branch current issue

Discussion in 'Homework Help' started by acelectr, Sep 1, 2010.

  1. acelectr

    Thread Starter Member

    Aug 28, 2010
    73
    0
    Hi guys. I worked a lot about this branch current problem on this link http://www.allaboutcircuits.com/worksheets/dcbranch.html I couldn't do the second question with the 5mA current source. The result that shown is 1.9mA but I'm keep getting 2.72 or somethng. I'm skipping something and I can't find it. :(:(

    Appreciate any help
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    But 1.9mA is correct answer.
    So maybe you show as your solution.
     
  3. acelectr

    Thread Starter Member

    Aug 28, 2010
    73
    0
    No 1.9 is not what I've found. that is the problem. in short how can one could solve this problem:) pls I worked on it a while and still couldn't get in to a conclusion. waiting for backup...
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    But if you show your approach to the problem then we can tell you in which part you made error.
    And I to check the answer, written nodal equation.
    5mA = (6V - Va)/1K + (7.2V - Va)/1K
     
  5. acelectr

    Thread Starter Member

    Aug 28, 2010
    73
    0
    firstly what I did was to choose the top middle node and determine 3 current directions arbitrarily and tried to apply simultaneous equation method. Well as it is written simultaneous equation is not really needed in order to solve the problem, because the lack of unknown variables. what ever later I applied the kirc's voltage law and see one of the resistors should has a 6 voltage drop across it as the other one has 7.2. I think this is the point where things don't make sense. Because applying ohm's law according to those values didn't really help me. But I also traced each loop and this is what I get. Where do I make the mistake? Though the nodal equation that you wrote, kinda make sense but if I would get an explanation about the contradiction that I'm facing with I would really have a relief.
     
  6. acelectr

    Thread Starter Member

    Aug 28, 2010
    73
    0
    firstly what I did was to choose the top middle node and determine 3 current directions arbitrarily and tried to apply simultaneous equation method. Well as it is written simultaneous equation is not really needed in order to solve the problem, because the lack of unknown variables. what ever later I applied the kirc's voltage law and see one of the resistors should has a 6 voltage drop across it as the other one has 7.2. I think this is the point where things don't make sense. Because applying ohm's law according to those values didn't really help me. But I also traced each loop and this is what I get. Where do I make the mistake? Though the nodal equation that you wrote, kinda make sense but if I would get an explanation about the contradiction that I'm facing with I would really have a relief.
     
  7. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    Ok as per the picture I assume some notation
    R1 = The 1000 ohms on 6V side
    E1 = R1's voltage drop
    I1 = Current through R1 which we will find
    V1 = 6V
    R2 = 1000 ohms .....
    E2 = Voltage drop of R2
    I2 = Current through R2 i.e.. (5-I1 given)
    V2 = 7.2V
    I3 = 5mA
    Let E be the voltage drop across the stuff through which I3 is flowing
    Now
    KVL for left side
    E1 + E - 6 = 0 ------- eq1
    or I1*1000 + E - 6 = 0
    KVL for right side
    E2 + E - 7.2 = 0 ------- eq2
    or 5 - I1 * 1000 + E - 7.2 = 0

    Now we got two equation
    I1*1000 + E - 6 + 0 = 0 --- eq1
    -I1*1000 + E -7.2 + 5 = 0 --- eq2
    Solving for E we get
    E = 4.1
    Putting the value of E in equation 1 we get
    E1 + E - 6 = 0
    E1 = 1.9
    By ohms law
    I = V/R
    I1 = 1.9/1000
    I1 = 0.0019 A i.e.. 1.9 mA
    and
    I2 = 5 - I1
    I2 = 3.1 mA

    Good luck
     
    acelectr likes this.
  8. acelectr

    Thread Starter Member

    Aug 28, 2010
    73
    0
    Thanks a lot man, I was waiting a while for this. Appreciate it. Though I thought that there was only an extra 5mA current and no resistor or no existence of any voltage there. Also I am kind of confused that here adding this 5mA current source reduced the currents at all points in the circuit. Kind of confused about these parts but thanks again;)
     
Loading...