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Series Wiring Problem

  1. Jun 22, 2009 #1
    Two resistances R1 and R2, are connected in a series across a 12-V battery. The current increases by .20 A when R2 is removed, leaving R1 connected across the battery. However the current increases by just .10 A when R1 is removed, leaving R2 connected across the battery. Find(a)R1 and (b) R2.

    V=12v

    2. Relevant equations
    R1+R2=Rs V1+V2=V

    I=V/R


    3. The attempt at a solution

    I+20= V/R1 I+10= V/R2

    I believe the two equations above would give me my answers but I'm stumped on how to go about finding the unknowns. I know V=12 but I can't figure out how find the current or resistances.

    Can somebody point me in the right direction?
     
  2. jcsd
  3. Jun 22, 2009 #2

    LowlyPion

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    Looks like you need 3 equations for the 3 unknowns.

    V = I*R looks like it yields all 3.

    (R1 + R2) = V/I

    R1 = V/(I + .1)

    R2 = V/(I + .2)

    Then solve right?
     
  4. Jun 23, 2009 #3
    Using the three equations makes sense, but I can't figure out how to find the unknowns. All I know is V, and I need to know I to be able able to find R1 and R2.

    my idea was to subsitute I=V/R into the equations and then try to solve for R1 and R2.
    so R1= V/(V/Rs)+.1

    but I don't know what Rs is either.
     
  5. Jun 23, 2009 #4

    LowlyPion

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    Don't use Rs as that is by definition R1 + R2.

    Otherwise, if you must, then you have 4 equations, with the 4th unknown now Rs where Rs = R1 + R2
     
  6. Jun 23, 2009 #5
    ok but if I plug R1 + R2 into the equation I still have too many unknowns.

    R1=V/(V/R1 + R2) +.1

    In that equation I still don't know how to find R1 or R2.
     
  7. Jun 23, 2009 #6

    negitron

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    Disregard; I misread the problem question.
     
    Last edited: Jun 23, 2009
  8. Jun 23, 2009 #7

    LowlyPion

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    You only have 3 unknowns : I, R1, R2, because V is 12.

    (R1 + R2) = V/I

    R1 = V/(I + .1)

    R2 = V/(I + .2)

    Rewriting you have

    12 = I*R1 + .1R1

    12 = I*R2 + .2R2

    12 = I*R1 + I*R2

    Adding equation 1 to equation 2 and subtracting equation 3 yields

    12 = .1R1 + .2R2 or, ...

    120 = R1 + 2R2

    Surely it's down hill from there.
     
  9. Jun 23, 2009 #8
    Here is my attempt at the solution:

    (R1 + R2) = V/I R1 = V/(I + .1) R2 = V/(I + .2)
    SO:
    (V/(I+.1)) + (V/(I+.2))=V/I

    (12/(I+.1)) + (12/(I+.2))= 12/I
    24I^2 + 3.6I = 12I^2 + 3.6I +.24
    12I^2 = .24
    I=.02 A

    R1 = V/(I+.1) = 12/(.02+.1) = 100 OHMS
    R2 = V/(I+.2) = 12/(.02+.2) = 54.5 OHMS
     
  10. Jun 23, 2009 #9

    LowlyPion

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    i = (√2)/10
     
  11. Jun 23, 2009 #10
    I = .14 A

    R1 = 50 ohms R2 = 35.3 ohms

    Thank you for all your help.
     
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