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Electrical circuits with resistors and variable voltage sources

  1. Nov 22, 2016 #1
    1. The problem statement, all variables and given/known data
    The following picture illustrate how the current in the circuit changes when you change ##E_2##. Where slope 1 represents the current flowing through ##E_1## and slope 2 represents the current flowing through ##E2##
    hfs54qD.png

    2. Relevant equations
    Usual circuit laws :P

    3. The attempt at a solution
    I think there is inconsistency in data.
    It didn't give me the direction in current so I assumed the following.
    If we look at the point where ##E_2## is equal to 5 ( 5 , 0.1). There is only one possibility of both current being upward so it can flow in R2
    So we now know that the positive values must be upward.
    and we can easily know R2
    ## R_2 = \frac{E_2}{0.2} = \frac{5}{0.2} = 25 ##
    Now if we look at point 10. We find that the current flowing in ##E_2## is 0.4 and upward
    The current in ##E_1## is -0.2 which means 0.2 downward. We conclude that the current in ##R_2## is 0.2
    ## R_2 = \frac{E_2}{0.2} = \frac{10}{0.2}= 50 ##
    Here is the inconsistency.
     
  2. jcsd
  3. Nov 22, 2016 #2
    That looks right - that the I2 current is 0.1 A for E2 = 5V. So what does that tell you about the value of R2?

    Now it looks like you have assigned a different current (0.2 A) through R2 for the same E2 voltage of 5 V. How can that be?

    Edit: Corrected "E1" to "E2" in second to last sentence.
     
  4. Nov 22, 2016 #3

    haruspex

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    That assumes the currents through E2 and R2 are the same.
     
  5. Nov 22, 2016 #4
    Do you happen to have a better representation of the plot? It is very difficult to read, making me question my ability to read data points off of it accurately.
     
  6. Nov 22, 2016 #5
    Well look at E2 0.4 A going to the node and 0.2 A coming out so the current in R2 must be equal to 0.2 A


    Not sure what you mean :/, There are two points with the same current going through R2 but different voltages how come that can be true?

    Nope, That is the only picture I got. :C sorry
     
  7. Nov 22, 2016 #6
    I'm sorry. Once again my failure to read the problem statement carefully has caused my misunderstanding. I thought I1 and I2 were currents through R1 and R2. My mistake. Disregard what I said above.
     
  8. Nov 22, 2016 #7
    I blame the picture, It is just unclear. :D
     
  9. Nov 22, 2016 #8
    Thank you for your kindness, but the text was pretty easy to read - if only I had read it.. :)
     
  10. Nov 22, 2016 #9
    Biker, how did you come up with this equation? Are you reading I1 = I2 = 0.1 for E2 = 5 V and adding them to get the R2 current?
    If 5V is the E2 voltage where I1 and I2 intersect, it looks like it could be somewhat less than 0.1 Amps, but it's hard to tell for sure.
     
  11. Nov 22, 2016 #10
    Yes exactly, But the difference between 50 ohms and 25 ohms is pretty large, isnt it?
     
  12. Nov 22, 2016 #11
    Yes, it is pretty large. But then again, if the intersection of I1 and I2 is really at E2 = 5V, I don't think it would be much of a stretch to say that those currents intersect at, say, 0.06 Amps, or even 0.05 Amps. Because if you look at the exact point where I2 is at 0.1 Amps, I1 appears to be some distance below it. The plot is just too vague to get very accurate information from it.
     
  13. Nov 23, 2016 #12

    haruspex

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    Sorry, not following your logic. What specific data points are you making use of? From your original post, looks like you are using E2=5V, but in that case where are you getting the 0.4A from?
    The sketch of the graphs is rather hard to read numbers off. I took the following:
    E2=2, I1=.2
    E2=10, I1=-.2
    E2=4, I2=0
    E2=10, I2=.42
    and got R2=8.
     
  14. Nov 23, 2016 #13
    I'm pretty sure what he's doing there is reading the I1 and I2 currents at E2 = 10 V.
    At E2 = 10 V, I1 = -0.2 A and I2 = +0.4 A (from the plot).
    The current into R2 has to be the sum of those two currents, or IR2 = -0.2 + 0.4 = 0.2 Amps.
    Since the voltage across R2 is 10 V, R2 is calculated to be (10 V)/(0.2A) = 50 ohm

    Another data point I was looking at is where I1 crosses the 0 line. I read E2 to be approximately 5.8 V, and I2 to be approximately 0.13 Amps. At that data point, I2 is equal to the current through R2. That would result in R2 = (5.8)/(0.14) = 44.6 ohm
     
  15. Nov 23, 2016 #14

    haruspex

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    Ok, thanks.
     
  16. Nov 23, 2016 #15

    NascentOxygen

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    The inconsistency lies in your use of a graphical aid. These are straight line plots, and you must not lose sight of that.

    You decided the plots intersect at (5,0.1). Okay, we'll go with that.
    You also decided another relevant point is at (10,0.4). Okay, we'll use that, too.

    By those 2 points you have now fully defined these plots. If you need any more data points you must calculate them using geometry of a straight line. You should not attempt to read any further data off your scrappy thumbnail sketch because estimation errors in further readings will almost certainly give data not consistent with the plots defined by the 2 points you have already decided on.

    If you had an accurate graph this "problem" would not arise. I put it in quotes because it is not really a problem at all. These are straight line plots, and you should appreciate the need to be rigorously consistent in treating the variables relationship at all times as linear.
     
  17. Nov 24, 2016 #16

    epenguin

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    A rather hard thread to read. It is hard to read the diagram which led to me misreading the problem - please always use an app like DocScanHD to present a less murky one.
     
    Last edited: Nov 24, 2016
  18. Nov 24, 2016 #17

    gneill

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    Neither line represents the current through R2. They represent the currents through the sources.
     
  19. Nov 24, 2016 #18

    epenguin

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    :oops: I have deleted most of my post as I misunderstood the data and problem.

    It seems to me that significant points are where the graphs cross the horizontal axis. This gives you the ratio of the E's and the ratio of the R's.

    As far as I can see the in this complicated thread the OP has not given any general equation for the circuit to base an easy-to-follow reasoning on.
     
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