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Homework Help: Kirchhoff's Rule-Current in a Two Loop Circut

  1. Feb 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the current in each branch of the circuit of the figure below . Specify the direction of each.

    The top resistance is 22, middle is 56, and the bottom is 75.

    I'm predicting the current is going in this direction:

    2. Relevant equations
    ---Kirchoff's Rules (Junction rule & Loop Rule)

    3. The attempt at a solution

    My starting/reference point is at the 5V battery on the positive anode side; going toward the right.

    EQ1: -5v - 22i1 + 56i2 = 0

    EQ2: -5v - 22i1 + 75 i3 + 1v = 0

    EQ3: i2= i1 + i3

    The problem I'm having is the fact I have two variables to solve for; regardless how I manipulate. the equations.

    For instance:

    if you solve for i1 in EQ2: i1= (-4v + 73 i3)/ 22

    ..... now I plugged i1 into EQ1: -5V - 22 ((-4v + 73 i3)/ 22) + 56 i2 = 0

    -As you can see, I still have two unknowns: i3 and i2. The Junction rule eq (EQ3) wouldn't help either. It would just replace i3 with i1 again. So with that said, what am I doing wrong/missing? I have all 3 loops covered in my two equations, so I'm kind of lost. Any help would be appreciated.
  2. jcsd
  3. Feb 9, 2014 #2


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    Staff: Mentor

    Hi Hayliee30! Your eqn 3 isn't right. There's a sign problem.

    Next time, please draw the circuit then photograph or scan it. While I can see you are good at ASCII art, in circuit dagrams ASCII just doesn't cut it!

    You have found i1 in terms of i3, so replace i1 in the other pair of equations, leaving you with a pair of equations in two unknowns, i2 and i3. Two equations, in two unknowns, you can solve this!
  4. Feb 9, 2014 #3
    If you solve for i1 in equation 2, you need to use the result to remove i1 from both equation 1 and equation 3,
    to get 2 equations with two unknowns.

    It's easier to start with EQ3, wich is already solved and use it to eliminate i2 from EQ1. EQ2 already lacks i2.

    It's even easier to use nodal analysis. The left node of the circuit will be at 0, the right node at U.
    Compute the currents i1,i2,i3 as a function of U. i2 = i1 + i3 will get you a single equation for the single unknown U.
  5. Feb 9, 2014 #4
    NascentOxygen, sorry! Next time I will upload a picture with it. As for the help, thanks guys! I forgot about subst. i1 into both equations. The only question I have remaining is for EQ3. Where is the sign problem exactly? The way I have it set up, it's kind of "awkward" for the junction rule. Is it i2 = i1 - i3? That wouldn't really make any since either. I dont't see how i1 is the sum of currents i2 and i3 based on the current flow I drew
  6. Feb 9, 2014 #5


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    Staff: Mentor

    There are a couple of ways to "say aloud" Kirchoff's current law. But I think the least confusing :tongue: is "the sum of all currents into a node = 0"

    Obviously, if a current has been already designated as leaving the node, you now attach a negative sign when counting its contribution to the current entering the node.
  7. Feb 9, 2014 #6
    Ah, that makes sense! Thank you so much, NascentOxygen!
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