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Circuits problems

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    See image attached. Please ignore all work you see around the problem. :]
    I need help with 9, 10, and 11 in this picture.


    2. Relevant equations

    V = IR

    3. The attempt at a solution

    9. [tex]R_{eq} = R + (\frac{1}{2R} + \frac{1}{2R})^{-1} = 2R = 24[/tex]
    [tex]V' = (0.5)(\frac{1}{2R} + \frac{1}{2R})^{-1} = 6 V[/tex] (this is the voltage drop across the parallel set of resistors of 2R and 2R)
    [tex]I_{2} = I_{3} = \frac{6}{2R} = 0.25[/tex] (this is the current across each of the parallel branches, and they are only equal because they have the same resistance and voltage]
    [tex]I_{1} = I_{2} + I_{3} = .25 + .25 = .5[/tex]
    [tex]V_{i} = I_{1}R = (.5)(12) = 6[/tex]
    [tex]E = V' + V_{i} = 6+6 = 12[/tex]

    Correct answer is actually 24 V.


    10. [tex]R_{eq} = (\frac{1}{250} + \frac{1}{300})^{-1} = 136.364[/tex]
    [tex]V = IR_{eq}[/tex]
    [tex]24 = I(136.364)[/tex]
    [tex]I = 0.176[/tex]
    [tex]V' = (0.176)(136.364) = 24[/tex] (this part is weird, is it because the whole circuit's in parallel anyway so there's really no voltage drop across anything?)
    [tex]I = \frac{V}{R} = \frac{24}{300} = 0.08 A[/tex]

    Correct answer is actually 40 mA.

    11. Taking both the loops clockwise:
    [tex]0 = -10I_{1} -20I_{2} - 5I_{1} + 50[/tex]
    [tex]0 = 20I_{2} - 10I_{3} - 40[/tex]

    Stuck from here.


    Any help or pointers would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2009 #2

    tiny-tim

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    Hi clairez93! :smile:
    Yes, I2 = I3 = I1/2,

    but where does 6/2R come from? :confused:

    Use I = I3 = I1/2 and apply Kirchhoff's rules to the outer circuit. :wink:
     
  4. Mar 21, 2009 #3
    I used I = V/R, and the voltage across it I thought was 6, and then the resistance is 2R, so I = 6/2R. Is that not correct?
     
  5. Mar 21, 2009 #4

    tiny-tim

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    why? :confused:
     
  6. Mar 21, 2009 #5
    From this.
     
  7. Mar 21, 2009 #6

    tiny-tim

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    ah! but 0.5 only the current through one of the 2Rs :redface:
     
  8. Mar 21, 2009 #7
    Aha! I see! so then the voltage drop across is really 12 V.
    And therefore I2 = 12 / 2R = 0.5 = I3
    And therefore I1 = I2 + I3 = 1
    And therefore the voltage across R is IR = 1*12 = 12.
    And therefore E = 24!
    Thanks!

    Now i am still having troubles with 10 & 11.
     
  9. Mar 21, 2009 #8

    tiny-tim

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    10: I make the same as you … but much moire quickly, just by looking at the right-hand loop :confused:

    11: use I1 = I2 + I3 :wink:
     
  10. Mar 21, 2009 #9
    Are my loop equations correct? I just want to make sure before I solve.
     
  11. Mar 22, 2009 #10
    Okay, so my loop equations are wrong, I believe because I am getting odd numbers.
     
  12. Mar 22, 2009 #11

    tiny-tim

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    Show us? :smile:
     
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