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Which way is the current flowing?

  1. Feb 25, 2007 #1
    Please click on the image to make it bigger



    [​IMG]

    I'm having real difficulty picturing how the current is flowing in this circuit with the given solution.

    From the 5v source into node [tex] v_2[/tex] is the current flowing from left to
    right into this node?

    These are the nodal equations that I would use to solve this problem, however they are wrong. Why?

    At node two I would have the current flowing from left to right from the source into this node. I would have the 6mA flowing into this node. Finally I would have a current flowing out of this node from left to right. So I have 2 currents entering this node and one leaving

    [tex] \frac{V_2-5}{200} + 6mA = \frac{V_0-V_2}{6000}[/tex]


    at node two

    [tex]\frac{V_0-V_2}{6000} = \frac{V_0-(-6)}{3000}[/tex]
     
    Last edited: Feb 25, 2007
  2. jcsd
  3. Feb 25, 2007 #2
    Before attempting to solve this kind of circuits, you should probably start by replacing voltage sources with norton equivalents. Basically you're performing transformation from voltage sources -> current sources & when you're done you'll immediately see that this problem is so easy that it can be solved by using only current divider rule

    P.S Voltage source -> current source conversion

    [tex] V_{1} = I_{1}\cdot R[/tex]
    [tex] I_{1} = \frac{V_{1}}{R}[/tex]


    BTW. Another method is to apply superposition theorem to each source and then compute the result. Good Luck!
     
  4. Feb 25, 2007 #3
    Ok, but how can I solve it this way?
     
  5. Feb 25, 2007 #4

    I just told you! ;) You're just modifying circuit to make it easier for you to solve it. Looks like you're not familiar with Norton theorem. Now, take a look at 5V source together with 2k[tex]\Omega[/tex] resistance. Replace the voltage source with a current source where current will have the magnitude equal to [tex]\frac{5}{2k\Omega}[/tex] and a resistor in parallel with that source with resistance equal to [tex]2k\Omega[/tex]. Do the same thing for 6V source together with [tex]3k\Omega[/tex] resistance, redraw the circuit and solve for voltages/currents that you need to obtain.
     
    Last edited: Feb 25, 2007
  6. Feb 25, 2007 #5
    I understand what you are saying,but I have to solve it using nodal analysis. I just want to know which way the currents are flowing into and out of the nodes.


    Yes it can be done how you are suggesting and it makes the problem much simpler, but that is not the point of this exercise for me.
     
  7. Feb 25, 2007 #6
    Ok, first of all, there is only one "real" node here and it lies between 6mA source [tex]2k\Omega[/tex] and [tex]6k\Omega[/tex]. So you start by writing the nodal loop equation at this point:

    Vk is node voltage so (I'm assuming that the current that flows into the node has negative sign, and current flowing out the node has positive sign):

    [tex]\underbrace{\frac{Vk-5}{2k\Omega}}_{sign determines the direction}-6mA+\underbrace{\frac{Vk+6}{9k\Omega}}_{sign determines the direction} = 0[/tex]
    solving this equation will yield something about 12.8 V, knowing that every other current can be found.
     
    Last edited: Feb 25, 2007
  8. Feb 25, 2007 #7
    This is where my confusion lies. In your equation the 6mA is leaving the node (k) and the other currents are entering (+ entering, - leaving). why is this?

    If I can just get past this hole in my thinking I'll be ok I think
     
  9. Feb 25, 2007 #8
    As I've mentioned before, I'm using the following strategy when dialing with nodal analysis:

    1. Current entering the node has negative sign! In this situation the 6mA source points towards the node, so the sign is negative.
    2. I do not no anything about the currents to the left and right of the node, so i will let Vk determine the direction.
    3. I'm adding up everything, according to Kirchoff's rule that says sum of currents at the node is equal to zero.
     
  10. Feb 25, 2007 #9
    You have been very helpful. I appreciate your time and effort.
     
  11. Feb 25, 2007 #10
    no problem ;)
     
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