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Using Kirchhoff's Rules

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data
    What must the emf ε in the figure be in order for the current through the 7.00 ohm resistor to be 1.75A? Each emf source has negligible internal resistance.


    2. Relevant equations
    Kirchhoff's Rule:
    [itex]\sum I=0[/itex] junction rule

    [itex]\sum V=0[/itex] loop rule

    3. The attempt at a solution
    I have drawn my current directions as shown in the attachment below.

    By the junction rule, I know that [itex]I_2 = I_3+I_1[/itex] and it is given that [itex]I_2=1.75A[/itex].

    I drew a loop clockwise around the entire thing and came up with the following equation:

    [itex]24V-(7 ohm)(1.75A)-(I_1)(3 ohm) = 0[/itex]

    so that I have

    [itex]I_1 = (24V-12.25V)/3= 3.92A[/itex]

    and then

    [itex]I_3 = I_2 - I_1 = 2.17[/itex]

    Drawing loop 2 clockwise within the right inner loop I have the following equation:

    [itex]-(7 ohm)(I_2) - (I_3)(2 ohm) + ε = 0[/itex]

    solving for ε, I get

    [itex]ε = (12.25V)+(4.34V) = 16.59V[/itex]

    Also, if I draw a clockwise loop inside the right inner loop, I have the following equation:

    [itex]-ε + (I_3)(2 ohm) - (I_1)(3 ohm) + 24 V = 0[/itex]

    which simplifies to

    [itex]ε = 16.58V[/itex]

    I can find no mistakes (and I have checked over my work a few times) but it is the wrong answer. Could anyone please help me? Thanks!
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2012 #2

    SammyS

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    That should be [itex]I_3 = I_2 - I_1 = -2.17\text{A}[/itex]
     
  4. Nov 6, 2012 #3
    Thank you soo much! I can't believe I missed that - I was freaking out thinking I had the completely wrong idea. Thanks!
     
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