Ques. about Checking Solution.Node Analysis

Discussion in 'Homework Help' started by freemindbmx, Jun 29, 2014.

  1. freemindbmx

    Thread Starter Member

    Mar 5, 2014
    72
    0
    Im having some trouble in explaining/understanding the "checking your work" after you solve your problem, for real world applications.

    I posted the problem below.

    Node analysis-Current leaving is positive

    V1:
    KCL @ V1 = I1+I2+I3=0

    ((V1-21V)/5Ω)+(V1/10Ω)+((V1-V2)/5Ω)=0

    V1(1/5Ω+1/10Ω+1/5Ω)-V2(1/5Ω)=21/5V

    V2:
    KCL @ V2 = I4+I5+I6=0

    ((V2-V1)/5Ω))+(V2/10)+((V2+10.5V)/5Ω))=0

    -V1(1/5Ω)+V2(1/5Ω+1/10Ω+1/5Ω)=-10.5/5V

    I solve for V1 by using the Nodal Equation of V1.
    V1=(8.4+V2(0.4)) Then plug this into Node Equation of V2.

    V2=-1V Then plug this into V1=(8.4+V2(0.4)) V1=8V

    Now to check my work.
    I want to test and see if the voltage's that involve the 21V dependent source.Do in fact add up to 21V.

    V1+I1(5Ω)=21V
    Now do I use my original KCL of V1 which says I1+I2+I3=0 and solve for I1. I1=-I2-I3 Or do I just add up I1=I2+I3 since current leaving must be equal to current entering?

    Plugging in with I1=-I2-I3
    8V+(-8V/10Ω)-((8+1)/5Ω))=-5V Doesn't equal 21V
    But when I just plug in I1=I2+I3 I get 21V
    So how do I put this into words? And I just use the Current Continuity which is current leaving=current entering and not my original KCL of V1?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    hi, I think that you mix up the KCL with KVL

    I1 = (8V - 21V)/5Ω = -2.6A and this means that I1 is flow in the opposite direction to the one on the drawing .

    I2 = 8V/10Ω = 0.8A

    I3 = (8V - (-1V))/5Ω = 1.8A

    So we have

    -2.6A + 0.8A + 1.8A = 0A so every thinks is ok.

    Or

    I1 = -I2-I3 = -0.8A - 1.8A = - 2.6A

    Now we can do KVL But now we have to kept in mind that I1 is flowing from 21V voltage source into V1 node. And this is why
    V1 = -I1*5Ω + I2 *10Ω = -(-2.6A)*5Ω + 0.8A*10Ω = 21V
    And from this we can conclude that I1 = I2 + I3 but now I1 = (21V - V1)/5Ω
     
  3. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hello there,

    You can always check the individual branches to make sure they all work out ok. If you solved for the node voltages then you know the current through that branch because you know the resistance and then you can sum the currents at each node to make sure the sum is zero at each node.

    If you have node 1 voltage of 2v and node 2 voltage of 1v and there is a resistor of 10 ohms between those two nodes, then you know that the current is I=(2-1)/10=1/10=0.1 amps. Doing that for each branch you can then sum them at each node and should come up with zero for all the nodes if the original analysis was correct.
     
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