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Homework Help: Find Resistance Given Voltage

  1. Jan 15, 2010 #1
    Hey, I would really appreciate any help with this question.

    1. The problem statement, all variables and given/known data

    Examine the following circuit diagram and state the value of a) V2 b) I2 c) R1 d) R2 e) RT

    http://img136.imageshack.us/img136/3100/90933761.png [Broken]

    One solid circuit and no other information is given.

    2. Relevant equations

    V = IR

    In series, RT = R1 + R2 + ... + Rn
    In parallel, 1/RT = 1/R1 + 1/R2 + ... + 1/Rn

    3. The attempt at a solution

    I've solved for a) V2 = 15V and e) RT = 4 ohms
     
    Last edited by a moderator: May 4, 2017
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  3. Jan 15, 2010 #2

    vela

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    You haven't provided enough info.
     
  4. Jan 15, 2010 #3
    Thanks for the reply.

    That was all the info provided in the question. That's what I thought myself, but this assignment's due tomorrow and I thought I'd get some feedback on what other people thought.

    Again, thanks
     
  5. Jan 15, 2010 #4

    vela

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    Well, you can solve for some of those quantities, but not all. I can say that your answer for V2 and RT are wrong.

    How did you calculate those?
     
  6. Jan 15, 2010 #5
    For V2: In a series circuit VT = V1 + V2 + ... + Vn
    In a parallel circuit VT = V1= V2 = V3 = ... = Vn
    I treat V2 and V3 as a single entity, since it is parallel within a series circuit.
    Thus, VT = V1 + 2V2 = V1 + 2V3
    40 = 10 + 2V2, 30 = 2V2, V2 = 15V
    40 = 10 + 2V3, 30 = 2V3, V3 = 15V

    RT = VT/IT = 40/10 = 4 ohms

    can you see what i did wrong?
     
  7. Jan 15, 2010 #6

    vela

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    V2 and V3 don't add because they're not in series. When you use Kirchoff's voltage law, either V2 (if you use the inside loop) or V3 (if you use the outside loop) appears, but not both.

    I'm not sure what I was thinking before, but I was wrong about RT. Your answer is correct.

    You should be able to calculate R1, but to get R2 and I2 requires more information.
     
  8. Jan 15, 2010 #7
    Ah, thanks again
     
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