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Kirchhoff's law

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    find the voltage between a and d and terminal of each batteries
    Untitled.png


    2. Relevant equations
    Junction Rule and Loop Rule

    3. The attempt at a solution
    I got three equation
    JR : I3 = I1+I2
    LR :
    45-I3-47I3-34I1 = 0
    45-I3-47I3+85-I2-18I2 = 0
    I can't solve these variables :(
     
  2. jcsd
  3. Oct 16, 2016 #2

    cnh1995

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    There are three equations with three unknowns. You can use the substitution method.
     
  4. Oct 16, 2016 #3

    FactChecker

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    You have 3 linear equations in 3 unknowns (I1, I2, I3). So the equations can be solved unless they are self-contradictory.
    I assume that you have not dealt with matrices yet, but you can still solve these.
    1) Use the first equation to replace every I3 in the other 2 equations.
    2) Solve the modified second equation for I2 in terms of I1.
    3) Use that result to replace all the I2s in the modified last equation.
    4) Solve the modified last equation for I1. That should give you a number for I1.
    5) Plug the I1 number into the result of step 2 to get a number for I2.
    6) Plug the I1 and I2 numbers into the first equation to get a number for I3.
    So now you have the values of all I1, I2, I3.
     
  5. Oct 17, 2016 #4
    ok, I give up, is there other way to solve this?
    some says I can actually use ohm's law directly

    Vad=(ΣI) (Σ1/R)
     
  6. Oct 17, 2016 #5

    gneill

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    There are other methods of analysis, some of which will yield fewer equations to solve. But you haven't learned of these methods yet, and they won't be taught until you've learned the fundamentals of Kirchhoff's laws. All other methods are based upon this foundation.

    But most importantly, in general all methods involve solving simultaneous equations when the circuit is more complex than a pair of nodes. You won't be able to avoid this; You will have to become familiar with solving simultaneous equations.

    Regarding what "some says", you can't use Ohm's law directly to solve this circuit, but there is a method that incorporates Ohm's law along with Kirchhoff's laws. An analysis technique that you will eventually learn of is called Nodal Analysis. It is based upon a clever use of Kirchhoff's laws to find the potentials at the nodes of a circuit, and involves writing equations for each essential node in a circuit (you'll learn what an essential node is, too), one equation for each node, and solving the simultaneous equations to find the potentials at each node with respect to a reference node. Again, simultaneous equations are involved in general, but in the case of this particular circuit there is only one essential node!

    So for this circuit in particular it is possible to write a single equation with one unknown to find the potential of the one essential node with respect to a reference node. From there you could find the individual currents in the branches connecting the essential node to the reference node. Rather than solving simultaneous equations you would solve a single equation for one node potential and then do several individual calculations to find currents and other potentials around the circuit. It would be more work overall to find the branch currents, but would involve no simultaneous equations (for this particular circuit).
     
  7. Oct 17, 2016 #6

    FactChecker

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    You can do it by hand. If you study matrices and linear algebra, there are ways to automate the process.
    Modify Equ 2: 45-I3-47*I3-34*I1 = 0;
    simplify: -48*I3 - 34*I1 = -45;
    use equation 1: -48*(I1 + I2) -34*I1 =45;
    simplify: -82*I1 - 48*I2 = -45;
    Modify Equ 3: 45-I3-47*I3+85-I2-18*I2 = 0;
    simplify: -47*I3 - 19*I2 = -130;
    use equation 1: -47*(I1 + I2) -19*I2 = -130;
    simplify: -47*I1 - 66*I2 = -130;
    I2 = (-45 + 82*I1) / -82 = 45/82 - I1 = 0.55 - I1
    Modified last equ: -47*I1 - 66*I2 = -130;
    replace I2s: -47*I1 - 66*(0.55 - I1) = -130;
    simplify: 19*I1 + 36.3 = -130
    19*I1 + 36.3 = -130;
    19*I1 = -166.3;
    I1 = -166.3/19 = -8.75
    I2 = 0.55 - I1 = 0.55 - (-8.75) = 9.30
    I3 = I1 + I2 = -8.75 + 9.30 = 0.55
    I1 = -8.75;
    I2 = 9.30;
    I3 = 0.55

    You will have to double check all the steps and calculations. It is always good to plug the final answers into the original equations to check that they work and there were no calculation mistakes.
     
  8. Oct 17, 2016 #7

    gneill

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    I can affirm that these values are not correct :smile:

    And FactChecker, you should be aware that doing the work for the member requesting help is against the rules. We can only guide the member to a solution via hints, suggestions for research, pointing out errors, and so forth. They have to do the heavy lifting themselves.
     
  9. Oct 17, 2016 #8

    FactChecker

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    Sorry. I will limit my answers more in the future. I wanted to give a detailed outline of the substitution method. I knew that the answer was wrong and said that he should check the work so that he would go through it in detail and correct it. I know the correct values.
     
  10. Oct 17, 2016 #9

    gneill

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    A better approach would be to present a correct but unrelated example, or steer the member to a tutorial on the web.
     
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