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Tricky(?) resistance question in a series circuit

  1. Jul 24, 2007 #1

    exi

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

    Two resistances, R1 and R2, are connected in series with a 12V battery. When R2 is removed (leaving R1 only), the current increases by 0.450 A.

    When R1 is removed (leaving R2 only), the current increases by 0.225 A.

    What are the two resistances?

    2. Relevant equations

    [tex]V = IR[/tex]

    3. The attempt at a solution

    I've tried to solve this algebraically, but I keep ending at a point with three unknowns and no realistic way to solve them. The fact that the current bumps are at a 1:2 ratio suggests that something similar should happen with the resistors, but hell if I know where to start.
     
  2. jcsd
  3. Jul 24, 2007 #2

    G01

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    Homework Helper
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    What are the three unknowns? What equations are they in? Can you please show me what you have so far? According to the rules you have to show some work to get helped here, and it'll help me know where exactly your stuck in this problem.
     
    Last edited: Jul 24, 2007
  4. Jul 24, 2007 #3

    exi

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    Seems like I'm making this a little excessively difficult, but I'm going with the assumption that the whole circuit is V=IR.

    So from that, in scenario #1, V = (I+0.450)(R-R2), which factors out to:
    12 = IR - IR2 + 0.450R - 0.450R2

    Scenario #2, V = (I+0.225)(R-R1), or:
    12 = IR - IR1 + 0.225R - 0.225R1.

    With three unknowns in each equation, it just feels like I'm going about this the wrong way.
     
  5. Jul 24, 2007 #4

    berkeman

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    Staff: Mentor

    Try focusing on the main parts:
    Remembering that V=12V, and R-R2=R1, R-R1=R2, and writing a 3rd equation that relates 12V, I and R1 and R2, do you have enough equations to solve for the unknown resistances and currents now?
     
  6. Jul 24, 2007 #5

    G01

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    It seems like berkeman has already helped you, but I did say I would help and I am a man of my word!:approve: So, even if I am just echoing berkeman, here goes.

    You have two equations and three unknowns. It looks to me like what you need is another equation, or a way to get rid of one of the unknowns. I'm going to go for the former. Here's my hint: What about the original circuit with two resistors? Can you analyze this to get more information?
     
  7. Jul 24, 2007 #6

    exi

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    hrm... I'm really not catching on to the best way to simplify this a bit more (V = IR? 12 = I(R1+R2)? etc...) Of an entire assignment of similar questions, this is the only one I'm missing.
     
  8. Jul 24, 2007 #7

    G01

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    Ok you said it!

    You now have:

    12=I(R1+R2)

    12 = IR - IR2 + 0.450R - 0.450R2

    12 = IR - IR1 + 0.225R - 0.225R1

    Here are 3 equations with 3 unknowns. You should be able to solve for all three now.
     
  9. Jul 25, 2007 #8

    exi

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    Gotcha, thanks.
     
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