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Possible resistances using 3 resistors?

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data

    You have 3 resistors 1000ohms each. List all the resistances you can achieve using this set.

    2. Relevant equations

    R in series = R1 + R2 + R3....etc

    1/R in parallel = 1/R1 + 1/R2 + 1/R3...etc


    3. The attempt at a solution

    If they are all in series you have 3000ohms. If they are all in parallel you have 333.333ohms. If two are in parallel while one is in series, you have 1500ohms. If one is in parallel while two are in series, then I'm getting that the Req will be 3000ohms. I'm not sure if that last part is correct. If it is, then the possible resistances are 333.33ohms, 1500ohms, and 3000ohms.
     
  2. jcsd
  3. Feb 25, 2014 #2

    ehild

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    The resistance (for two series, one parallel) is not correct.

    ehild
     
  4. Feb 26, 2014 #3

    NascentOxygen

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    Does it specify that you must use all 3 resistors?
     
  5. Feb 26, 2014 #4

    lewando

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    Or can you connect a resistor to itself?
     
  6. Feb 26, 2014 #5
    I dont think you can have 1 resistor in parallel with itself. So for the last one you have one in parallel with the 2 in series.
     
  7. Feb 26, 2014 #6

    lewando

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    I meant connect one terminal of a resistor to the other terminal of the same resistor. Forming a circular resistor, effectively shorting it out of existence.
     
  8. Feb 26, 2014 #7
    How would you calculate the resistance for two series and one parallel? I'm doing it as Req = R1 + R2 + (1/R3)^-1
     
  9. Feb 26, 2014 #8
    I was responding to the OP's post lol. He was trying to have 'one in parallel with 2 in series' but what he was doing was having one resistor in parallel with itself ([itex]\frac{1}{R}[/itex]-1) or R+R+R which was wrong.
     
    Last edited: Feb 26, 2014
  10. Feb 26, 2014 #9
  11. Feb 26, 2014 #10
    So two in series and one in parallel would be equal to R1 + (1/R2+1/R3)^-1 ?

    So all in all you can have:

    One resistor alone = 1000ohms

    Two resistors in series = 2000ohms

    Two resistors in parallel = 500ohms

    Three resistors in series = 3000 ohms

    Two in parallel, one in series = 1500ohms

    Two in series, one in parallel = 1500ohms

    All in parallel = 333.33 ohms

    Is this all correct and am I missing any others?
     
  12. Feb 26, 2014 #11

    ehild

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    Is not that two parallel and one series with the parallel ones? Look at the figure: What is the equivalent resistance ?

    ehild
     

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    Last edited: Feb 26, 2014
  13. Feb 26, 2014 #12
    you are doing the same thing for both (2 in parallel, 1 series) so getting the same answer. did you see the link from my previous post? I thought it did a good job explaining it.
     
  14. Feb 26, 2014 #13
    It looks like the Req would be 3000ohms

    The link brought me to a picture of the cover of a Java textbook. However, isn't this two in parallel and one in series? 195px-Resistors_in_series_and_parallel.svg.png

    Wouldn't the Req of that be 1500ohms?
     
  15. Feb 26, 2014 #14
    Wouldn't the resistance of this also be 1500ohms? (2 in series 1 in parallel)

    http://www.cheng.cam.ac.uk/research/groups/electrochem/JAVA/impedance/figure/sparal1.gif
     
  16. Feb 26, 2014 #15
    I think I see now. For two in series and one in parallel you need to find the Req of the 2 in series which is 2000ohms. Then you do (1/2000 + 1/1000)^-1 which is 666.67 ohms.

    That leaves 7 possible combinations with three resistors:

    One resistor alone = 1000ohms

    Two resistors in series = 2000ohms

    Two resistors in parallel = 500ohms

    Three resistors in series = 3000 ohms

    Two in parallel, one in series = 1500ohms

    Two in series, one in parallel = 666.67ohms

    All in parallel = 333.33 ohms
     
  17. Feb 26, 2014 #16

    NascentOxygen

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    That's looking better.
     
  18. Feb 28, 2014 #17
    thats weird, when I see the image it is of a circuit! It must be a browser issue or something. I attached the image below that I was talking about (this link should work). anyway it looks like you already figured it out. good job.

    http://s29.postimg.org/vtkoxgodz/combination.png [Broken]
     
    Last edited by a moderator: May 6, 2017
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