarly, you can set Vc to be any voltage you like with proper adjustment of the "rest of the world's" resistor and source values. The choice of a resistor and voltage source for the "rest of the world" was arbitrary and for simplicity. You could have other constructs there and still vary Vc however you like, always with the constrain that it has to suck 2 amps.

    Can you see how this is very unrealistic, and rather silly, in the real world?

    Can you see how the 5 volt source is irrelevant to the problem, and could have any value or just not be there, and cause no change in Vc?
     
    Last edited: Sep 16, 2014
  5. Sep 16, 2014 #4

    CWatters

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    Is that really the problem statement word for word? I suspect not because it can't easily be answered.

    At best you can say...

    VC = 5V + 4V + whatever the voltage is across the 2A current source.

    The latter term is unknown.

    The current source will adjust it's voltage so that KVL holds. That depends on Vc. If the current source is an ideal current source then Vc could be anything. It could be negative or huge (eg-1,000,000V). The current source will do whatever it takes to make KVL hold. That allows Vc to be anything.
     
  6. Sep 16, 2014 #5
    Vc can't be just anything. Vc depends on what's inside the box.

    If the circuitry in the box is represented by its Thevenin equivalent: a voltage source Vth in series with a resistance Rth, then Vc = Vth - 2*Rth.

    A combination of a current source in series with a resistor and a voltage source behaves just like the current source alone. Vc won't depend on the 2Ω resistor and the 5V source; it depends only on what's in the box; not only is the 5V source irrelevant, so is the 2Ω resistor.

    If nothing is in the box, just a pair of terminals hanging in the air, then Vc→∞, and Vc will be whatever it takes to sustain the arc across the terminals.

    For some really kinky circuit elements, see:

    http://en.wikipedia.org/wiki/Norator

    http://en.wikipedia.org/wiki/Nullator

    http://en.wikipedia.org/wiki/Nullor
     
  7. Sep 16, 2014 #6

    rude man

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    You are right, C.W. I misinterpreted the current source as an ammeter ... :redface:
     
  8. Sep 17, 2014 #7

    CWatters

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    Now that you mention it perhaps it is an ammeter. It would make some sense given the odd wording of the question.
     
  9. Sep 17, 2014 #8

    phinds

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    Yeah, but that IS the traditional symbol for a current source, whereas the ammeter is usually just an A in a circle
     
  10. Sep 17, 2014 #9

    NascentOxygen

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    If the needle pointed to 1 o'clock it would be acceptable as an ammeter. Which makes me wonder has OP's effort at reproducing the figure from the textbook erred on this seemingly-small detail.

    What do you say, anand raj?
     
  11. Sep 17, 2014 #10

    phinds

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    Yeah, that could be it. It's probably an easy mistake for a beginner to make. I'm still leaning towards the current source.

    What do you say, anand raj?
     
  12. Sep 17, 2014 #11

    NascentOxygen

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    Though even if it is established that it's an ammeter, we are still left with an unanswerable question.

    What's in the box? Perhaps anand raj could tell us if there is some overlooked detail that should be shared. I'm thinking the box represents a variable load. But only because if it were to represent a fixed load, the question would seem trite.

    Nominating an ammeter reading of 2A seems pointless and confusing if the load is intended to be variable, though.
     
  13. Sep 17, 2014 #12

    rude man

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    If it's an ammeter it doesn't matter what's in the box, long as it results in a current of 2A:

    Some examples:
    1. a 9V battery with zero internal resistance
    2. a 12V battery with an internal resistance of 1.5 ohms
    3. a 1000V battery with an internal resistance of 495.5 ohms

    etc. etc.
     
  14. Sep 17, 2014 #13

    NascentOxygen

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    If it's an ammeter icon, it may not tell us the direction of the current. Besides, I see nothing here that suggests the box contains a voltage/current source of any description.

    EDIT: Reviewing, the subject header says there's a current source in series, so unless that's OP's mistaken interpretation, we can narrow it down to 2 correct answers, according to current direction.

    My thinking is that this clumsy question is intended to be simply a review of dealing with internal resistance of the 5V source. Perhaps the original homework question may be clearer on this, but we simply don't know until OP reappears.

    @OP: Was this a multiple-choice question? If so, what are the answers offered?
     
    Last edited: Sep 17, 2014
  15. Sep 17, 2014 #14

    rude man

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    I could haved sworn I saw an arrow pointing from left to right. :wink:
     
  16. Sep 17, 2014 #15

    NascentOxygen

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    So? No one modifies circuit icons for each instance.

    Let's look at the polarity of the voltage that's driving that current. Oh, it has an arrow at both ends! So I'm none the wiser.

    EDIT: I believe there is indeed a "+" sign hiding under the upper arrow head, indistinguishable, but I'm willing to imagine I can see one there to match the "-" sign I'd hitherto overlooked near the lower arrow head.

    The possibilities are beginning to show a definite tendency towards the convergent ...
     
    Last edited: Sep 17, 2014
  17. Sep 18, 2014 #16

    CWatters

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    I don't think much more can be said until the OP replies.
     
  18. Sep 23, 2014 #17
     
  19. Sep 23, 2014 #18
     
  20. Sep 26, 2014 #19

    NascentOxygen

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    Ah, all clear now. :confused:
     
  21. Sep 27, 2014 #20
    I have solved this problem somewhere and its a MCQ . One of the options was none of the above which was correct since data is inadequate as the voltage across current source is unknown.
     
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