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Designing a device off of given voltage and amperage values?

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    You are working with a device in a black box which has two terminals, which are in turn, part of a larger circuit. You measure the voltage and the current at the terminals and record the following:

    vi (V) | i1 (A)
    100 | 0
    120 | 4
    140 | 8
    160 | 12
    180 | 16

    a: construct a circuit model for this device using a current source and a resistor
    b: How much power will the device deliver to a 20 ohm resistor?

    2. Relevant equations

    ????
    Ohms Law?

    3. The attempt at a solution

    I haven't ever seen anything like this before in my class, my textbook offers no help that I can think of to even begin with this.

    All I know is that at V=100, A = 0, Resistance would be unavailable due to division by 0.
    V=120, A = 4, R=30
    V=140, A=8, R = 35/2
    V=160, A=12, R = 40/3
    V=180, A=16, R = 45/4

    I am absolutely lost here guys.. I don't know if I have a hard teacher (Or a lousy teacher), or if I should take the humility route and wonder if I'm not meant to go for EEing because I cannot figure this out. But then again, this entire homework he has assigned us has been chalked full of material he hasn't covered that I had to research on my own.
     
  2. jcsd
  3. Feb 8, 2013 #2

    gneill

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    You're told that there are two components, a resistor and a current source. Can you tell what the connection arrangement must be? How many ways is there to connect them? What characteristics of these connection arrangements might support or contradict the observed behavior?
     
  4. Feb 8, 2013 #3
    With only 1 current source and only 1 resistor? I have absolutely no idea. Hell, even with multiple resistors and whatever, I have absolutely no idea. Nothing is a constant in this scenario.
     
  5. Feb 8, 2013 #4

    gneill

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    One current source. One resistor. How many distinct ways is there to connect them together?
     
  6. Feb 8, 2013 #5
    One way. But that doesn't solve the mystery of the non-constant resistor.
     
  7. Feb 8, 2013 #6

    gneill

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    Well, you don't know that for sure yet.

    But there are two ways they can be connected. Which way is possible, and which way is not? (and why?)
     
  8. Feb 8, 2013 #7
    I honestly don't see more than one way it could be connected, and that is a simple circuit in which the current flows from one end of the current source through the resistor, and back into the current source (I suppose it COULD be two circuits if you reverse polarity, but polarity doesn't really matter in this scenario as far as I know....)
     
  9. Feb 8, 2013 #8

    The Electrician

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    Have you studied Thevenin and Norton's theorems? You may have to do some more research on your own; it's a good way to reinforce the learning experience.

    When you have a two terminal device (your black box), you can learn something about it by determining the I-V characteristic. You do this by applying a voltage to the terminals, measuring the current into the terminals and plotting the measured data.

    Imagine that your black box contained a 5 ohm resistor and you applied some voltages, measured the currents and plotted the results. You might get something like this:

    attachment.php?attachmentid=55514&stc=1&d=1360366632.png

    Now, make a similar plot of your problem data:

    attachment.php?attachmentid=55515&stc=1&d=1360366722.png

    Since by now, you've looked up Norton equivalents, what does the fact that your problem data doesn't go through the origin tell you?
     

    Attached Files:

  10. Feb 8, 2013 #9

    gneill

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    Two ways:

    attachment.php?attachmentid=55516&stc=1&d=1360367165.gif

    Which way can be ruled out?
     

    Attached Files:

  11. Feb 8, 2013 #10
    Wait... where in the heck is the voltage source coming from? I thought it was only a resistor and a current source? And I really hate to play stupid.. but once again, I have no idea. I have NEVER seen ANYTHING like this in class or in my book.

    I'm going to rule out that the resistor in series version. Seems to me the resistor in parallel would be the only one to give varying gaps in numbers like that...
     
    Last edited: Feb 8, 2013
  12. Feb 8, 2013 #11

    The Electrician

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    You need to look up Thevenin and Norton equivalents and you will understand.
     
