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Kirchoff's Voltage Law and Current Law -- Help Please

  1. Apr 22, 2017 #1
    1. The problem statement, all variables and given/known data
    Ub8ZocX.png

    2. Relevant equations
    Using Loop 1, 2 and 3

    3. The attempt at a solution

    upload_2017-4-22_20-22-40.png
     
  2. jcsd
  3. Apr 22, 2017 #2
  4. Apr 22, 2017 #3

    ehild

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    You have to give I1, I2, I3 in terms of the resistances R1,R2,R3 and the Emf -s ε1 and ε2. The equations for loops 1 and 2 are correct, and adding the third equation I1=I2+I3 is enough to solve for the three currents.
     
  5. Apr 22, 2017 #4

    berkeman

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    Welcome to the PF.

    Things look okay so far to me. You still need to plug the equation for I2 into your equation for I1, and then the equation for I3 also gets combined. The result they want you to show only has the values of the voltage sources and resistances in it.
     
  6. Apr 22, 2017 #5
    I tried so many times but just can't get it with voltages and resistances. Can you show me step by step?
     
  7. Apr 22, 2017 #6

    gneill

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    You need only two of the KVL loops and the one KCL equation to solve the circuit.

    Your circuit has only two independent loops. Once you've written KVL for loops 1 and 2, every component has been "touched" by one or the other or both equations, and no new information can be gained by writing another loop equation; the equation set will simply become overdetermined.

    To see that this is so you might cast your three loop equations into matrix form:

    upload_2017-4-22_16-5-43.png

    If you take the determinant of the impedance matrix (the square matrix), you'll find that it is zero, and hence no unique solution can be found from them.
     
  8. Apr 22, 2017 #7

    vela

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    Only two of the loop equations are useful. The third one isn't independent; if you add the equations for loop 1 and loop 2 together, you get the equation for loop 3. So solve the system consisting of two of the loop equations and the KCL equation.

    At this point, it's algebra. You've likely solved this kind of problem before: 3 equations, 3 unknowns. It just doesn't look familiar because you're not using the variables x, y, and z.
     
  9. Apr 22, 2017 #8
    I can't do the algebra its just not working for me! I just need a full method step by step in how you get the final answer with voltages and resistances
     
  10. Apr 22, 2017 #9

    vela

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    Can you solve a problem like the following? (Not asking you to solve it, but do you know what to do?)
    \begin{align*}
    2x + y &= 6 \\
    -x + 2y &= 0
    \end{align*}
    How about this?
    \begin{align*}
    2x + 3y &= 2 \\
    4y-2z &= 0 \\
    -3x + 5z &= 10
    \end{align*}
     
  11. Apr 22, 2017 #10
    WHAT?? NO IDEA! Thought simultaneous equations but no
     
  12. Apr 22, 2017 #11

    vela

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    Yes, they're simultaneous equations. You've solved them before. If you don't remember how, you need to go back and review the methods.

    No one here is going to work it out step by step for you because that's against forum rules. Berkeman already gave you a suggestion on how to finish the problem above, but you need to do the actual work. Don't be afraid to make mistakes. Be brave.
     
  13. Apr 22, 2017 #12

    kuruman

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    Here is another strategy. You have three equations. One of them is KCL ##I_1=I_2+I_3##. OK, replace all occurrences of ##I_1## with ##(I_2 + I_3)## in the two KVL equations. You will get two equations having unknowns ##I_2##, ##I_3##. Solve one of them to find ##I_2## in terms of ##I_3##. Replace the expression for ##I_2## that you get in the other equation. This will give you a single equation with ##I_3## as the only unknown. Solve it. Then go back and find ##I_2## and finally go back and find ##I_1##. There are less cumbersome ways of getting to the answer as other have suggested, but this is straightforward and easy to understand.
     
  14. Apr 22, 2017 #13
    That is really long and I want to know a quicker way please
     
  15. Apr 22, 2017 #14

    kuruman

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