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Finding load current using kirchoffs

  1. Apr 11, 2013 #1
    Hi there attached is a task where i had to calculate the value of the load using kirchoffs law I would be grateful if someone can check this to see if its correct, its part of an assignment question.
     

    Attached Files:

  2. jcsd
  3. Apr 11, 2013 #2

    gneill

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    Staff: Mentor

    Looks okay to me.
     
  4. Apr 11, 2013 #3
    Yup, looks right. Nice work.
     
  5. Apr 12, 2013 #4
    You did get the correct final answer, but I wouldn't call it nice work.

    Under the heading "Loop B E F C" you have a number of algebra errors.

    Your next to last equation has inconsistent units, a problem inherited from above.

    Furthermore, in the next to last equation, you didn't divide the 3V by 4Ω without which you can't get a numerical value of .851

    Don't turn this paper in.
     
  6. Apr 12, 2013 #5
    So where have I gone wrong? It really is bugging me because I know my answer is right as I have done this circuit with nortons theorum and it gave me 0.444a, but with this was a stab in the dark with kirchoffs but the paper requires me to use kirchoff I'm all over the place:(
     
  7. Apr 12, 2013 #6
    The first equation you have after the heading "Loop B E F C" is:

    2v = 6ΩI2 + 1ΩI1(I2-I1)

    The I1 in red shouldn't be there. Then next you have:

    2V = 1ΩI1 + 7ΩI2 but it should be:

    2V = -1ΩI1 + 7ΩI2

    Next you have :

    3V = 4ΩI1 - 1ΩI2 which you brought down from Loop A B C E; it is correct.

    But next you rewrote equation (2) introducing a typo. The coefficient of I1 should be -1Ω (although you had +1Ω; you didn't even copy what you had correctly, but it was wrong anyway. But, at the very least you should be able to copy it without introducing a typo). You have:


    2V = -4ΩI1 + 7ΩI2

    but it should be:

    2V = -1ΩI1 + 7ΩI2

    Then you copied equation 1 again:


    3V = 4ΩI1 - 1ΩI2

    Next, you multiplied your incorrect version of equation 2 by 4:

    8V = -16ΩI1 + 28ΩI2

    but it should actually be:

    8V = -4ΩI1 + 28ΩI2

    Apparently, what you did at the end was to add these two equations:

    3V = 4ΩI1 - 1ΩI2

    8V = -16ΩI1 + 28ΩI2

    and got as a result:

    11V = 27ΩI2

    But if you had done the algebra correctly, you would have gotten:

    11V = -12ΩI1 + 27ΩI2

    which would have given you a wrong answer.

    The two equations you should have added are:


    3V = 4ΩI1 - 1ΩI2 This is the correct equation (1) from Loop A B C D

    8V = -4ΩI1 + 28ΩI2 This is the correct equation (2)

    Now, if you add these two equations, you get:

    11V = 27ΩI2 which is correct. You somehow got this even though your algebra was incorrect.
    -----------------------------------------------------------------

    Next, you aren't keeping track of your units properly.

    Under the heading "Sub I2 into equation (1)", you have:

    3V = 4ΩI1 - 1ΩI2 (0.407a)

    which then becomes:

    3V = 4ΩI1 - 0.407a

    but the second term on the right hand side isn't amps; you got it by multiplying 1Ω times .407 amps which gives .407 volts, not .407 amps.

    The units problem still exists in the next equation:

    3V + 0.407a = 4ΩI1

    It should be 3V + 0.407V = 4ΩI1

    Then you would have:

    I1 = 3V/4Ω + 0.407V/4Ω = 0.851a

    I think you need to practice your algebra skills some more, and exercise greater care with your units.
     
  8. Apr 12, 2013 #7

    gneill

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    Staff: Mentor

    The Electrician is correct. I paid more attention to your initial loop equations and the final expressions for the result values than to the mechanics of the math lying in between. Shame on me :redface:

    Clean up your algebra as The Electrician suggests.
     
  9. Apr 12, 2013 #8
    First of all let me thankyou guys for your help, You are right electrician my algebra is attrocious:( But I've had another go hopefully it is right this time thankyou one more time.
     

    Attached Files:

  10. Apr 12, 2013 #9
    Under the heading "Loop A B C D", you are not handling your units in a consistent way; you should have:

    3v = 3ΩI1 + 1Ω(I1 - I2)
    3v = 3ΩI1 + 1ΩI1 - I2
    3v = 4ΩI1 - 1ΩI2

    and under the heading "Loop B E F C", the 5th and 6th equations down, you should have:

    3v = 4ΩI1 -1ΩI2 You copied this wrong; the 4 is not negative.
    8v = -4ΩI1 + 28ΩI2 You forgot to multiply the -1 by 4.

    You've got to be more careful or you're going to have trouble come exam time!
     
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