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Node Voltage (is my answer correct )

  1. Aug 20, 2012 #1
    http://www.flickr.com/photos/84781786@N03/7822251544/ [Broken]
    Find the node VOLTAGE for V1,V2,V3,V4

    2. Relevant equations
    KVL


    3. The attempt at a solution

    V1 = 6V
    V4= 16V
    V2= V2/1.5k - 6/1.5k + V2/ 1k + V2/3.3k - V3/3.3k
    6/1500 = V2/5.8k - V3/3.3k
    V3 = V3/2.2k + V3 / 4.7k + V3/3.3k - V2/3.3k - 16/2200
    16/2200 = - V2/3.3k + V3/ 10.2k

    But I'm getting 2 negative answers
    V2= -34.6 V V3= -32.9
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Aug 20, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Why is that V2 = ? Surely you're forming a sum of currents at the node, which should yield a (zero) current?
    How did you arrive at the V2/5.8k term?
    Again, the "V3 =" doesn't make sense. Perhaps you're just being a tad sloppy with terminology? The "=" sign is reserved for use in equations. Did you mean to indicate that "This is the expression for node V3?"

    Also again, can you explain how you arrived at the term V3/10.2k?
     
  4. Aug 21, 2012 #3
    I agree with Gneill - Your equations are a bit off - I think you mean:

    V1: V1 = 6V
    V2: V2 = 16V
    V3: 0 = V2 / 1.5k - V1 / 1.5k + V2 / 1.0k + V2 / 3.3k - V3 / 3.3k
    V4: 0 = V3 / 3.3k - V2 / 3.3k + V3 / 4.7k + V3 / 2.2k - V4 / 2.2k

    Note that using the node voltage method used Kirchov's Current Law which states that all the currents going into a node must equal zero. Using the node voltage method you are simply summing the currents into each node and setting them equal to zero - not V3 or V4.

    If you solve these equations you will come up with the correct answer... it took me a couple tries - my arithmetic is rusty, I guess... but I think I just fat fingered a value into my calculator. I did finally arrive at the correct answer, which I verified with software.

    Hope this helps.

    Also note that:
    V2 / 1.5k + V2 / 1.0k + V2 / 3.3k does NOT equal V2 / 5.8k
    http://www.mathsisfun.com/fractions_addition.html
    I think there was some confusion there as well which could have been the root of your negative number solution.

    On the plus side - good for you for realizing there was no way the answer could be negative without a negative voltage source.
     
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