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Intro to Elec node analysis help

  1. Aug 29, 2005 #1
    Hi Guys,

    I am new to Electrical Engineering and am trying to figure out the voltages for (1) (2) (3) in the link below, assuming that a 5V dc power supply is connected to the terminals on the left. I am new to Elec Engineering and am struggling a bit. I thought to try Node Voltage analysis to get the answers. Obviously (1) =5V

    However for (2), I am not sure how to figure out the equations.

    Link is

    http://photobucket.com/albums/a339/Eliatamby/?action=view&current=SV400003.jpg

    I was thinking the KCL equation at node (2) would be:

    (v1-v2)/100 = v2/3900 + (v2-v3)/220

    Also, I have a general question. For the current entering node (2) can't I get uses Ohm's Law at the 100Ohm resistor? E.g. 5/100 so the current entering v2 would be 0.05 A. Then I could sub this into the (v1-v2)/100 equation to find out the value of v2? Would that work?

    Thanks for your help, please let me know if this is not clear
     
  2. jcsd
  3. Aug 29, 2005 #2
    Ok I came up with two equations. The first being the KCL at node 2 (slightly modified from above)

    (v2-v1)/100 = v2/3900 + (v2-v3)/220

    The second is the KCL at node 3 :

    (v3-v2)/220 = v3/1000 +v3/1200

    I simplified the first equation to 223v2 = -195v3 + 2145

    I simplified the second equation to v3=300v2/179

    When I sub equation 2 into equation 1, I get 3.9 for v2 which seems about right. When I sub that value into equation 2 for v3, I get around 6, which doesn't seem right at all.

    I tested the circuit in my lab a last week and got v2=4.3 and v2=3 (approx). Are there things wrong with my equations? Am I not using KCL (or node analysis) correctly?

    Thanks for any help, I am really stuck
     
  4. Aug 30, 2005 #3
    I actually think I have got it!!!

    I realised after going through my textbook that I am not denoting the voltage across the resistors consistantly.

    I chose the equations:

    (v1-v2)/100 = (v2-v3)/220 + v2/3900 (1)

    (v2-v3)/220= v3/1000 + v3/1200 (2)

    In retrospect this is a such a simple thing, but something I just overlooked because I never thought properly about it. But solving these two equations gave me figure which were with 0.2 V of what I measured myself.

    No subject makes me feel so slow as Elec1011... :frown:
     
    Last edited: Aug 30, 2005
  5. Sep 2, 2005 #4
    using nodal analysis, i arrived with 2 equations:

    -(V2-V1)/100 - (V2-V3)/220 - V2/3900 = 0 <equation1>

    -(V3-V2)/220 - V3/1000 - V3/1200 = 0 <equation2>

    also:

    V1 = 5 <equation3>

    since the 3 equations are reducible to 2 unknowns, therefore

    -(V2-5)/100 - (V2-V3)/220 - V2/3900 = 0 <equation4>

    -(V3-V2)/220 - V3/1000 - V3/1200 = 0 <equation5>

    then i manipulated equations 4 and 5 and arrived with equations 6 and 7

    -12700V2 + 3900V3 = -42,900 <equation6>

    60/11 V2 - 421/55 V3 = 0 <equation7>

    solving equations 6 and 7 sumultaneously and using my casio fx-991ES calculator, i arrived with

    V2 = 4.324203318 V

    and

    V3= 3.081380037 V

    since the values i got were close enough to the measured value you obtained in your lab, i therefore conclude that my answer is certified true and correct<i guess!!!Ü>

    we just had our exam for the nodal and mesh analysis about 2 to 3 weeks ago that's why the topic is still fresh from my memory

    by the way, do you have examples about norton and thevenin's theorem???

    if you do, kindly post some<like this problem u posted> because i only have some examples...thanks!
     
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