Loop analysis homework problem

Discussion in 'Homework Help' started by jut, Oct 5, 2007.

  1. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    I am to use the loop analysis method to solve for an unknown current in the following the three problems, which are all the same... you just have to use different loops in each problem.

    I was able to get #9 correct, but could not get the same answers in #10 and #11. The loops with current sources seem to be screwing me up.

    [​IMG]

    Loop analysis method is essentially as follows; draw loops currents, assign polarities, write KCL equations, solve.

    For #9, my loop equations are:
    Loop 1: i1 = 2
    Loop 2: 40 - 4*i2 + 4*i1 - 20 - 2*i2 - 4*i2 + 4*i3 = 0
    Loop 3: 10 +4*i3 - 4*i2 = 0

    ... right :)

    Using matrix algebra to solve, I get... (These results were confirmed with multisim)
    i1 = 2
    i2 = 3
    i3 = 1/2


    For #10, my loop equations are:
    Loop 1: i1 = 2
    Loop 2: i2 = 2
    Loop 3: 10 +4*i3 - 4*i1 = 0

    ... wrong :(

    For #11, my loop equations are:
    Loop 1: i1 = 2
    Loop 2: 40 - 4*i2 + 4*i1 - 20 - 2*i2 - 4*i2 = 0
    Loop 3: i3 = 2

    ... wrong :(

    Any help would be great, thanks for reading.
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    You are adding up the voltages around the loops and setting them equal to zero. You need to know the voltage across the current source. Does that help?
     
  3. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    You can't assign a voltage to a current source. You can only say the current in a loop is equal to the current sources in that loop. Or at least I thought....... doing that is not working for problem 10 and 11.
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Since when using the loop method, what you do is add all the VOLTAGES around the loop, you must select a voltage to be across the current source, but it can be an unknown. I selected the symbol Vc to represent the voltage across the current source, with + at the top. Then the loop equations can be written:

    Loop1:
    -Vc + 20 + 2*I1 + 4*i1 - 4*i3 = 0

    Loop2:
    -Vc + 4*i2 + 40 = 0

    Loop3:
    -4*i1 + 4*i3 +10 = 0

    Set up your matrix equation:

    [ 6 0 -4 ] [i1] ...... [Vc - 20 ]
    [ 0 4 0 ] [i2]...=...[ Vc - 40 ]
    [ -4 0 4 ] [i3].......[ -10 ]

    get a solution vector:

    [ Vc - 30 ]
    [ ------- ]
    [.....2.....]
    [............]
    [ Vc - 40 ]
    [-------- ]
    [......4.....]
    [............]
    [ Vc - 35 ]
    [---------]
    [.....2......]

    I hope this makes sense. The 3 elements of the solution vector are:
    (Vc - 30)/2 and (Vc - 40)/4 and (Vc - 35)/2.

    Take the first two elements of the solution vector and set up the equation:

    (Vc - 30)/2 + (Vc - 40)/4 = 2

    because we do know that i1 + i2 = 2. That is, the 2 amps out of the current source has to equal the sum of i1 and i2. Solve this equation and get Vc = 36. Substitute 36 for Vc in the solution vector and get i1 = 3, i2 = -1 and i3 = .5 Problem solved.
     
  5. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    hi jut,
    i think where u went wrong was u took i=2A in loops which had the 2A source in common,
    the 2A is the current flowing thru the current source(like current flowing thru an element) it is different from the loop current.
    the current thru the 2 A current source should be 2A=i1 + i2 ----for 10
    and 2A=i1 + i3. -------for 11.
    it think the method posted in the above post is a good one.
     
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