just starting circuits....need a little help

Discussion in 'Homework Help' started by sideline81, Dec 2, 2011.

  1. sideline81

    Thread Starter New Member

    Dec 1, 2011
    4
    0
    I'm just starting to learn circuits so I may just be missing something small here, but I need some help.

    I'm given a single-node pair circuit and need to find iA, iB, and iC. I think whats throwing me off is that there is a dependent current source in the middle of the circuit and I'm not sure how I should handle it.

    Any help would be great.
    Thanks
     
  2. syed_husain

    Active Member

    Aug 24, 2009
    61
    5
    lets assume, the bottom node is grounded. you can make ground anywhere you like but sometimes for certain problems choosing ground makes calculation a lot easier. for your problem you have only two options either top or bottom.

    i will give you a hint. if you follow the current direction you can write according Kirchoff's Current Law (KCL) at the top of the node:

    Code ( (Unknown Language)):
    1. [tex]i_{A}+i_{B}+i_{C}+2 = 5.6[/tex]
    and
    Code ( (Unknown Language)):
    1. [tex]i_{B} = 0.1*V_{x}[/tex]
    the rest is up to you. if you have still problem post your working.
     
  3. sideline81

    Thread Starter New Member

    Dec 1, 2011
    4
    0
    That's the setup I was using and came up with 13.5V

    Here is how I got 13.5V

    5.6 = iA + iB + iC + 2
    5.6 = V/18 + .1V + V/9 +2
    100.4 = V + 1.8V + 2V + 36
    64.8 = 4.8V
    V = 13.5

    Then to find iA i calculated 13.5V/18ohms = .75A, but that isn't right.

    not sure what I'm doing wrong
     
  4. syed_husain

    Active Member

    Aug 24, 2009
    61
    5
    i am also getting the same answer as yours. are you sure about the answer? can you make sure the sum of currents in your answer (not what you calculated) is 5.6?
     
  5. sideline81

    Thread Starter New Member

    Dec 1, 2011
    4
    0
    I get 1A = .75A , iB = 1.35A , and iC = 1.5A
    The problem is an online assignment and when I try those answers it says they are incorrect. When I add up all the currents, they do equal 5.6, which I would normally take to mean they are correct. I don't know what else to try
     
  6. syed_husain

    Active Member

    Aug 24, 2009
    61
    5
    i think i got the solution. the equation

    Code ( (Unknown Language)):
    1. [tex]i_{A} +i_{B}+i_{c}+2=5.6[/tex]
    from this we can write following(i took the ground connection at bottom):
    Code ( (Unknown Language)):
    1. [tex] \frac {-V_{x}}{18} +0.1*V_{x} -\frac{V_{x}}{9} +2 = 5.6 [/tex]
    in previous post i did not notice the polarity sign of Vx given.
    now
    Code ( (Unknown Language)):
    1. [tex]V_{x} = - 54 V[/tex]
    ,
    Code ( (Unknown Language)):
    1. [tex]i_{A} = 3 A,i_{B}= -5.4 A, i_{C} = 6 A[/tex]
    hope this will help

    cheers
     
  7. sideline81

    Thread Starter New Member

    Dec 1, 2011
    4
    0
    ha...I didn't catch that at all. That was the problem though. Thanks a lot for you help!
     
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