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Series Wiring

  1. Jul 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Two resistances, ##R_1## and ##R_2##, are connected in series across a 12 V battery. The current increases by 0.2 A when ##R_2## is removed, leaving ##R_1## connected across the battery. However, the current increases by just 0.1 A when ##R_1## is removed, leaving ##R_2## connected across the battery. Find ##R_1## and ##R_2##.

    2. Relevant equations
    Ohm's law.

    3. The attempt at a solution
    ##ΔI_1=0.2 A##
    ##ΔI_2=0.1 A##


    (1) ##I=\frac{V}{R_1}-ΔI_1##

    (2) ##I=\frac{V}{R_2}-ΔI_2##

    (3) ##I=\frac{V}{R_1+R_2}##

    From (1) and (2):
    (4) ##R_2=\frac{120R_1}{R_1+120}##

    From (2) and (3):
    (5) ##\frac{V}{R_2}-ΔI_2=\frac{V}{R_1+R_2}##

    Then I put (4) into (5) and get:
    ##R_1=0.503Ω##
    ##R_2=0.5009Ω##

    The correct answer is ##R_1=35Ω##, ##R_2=50Ω##.

    I fail to spot my mistake. Please help.
     
  2. jcsd
  3. Jul 10, 2015 #2
    Recheck eq. 4
     
  4. Jul 11, 2015 #3
    I see, it's suppose to be ##R_2=-\frac{120R_1}{R_1-120}##, but then I get two ##R_1## values:
    ##R_1=35.15##
    ##R_1=204.85##
    Both seem correct. How come the authors chose only one of them?
     
  5. Jul 11, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Calculate the value of ##R_2## associated with each choice of ##R_1##.
     
  6. Jul 11, 2015 #5
    In solving a quadratic eq you sometimes get extraneous roots. You must determine if it is a plausible soution. the 204.85 ohm solution does not satisfy eq 4 predicting a neg value for R2. So discard it.
     
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