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Circuit analysis

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

    Find the three currents in the image.

    2. Relevant equations
    Khirkoffs laws
    Sum of currents equal zero.
    The potentials around a closed loop add up to zero.
    The equations given in the solution are
    [tex]-2I_1 + 12I_2+6=0[/tex]
    [tex]-12I_2+8 = 0[/tex]
    [tex]I_1+I_2+I_3=0[/tex]

    3. The attempt at a solution

    When I tried to solve this I had the first equation to be
    [tex]-2I_1-12I_2+6=0[/tex], by doing a potential walk round the left subcircuit (?).
    Why should the second term be positive?
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2009 #2

    berkeman

    User Avatar

    Staff: Mentor

    By convention, when you have the current going down through a component, the top of the component will be + and the bottom will be -. Since you are going around the loop on the left against the direction labelled for I1, it is a negative voltage term. Since you are continuing around that clockwise loop on the left and go through the middle resistor in the same direction as the current arrow shown, it is a + voltage drop.
     
  4. Mar 1, 2009 #3
    Ok, but how do I know the current is going down in I_2, besides the arrow? If the problem didn't give the currents directions, I should still be able to solve it! Is it that the right battery has a higher voltage? If so, what if the problem didn't give the battery voltage in numerics, but instead gave, say, the voltage v_1 and v_2. Then I wouldn't know which one has the higher voltage...
     
  5. Mar 2, 2009 #4
    Ok, sorry. i just realised I was thinking wrong about the circuit, there is only one option for the direction of I_2.

    Thanks
     
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