1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Node analysis question

  1. Feb 7, 2007 #1


    User Avatar

    1. The problem statement, all variables and given/known data
    (I have attached a picture of the circuit)
    Using node voltage analysis in the circuit, find the current i drawn from the independent voltage source. Let V=3V, R1=1/2, R2=1/2, R3=1/4, R4=1/2, R5=1/4, I=0.5.

    2. Relevant equations

    3. The attempt at a solution
    I haven't seen one like this w/ the independent voltage source, is it solved the same way? I tried to perform KCL at each node and got:
    node 1: (V-v1)/R1 + v1/R2 + (v1-v2)/R3 = 0
    node 2: (v1-v2)/R3 + (v3-v2)/R4 = I
    node 3: (v3-v2)/R4 + v3/R5 = 0

    When I substituted for V, I, R1, R2, R3, R4 and R5 and entered in the matrix I got:
    Then I related i=(V-v1)/R1. I got an answer of 18.75 A. I thought the answer maybe sounded high and was wondering if I had messed up somewhere.

    Attached Files:

  2. jcsd
  3. Feb 7, 2007 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, but it's even better because an independent voltage source puts a constraint on the node voltages. The independent source is the reason that you have 3 node equations instead of 4.

    Taking the current through R1, R2, and R3 to be going to the left, and taking the current through R2 and R5 to be going down, I wrote down the node equations and solved the model to find that v1=1.375 V (I did this all by hand mind you, so let the buyer beware), which will give a more reasonable current. Check your equations to make sure that you were consistent with the directions of your currents.
  4. Feb 7, 2007 #3
    Keep in mind that the voltage source can be treated this way because it's connected to ground. If it wasn't, as in it is inbetween two nodes, you'd just have to use a supernode with a supplemental.
  5. Feb 7, 2007 #4
    You only need to perform KCL at the nodes with more than 2 branches. And in the figure, there are only 2 such nodes (excepting the ground), i.e., the one common to R1, R2 and R3; and another common to R3, R4 and the current source.

    There is a mistake here. KCL requires either (1) the sum of all currents leaving the node equals zero, or (2) the sum of all currents entering the node equals zero. What you wrote above (for node 1) is not consistent with (1) or (2).
  6. Feb 8, 2007 #5


    User Avatar
    Gold Member

    Indeed, recheck your current directions for node 1 (R1, R2, and R3).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook