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Homework Help: Using kirchoffs laws to find current

  1. Feb 8, 2010 #1
    Calculate the current in each of the resistors and indictate the direction

    of current flow.

    what is the potential difference between A & B

    I have attached the circuit & labelled the the circuit directions


    using kirchoffs laws -


    1) i3 = i1+i2

    FOR THE LOOP AFEBCD

    2) 20i1 - 10 + 1 - 5i2 = 0

    THEN FOR LOOP ABCD

    3)-10i3 + 1 - 5i2 = 0

    SUB FOR i3

    -10(i1+i2) + 1 - 5i2 = 0

    -10i1 -10i2 + 1 - 5i2 = 0

    4)-10i1 + 1 - 5i2 = 0

    SUBTRACT equation 2 - 4

    20i1-9 - 5i2 = 0
    -10i1 + 1 - 5i2=0

    10i1 + 10 = 0

    i1 = 1

    THEN SUBSTITUTE FOR i1 in equation 4

    -10 + 1 - 5i2 = 0

    9 - 5i2 = 0

    SO

    i2 = 1.8

    Then from E1 i3 = i1 + i2

    i3 = 2.8

    hopefuly this is correct, i'm not sure how i go about finding the potential difference between the two points ?
     

    Attached Files:

  2. jcsd
  3. Feb 8, 2010 #2
    can you please indicate what is the direction of each loop in your solution (clockwise or counterclockwise) .. so I can follow your solution and check ..
     
  4. Feb 8, 2010 #3

    ehild

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    You made a mistake from here

    -10i1 -10i2 + 1 - 5i2 = 0--->

    4)-10i1 + 1 - 15i2 = 0 : this is the correct equation.

    The potential increases from B to A as the current i3 flows from A to B. The change of the potential across the 10 ohm resistor is 10*i3: U(A)=10*i3+U(B). The potential difference between A and B is U(A)-U(B)=10*i3.

    ehild
     
  5. Feb 9, 2010 #4

    The left hand loop is clockwise running EFAB
    the right hand loop is counterclockwise running DABC

    ok if i use the above equation then i get stuck trying to find a value for i1



    4)-10i1 + 1 - 15i2 = 0

    equation 2 - 4

    20i1-9 - 5i2 = 0
    -10i1 + 1 - 15i2=0

    10i1 + 10 -20i2 = 0

    how do i proceed as I can’t get cancel out the i1 or i2 to get a result
     
    Last edited: Feb 9, 2010
  6. Feb 9, 2010 #5

    ehild

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    There are several ways to proceed. For example, you can multiply eq.(4) by 2 an then add it to eq (2):

    -20i1 + 2 - 30i2=0 (4a)
    20i1-9 - 5i2 = 0 (2) (2)
    _____________________

    -7-35i2=0

    go ahead.

    ehild
     
  7. Feb 9, 2010 #6
    Ok so rearrange

    -35i2 = -7

    i2 = 5

    then substitue to find i1

    10i1 + 1 - 15*5 = 0

    10i1 -59 = 0

    i1 = 5.9 ?

    i3 = 10.9

    i am really confusing myself now,
     
  8. Feb 9, 2010 #7

    ehild

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    All wrong. Amazing!


    You have the equation:

    -7-35*i2=0

    add 35i2 to both sides (the equation stays valid if you add, subtract, multiply or divide both sides with the same quantity.)

    -7=35*i2

    Divide both sides by 35:

    -7/35 = i2 --->i2=-0.2 A

    Could you follow?

    ehild
     
  9. Feb 9, 2010 #8

    rl.bhat

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    -35i2 = -7
    This is wrong. It should be
    35i2 = -7
     
  10. Feb 9, 2010 #9
    ok i follow that, sorry for all the mistakes but i am only learning !!!

    Now I need to get i1 so if I use E2

    20i1 – 9 – (5*-0.2) = 0
    20i1 – 9 – 1

    20i1 – 10 = 0

    Add 10

    20i1 = 10
    Divide by 20

    So i1 = 0.5A

    And therefore i3 = i1 + i2

    0.5A + -0.2A = 0.3A

    i have checked this though so i hope no more errors
     
  11. Feb 9, 2010 #10

    ehild

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    How old are you?

    20i1 – 9 – (5*-0.2) = --->

    5*(-0.2)=-1

    -(-1)=+1

    20*i1-9+1=0

    20*i1 - 8 =0

    20*i1 = 8

    i1 = 8/20

    i1 = 0.4 A



    ehild
     
  12. Feb 9, 2010 #11
    why does that matter - i'm 14 ?

    so i make mistakes isnt that why i asked for help, if i knew the answer i wouldnt be here would i ?

    thanks for the help (and making me feel bad) anyway
     
  13. Feb 9, 2010 #12

    ehild

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    Sorry, I did not want to make you feel bad. I asked your age because you know a lot of Physics for your age, so I wondered why do you make so silly mistakes in Maths. Forgive, me please...

    ehild
     
  14. Feb 9, 2010 #13
    Thats ok,
    Thanks for all the help :approve:
     
  15. Feb 20, 2010 #14
    Hi. I'd like to ask a question about this. In setting up the equations 2 and 3, I don't really get why the signs of the terms are what they are. For example, in equation2), 20i1 was positive but 5i2 was negative. And then in equation 3 they are both negative. When I had a go at this from scratch, I had the signs in equation 2 the other way around and needless to say i got different answers. I had the direction of the current just as victoriafello did too.
    Any clarification is appreciated.
     
  16. Feb 21, 2010 #15
    If you assign directions to the current, then going round them you have to respect the directions you've allocated. If you go round the loop through a resistor following the same direction as the current, the value will be negative (-i x R) because that's what a resistor does. If you go the other way round when calculating the loop, you have to make the sign positive. When you go through a cell or battery you have to respect the potential difference. When you go from the negative terminal to the positive terminal you use the positive V and negative V for the other direction.
    Is that clear?
     
  17. Feb 21, 2010 #16
    Thanks
     
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