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Determine the currents in this electrical network

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    This is quite an easy problem but I'm not sure if I set up my equations correctly, and I want confirmation.

    We are given the following circuit:

    attachment.php?attachmentid=467003&d=1443975912.png

    And our task is to determine the loop currents [itex] i_{1}, i_{2}, i_{3}, i_{4} [/itex].

    2. Relevant equations

    Kirchhoff's first law: sum of currents entering and leaving a loop = 0.
    Kirchhoff's second law: sum of electromotive forces = sum of potential difference of loop.
    Ohm's law: V = IR.

    3. The attempt at a solution
    Just apply Kirchoff's laws and Ohm's laws to get the following:

    loop 1:
    [itex] 50(i_{1} - i_{2}) + 30(i_{1} - i_{3}) = 120V [/itex]

    loop 2:
    [itex] 50(i_{2} - i_{1}) + 15i_{2} + 10(i_{2} - i_{3}) + 25(i_{2} - i_{4}) = 120V [/itex]

    loop 3:
    [itex] 30(i_{3} - i_{1}) + 10(i_{3} - i_{2}) + 20(i_{3} - i_{4}) + 5i_{3} = 120V [/itex]

    loop 4:
    [itex] 25(i_{4} - i_{2}) + 10i_{4} + 30i_{4} + 15i_{4} + 20(i_{4} - i_{3}) = 120V [/itex]

    I'm not concerned about finding the numerical values, I'm just wondering if I've set these equations up correctly. Are some of these supposed to be -120 volts on the right hand side? If so, how do I determine that?
     
  2. jcsd
  3. Oct 4, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Only one loop has voltage sources in it, and that's Loop 1. So the RHS of all but the first equation should be zero.

    Now, you'll want to be a bit careful about the voltage in Loop 1. Note that one lead is labeled +120 V and the other -120 V. How do you think that might come about? Pencil in what the "missing" circuitry might look like.
     
  4. Oct 4, 2015 #3
    In that case, RHS of all the equations should be 0 except for loop 1 which should be 240V (potential difference). Is that correct?
     
  5. Oct 4, 2015 #4

    gneill

    User Avatar

    Staff: Mentor

    Yup. Be sure to get its polarity right in your equation.
     
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