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Soving for resistor values and source current-image included

  1. Jul 2, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Question is in image included

    So, I need to find R2, R3, and Is. I was able to get the correct answer for R3 by doing R3= V/I (R3= 12/1 = 12) . however when I apply that equation to R2 ; R2= V/I. (R2=12/4 =3), I do not get the correct answer. The answer is supposed to equal 4 ohms. Am I using an incorrect method and just getting lucky finding R3?

    Once I find the value of the 2 resistors, I can figure out Rt and from there find Is.
     

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  3. Jul 2, 2016 #2

    CWatters

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    Method is correct but the current in R2 is not 4A.
     
  4. Jul 2, 2016 #3

    CWatters

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    Actually there is a way to find Is without working out R2 or R3 first.

    Have you studied KCL yet?
     
  5. Jul 2, 2016 #4
    Do you subtract 4 and 1 to get you 3? 12/3 would equal 4.. But i dont see why you would do that. Ive just been working from a textbook and haven't came across that.

    And yes, KCL states that the sum of the current going in equals the sum of the current going out. I havent learned beyond that though if there's anything else.
     
    Last edited: Jul 2, 2016
  6. Jul 2, 2016 #5

    CWatters

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    Yes. The current going into a node equals the current coming out. Google Kirchoff's Current Law.

    4A leaves the node at the top of R2 so 4A must enter. 1A comes from the R3 branch so 3A must come from the R2 branch.

    You can also use KCL to calculate Is.
     
  7. Jul 2, 2016 #6
    Wow... That makes alot of sense. I wouldnt have thought of that. I dont know why my textbook didn't show me to think like that. I wont forget it now though!

    It had me wondering why there were 2 different current values... Because according to kirchhoff's current law, the sum going in = the sum going out. Now it makes complete sense! Thank you!
     
  8. Jul 3, 2016 #7

    CWatters

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    Despite what I said above I find it's a bit dangerous to do "the sum going in = the sum sum out". It's better to define current going into a node as +ve and then make a sum to zero..

    In other words I prefer to write..
    +IR2 +1 +(-4) = 0
    then solve for IR2

    rather than write
    4 = 1 + IR2

    The reason is that later on when you get more complicated circuits it's very easy to make sign errors. In many cases you don't know the direction of one or more currents. In those cases you have to define a direction as positive, then later when you solve the simultaneous equations you might discover that one of them is negative. At that point you can refer back to your definition to work out what that means.

    Edit: I should add that this is my personal preference. Either way works but I find this way I make fewer sign errors.
     
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