Circuit Question

Discussion in 'Homework Help' started by Digit0001, Mar 19, 2011.

  1. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    Hi

    The two circuit i am having trouble getting the correct answers and could someone see if my equations are correct.

    Attachment: View attachment tutqns3.pdf

    Question 1:

    Mesh 1
    24ix + 30(ix - i2) = 12
    54ix - 30i2 = 12

    Mesh 2
    70i2 + 7ix + 30(i2 - ix) = 0
    100i2 - 23ix = 0

    Find ix by Mesh1 + Mesh 2 gives

    ix = 225.95mA book's answer is 254.8mA

    Question 2 equations
    Short circuit with current sources

    i1 = V1/10
    i2 = (V1 - V2)/8
    i3 = V2/5

    Node V1)
    4 + i2 = i3

    Node V2)
    2 = i1 + i2 + 4

    Open circuit with voltage source

    mesh:

    8i1 + 12 + 5i1 + 10i1 = 0

    P.S
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    For question 1 your two mesh equations are correct - you just haven't solved them correctly. The book answer is correct.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Question 2

    If I was applying superposition I would consider each source in isolation.

    So, for the 12V voltage source in isolation (current sources are considered open circuit). You have a voltage divider situation with the 5Ω in series with 18Ω and the 12V source. The 12V source would make Vo negative.

    Vo=-(5/23)*12=-2.609V

    Now find the Vo values for each of the two current sources considered in isolation. Essentially the current in the 5Ω can be found using the current divider rule for each situation.

    The book solution checks out OK.
     
    Last edited: Mar 20, 2011
  4. Digit0001

    Thread Starter Member

    Mar 28, 2010
    89
    0
    i have another two problems i am stuck on or have got the wrong answer to:
    View attachment tut3qns2.pdf

    Question 1
    using Figure 1
    Equation 1 Node V1:

    6 = V1/40 + V1-V2/10

    6=(V1+4V1-4V2)/40

    240 = 5V1-4V2

    Equation 2 Node V2:
    V1-V2/10 + 4Io = V2/20

    V1-V2+4Io = V2/20

    20(V1-V2+40Io) = 10V2

    20V1-20V2+800Io = 10V2

    20V1-20V2+800((V1-V2)/10) = 10V2

    100V1 - 100V1 = 10V2

    100V1 - 110V2 = 0

    Equation 1 - Equation 2
    240 = 5V1 - 4V2 multiply 20
    0 = 100V1 - 110V2

    4800 = 100V1 - 80V2
    0 = 100V1 - 110V2

    4800 = 30V2

    V2 = 160volts
    V1 = 176volts

    therefore Vo = 176 - 160 = 16volts

    Io= 16/10 = 1.6A

    Using Figure 1.1
    -V1/10 + 4Io + 1.5 = V1/20

    -V1 + 4V1/1o + 1.5 = V1/20

    20(-5V1+15) = 10V1

    -100V1 + 300 = 10V1

    -110V1 = -300

    V0=V1 = 2.72

    Io=0.272

    Therefore Total Vo = 16 + 2.72 = 18.72volts

    Total Io = 1.6 + 0.272 = 1.872A

    Question 2:
    Using Nodal Analysis
    Figure 3
    KCL
    ix + i1 = 10

    -V1/4 + -V2/8 =10

    -2V1 - V2/8 = 10

    80 = -2V1 - V2

    V1 = -40volts

    KVL
    4ix + V1 - V2

    4(-V1/4) + V1 -V2 = 0

    V2 = 0volts

    Using figure 3.1
    2Ix - 4Ix + 8Ix = 0

    14Ix = 0

    P.S
     
    Last edited: Mar 25, 2011
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