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KIRCHOFF's Law: Solving Circuit

  1. Dec 5, 2009 #1
    circuit.jpg

    according to circuit
    In Left Loop Current is I1 and Right Loop is I2

    Now Loop Left
    10-8I1-3I1-I2=0
    which becomes
    10-8I1-3I1+3I2=0
    11I1-3I2-10=0 ------------eq 1


    Now Loop Right
    15-6I2-3I2-I1=0
    15-6I2-3I2+3I1=0
    3I1-9I2+15=0
    I1-3I2+5=0 ---------------eq 2

    Deduct eq 2 from eq 1

    10I1-15=0
    I1= 1.5
    from eq 2 we have
    I2=2.1666

    Now I want to ask
    I= I1-I2
    or
    I=I2-I1
    or what and how to take decision what i have to do
     
  2. jcsd
  3. Dec 5, 2009 #2

    rl.bhat

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    Homework Helper

    When you consider the magnitude of the current it should be +ve. So I = I2 - I1
     
  4. Dec 5, 2009 #3
    Thanks Bhat i got your point, and one thing more what will be the direction current on both the loop
     
  5. Dec 5, 2009 #4

    Redbelly98

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    Staff Emeritus
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    Since you defined I1 and I2 to be clockwise for positive currents, and both currents are in fact positive, then the direction of each is _____?
     
  6. Dec 5, 2009 #5
    yes direction of current is positive in the diagram what if direction is counter-clock wise and one loop clock wise and othe one counter clock wise
     
  7. Dec 6, 2009 #6

    Borek

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    Staff: Mentor

    It doesn't matter how you define things, as long as you are consistent. When you assume one loop to be CW and the other CCW, current on the R3 will be sum of the currents in both loops. If you have assumed wrong (that is, currents don't go as you expect them) one of the currents in your final solution will be negative - and that's perfectly OK, that just means current flows in the opposite direction.
     
  8. Dec 6, 2009 #7
    Thanks Borek now i 'm clear about Kirchoff
     
  9. Dec 17, 2009 #8
    what is a comlex dc circuit
     
  10. Dec 17, 2009 #9

    cepheid

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    Gold Member

    Hello harroon1972, welcome to PF!

    If you have a separate question unrelated to what the OP (original poster) was asking, you should start a new thread for it. I'd be happy to help, but you need to be more specific. As it stands right now, it's not very clear what you're asking.
     
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