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Problem With Nodal Analysis Question

  1. Aug 11, 2014 #1
    1. The problem statement, all variables and given/known data

    https://courses.edx.org/static/content-mit-802x~2013_Spring/html/ps3_p3_fig1_a.png [Broken]

    $$ R_1 = 7 \Omega \quad
    R_2 = 49 \Omega \quad
    R_3 = 52 \Omega \quad
    R_4 = 91 \Omega \quad
    R_5 = 101 \Omega $$
    $$ V_1 = 5V \quad V_2 = 39V \quad V_3 = 44V$$

    2. Relevant equations

    $$ I = \frac{V}{R} $$

    3. The attempt at a solution

    I tried writing KCL for nodes B and M.

    I follow the direction of the current in the circuit diagram. Furthemore, current flows into node M in the branch with the R_5 resistor, and flows away from node M into the branch with the R_3 resistor on the side (I ignore node N because I assume that it is the same node as node M).
    Also, currents leaving a node are negative while currents that enter a node are positive.

    I ground node M.

    When I first see the negative terminal of a battery source, I count that as a voltage lift, otherwise, it is a voltage drop.
    $$
    \\
    \\
    \frac{(D-(B+V_3))}{R_2} + \frac{((M-V_2)-B)}{R_2} - \frac{(B-A)}{R_2} = 0
    \\
    \\
    \\
    \frac{(G-M)}{R_5}-\frac{(M-D)}{R_3} - \frac{((M-V_2)-B)}{R_2} = 0
    $$
    The first equation was for node B and the second was for node M.

    Some places where I might have messed up:
    I assume that G is +5V because it is next to the positive terminal of the battery and A is -5V since it is next to the negative terminal of the battery.
    For node M with the branch that has resistor R_4, I subtracted the voltage V_2 from node M since we hit the positive battery terminal which a represents voltage drop. While for node B, I added the voltage V_3 to B since we hit the negative battery terminal first, which represents a voltage lift from the battery.

    I then simplify the equation with those inferences and plug in the values for the the voltage and resistance of each branch:
    $$
    \\
    \\
    \frac{(D-(B+44))}{49} + \frac{((0-39)-B)}{91} - \frac{(B-(-6))}{7} = 0
    \\
    \\
    \\
    \frac{(6-0)}{101}-\frac{(0-D)}{52} - \frac{((0-39)-B)}{91} = 0
    $$
    I plug in the two equations into wolfram alpha to solve the system, and I get that
    $$B = -4.9V \quad D= -22.7V$$
    http://www.wolframalpha.com/input/?...-6)/7)+=+0,++6/101+-+(-d)/52+-+(-39-b)/91+=+0

    I checked to see if the node potentials for B and D were correct by solving for I_2.
    $$ I_2 = \frac{(B-V_3)-D}{R_2}
    \\
    \\
    I_2 = \frac{(-4.9-44)+22.7}{49} = -0.53A$$
    While the correct answer is 0.449A.

    Could someone help me see where I went wrong? Thank you.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 12, 2014 #2
    Firstly, there is only two essential nodes, so once you set node (M,N) to ground that leaves node B as the only other essential node. This means you only need to write ONE KCL equation to solve for one unknown nodal voltage. I think the problem you have here is that you need to write KVL equations. Ill start you off,

    Going from B->A->G->M, this is the branch current that connects the two essential nodes. We start at voltage VB, so

    VB-VR1+V1-VR5=0
    VB+V1=VR1+VR5
    Here I have the voltage drop across the resistors in terms of the source and the unknown,
    so for this branch of KCL i would have
    [itex]\frac{Vb+V1}{R1+R2}[/itex]+...+...=0

    Now do the same for B->M and B->D->M, I do my current direction from the Node VB to the reference node, and came out with -0.449A, so 0.449A is correct for I2.
     
  4. Aug 12, 2014 #3
    Is it possible to solve for the currents using KCL?
     
  5. Aug 12, 2014 #4

    gneill

    User Avatar

    Staff: Mentor

    Sure, but slightly indirectly. The KCL node equation is comprised of the sum of the individual branch currents entering/leaving the node. Once you solve for the node voltage you can break out each of those terms individually to give the individual currents.
     
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