  13. Feb 8, 2013 #12

    gneill

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    The problem statement says that the device is part of a larger circuit, the device itself being within the dotted line box. Voltage and current is measured, so the voltage source represents the voltage applied by the external circuitry.
    I understand. This happens. Generally it's meant to cause one to creatively apply the concepts already learned to a new situation. I do commiserate, having suffered the process myself lo those many years ago :smile:
    Right. The series version would allow only one, fixed current value to flow no matter what the input voltage. The parallel connection is more flexible in that regard, and being the only other option, it must be correct.

    If you call the resistor R and the current source Ix and write the equation for the current I in terms of the applied voltage V, you should be able to use some given data points to solve for R and Ix.
     
  14. Feb 8, 2013 #13
    I'm still not getting it... Am I supposed to come up with arbitrary values of like "120v times 4 amps is equal to Rx"?

    As for "The Electrician", and you suggesting to me that I read up on the "Thevenininin (how the heck do you say that guys name anyway?) and Norton Equivilencies", I don't understand where simplifying the circuits in this case will help, as I only have two components in my circuit right now anyway?
     
  15. Feb 8, 2013 #14

    The Electrician

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    It might be easier for you to understand the behavior of the black box to imagine a resistor in series with a voltage source. When there is 100 volts applied to the black box terminals, the current is zero. How can this happen? One way this could happen would be if there were a 100 volt source in the box connected in series with the resistor with such a polarity as to oppose the externally applied 100 volts. Do you grok this? (Sly reference to Heinlein's "Stranger in a Strange land".)

    Now, if you can understand how it could happen with a voltage source in series with a resistor, then understand that what you would have in the box would be a Thevenin equivalent.

    Next you need to understand that there is a Norton equivalent with similar behavior to the Thevenin version.

    Thevenin equivalents have a voltage source in series with an impedance; Norton equivalents have a current source in parallel with an impedance.
     
  16. Feb 8, 2013 #15

    gneill

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    Nope. You can write an equation for the circuit that gives I in terms of V, R, and Ix. Now, you are given sets of values for V and I, while R and Ix are unknowns. That's two unknowns. Use two of the given data points to obtain two equations in two unknowns.
    I think that what The Electrician is getting at is that if the black-box circuit is converted to its Thevenin equivalent (it's currently in Norton form), it may be more obvious how the observed behavior comes about, and how to go about solving for the component values.
     
  17. Feb 8, 2013 #16

    The Electrician

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    Yes. See post #14, posted simultaneously with post #15, a first for me!
     
  18. Feb 8, 2013 #17
    I'm going to give this a break... I've been at this homework for 7 hours now (past 3 on this particular problem) and I desperately need a break...

    If anyone is willing to work me through this problem step by step, I'd really appreciate it, but I feel like I'm still just as clueless as I was when I first began this problem. I hate to resort to "give me the answer", but I think that's the only way I'm going to understand it. It's a foreign concept to me, and I feel like every hint and pointer I get is at something I don't understand at all :(
     
  19. Feb 8, 2013 #18

    The Electrician

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    I got you 90% of the way there with this:

    Does this make sense to you?
     
  20. Feb 8, 2013 #19
    Somewhat, I understand why you'd get zero at the terminals (because its a short circuit?)
     
  21. Feb 8, 2013 #20

    The Electrician

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    There's no short.

    The problem statement says that the black box "...has two terminals, which are in turn, part of a larger circuit."

    What that means is that the "larger circuit" as it operates, is applying various voltages to the black box. When 100 volts is applied, no current flows. Applying 100 volts is not the same as applying a short circuit. There is a source of voltage inside the box which opposes the externally applied (by the larger circuit) voltage.

    If you imagine a 100 volt source in series with the resistor in the box, you can imagine that the polarity of that 100 volt source cancels out 100 volts worth of the externally applied voltage.
     
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