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Physics problem doubts - Circuits

  1. Dec 28, 2016 #1
    1. The problem statement, all variables and given/known data

    upload_2016-12-28_21-15-36.png

    2. Relevant equations

    Req= R1 + R2...Rn for resistors in series.
    Req= (1/R1 + 1/R2 +... 1/Rn)^(-1) for resistors in paralel
    And Ohm's Rules for circuits

    3. The attempt at a solution
    I did manage to solve it by using R3= 0 and thus finding R1= 2000 by using Ohm's Rules, then, using R3 going towards infinity, I find R23= R2, and from that I find R2 + R1= 6000. Since R1= 2000, R2= 4000, which is the correct answer. However, when I tried solving by using the point(5, 0.003) from the graph, I have the equation: R1 + (5R2/5+ R2) = 4000. Using R1 = 6000- R2 I get R2^2 -2000R2 - 10000=0. The positive root is 2000. That makes no sense since it inverts R1 and R2. Could someone tell me why it doesn't if I use that point?
     

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  3. Dec 28, 2016 #2

    mfb

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    The problem statement is inconsistent. As you figured out correctly, with the given numbers R1 and R2 are much larger than the scale of R3, but then R3 cannot have such a large influence with a few Ohm.
     
  4. Dec 28, 2016 #3

    SammyS

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    That is a correct analysis.

    However, There are a number of errors in the following.
    You should include enough parentheses to correct the equation R1 + (5R2/5+ R2) = 4000.

    It should be R1 + (5⋅R2/(5+ R2)) = 4000 , which is the equivalent resistance for the three resistors and which you apparently got from (12 V) / (0.003 A) .

    Solving this should not result in a quadratic equation, but it does give a negative value for R2 .

    As @mfb pointed out, there must be a problem with the statement of the problem.

    It seems the problem is that R3S must be 20kΩ rather than the 20Ω that is stated.

    You will also find that the current for R3 = 5kΩ is a bit less than 3mA .
     
  5. Dec 29, 2016 #4
    I do believe it is a quadratic, based on this calculation. So, does that mean problem is physically impossible if R3 is 20 ohms? Thanks for the huge help btw
    upload_2016-12-29_19-7-20.png
     
    Last edited: Dec 29, 2016
  6. Dec 29, 2016 #5

    SammyS

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    The values in the graph are physically impossible if r3s = 20 Ω .

    Using your approximate values from reading the graph, i = 3mA and R3 = 5 Ω, then the parallel combination of R2 and R3 would have to have an effective resistance of 2000Ω. The effective resistance of two resistors in parallel is less than the smaller of the two resistors involved. 2000 Ω is impossible if R3 is 5 Ω.

    5kΩ is more reasonable a value for R3.

    You will find that with the other two resistance values you found, that R3 = 4kΩ will give a current of exactly 3mA .
     
  7. Dec 29, 2016 #6

    mfb

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    Use the current at R3=0 to find R1, then you don't get a quadratic equation.
     
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