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Nodal Analysis

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    upload_2014-11-11_10-36-47.png
    Answer the following questions about the given circuit. These questions will walk you through the steps of the analysis process we call nodal analysis, which is applicable to all electric circuits.
    2.1 (3 pts) A quick Fermi-type problem before we start. Use Ohm’s Law to approximate the value of the currents


    2.2 (6 pts) Apply KCL at the labeled nodes: Using the currents as defined, write KCL at the labeled nodes. Do not change the direction of any currents from what I have shown! What is the voltage at the right hand end of resistor R3? (Remember that Node 0 serves as the voltage reference or 0V.) How many KCL equations do you have?


    2.3 (6 pts) Make the dictionary: Using Ohm's Law, write expressions for all six defined currents.


    2.4 (3 pts) Substitute the Ohm's Law expressions for the currents into the equations established in 2.4.


    2.5 (6 pts) Write the matrix equation corresponding to the equations you determined in 2.6.


    2.6 (3 pts) Use MATLAB or your calculator to solve the matrix equation for the node voltages.


    2.7 (3 pts) Now that you know the node voltages, solve for all six currents. Are the values of i3and i6y 0? Does this match your preliminary Fermi estimate?

    2. Relevant equations
    [itex] I = \frac{V}{R} [/itex]
    [itex] i = vG [/itex]


    3. The attempt at a solution
    I'm not really knowledgeable about circuits (this is just an introductory lesson). However, for the Fermi problem I was thinking that I could assume that the voltage is constant throughout the parallel parts of the circuit (is this a valid assumption). Furthermore, for part 2.2, I was thinking that I would need to write an Ohm's Law expression for each [itex] i_{a} [/itex]. Also, I know that [itex] G = \frac{1}{R} [/itex] so I was planning to account for G that way. I hope this is the right way to proceed, I was hoping someone can confirm. Thanks
     
  2. jcsd
  3. Nov 11, 2014 #2
    I don't believe this problem would be too difficult if it was just a regular circuit, but I am confused about how to account for the extended portion.
     
  4. Nov 11, 2014 #3
    I have some values for i, I wanted to see if I am on the right track:
    [itex] i_{6} = \frac{10-v{3}}{20} [/itex]
    [itex] i_{4} = \frac{v_{2}-v{3}}{1} [/itex]
    [itex] i_{1} = \frac{v_{2}-v_{1}}{41.61} [/itex]
    I got 41.67 by adding R2 and R1.
     
    Last edited: Nov 11, 2014
  5. Nov 11, 2014 #4

    ehild

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    Take care of the magnitude. 20 in the first equation is 20 MΩ and 1 in the second one is 1 mΩ.
    R1 and R2 are parallel resistors, their resultant is not 41.67 Ω.
     
  6. Nov 11, 2014 #5
    thanks for the response. I recalculated the resultant of R1 and R2 and found it to be approximately 10 ohms. Assuming, my equations for 'i' above are correct I found the following equations for each node. I was hoping you could let me know if I am on the right track:
    Node 3: [itex] \frac{10-v_{3}}{20 M \Omega} - \frac{v_{3}-0}{10 \Omega} - \frac{v_{3}-v_{2}}{1 m \Omega} = 0 [/itex]
    Node 2: [itex] \frac{v_{3}-v_{2}}{1 m \Omega} - \frac{v_{2}-0}{10 \Omega} + 1 = 0[/itex]
    Node 1: [itex] 1 - \frac{v_{2}-v_{1}}{10 \Omega} = 0 [/itex]
    Assuming that these equations are correct, my next step would be to formulate a matrix and plug it into MatLab, no?
     
    Last edited: Nov 11, 2014
  7. Nov 11, 2014 #6
    I'm thinking that maybe my answer is way off track here, I'm going to read up on the subject and see if I can reach a different conclusion
     
  8. Nov 11, 2014 #7

    ehild

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    Check node 2. I3 flows through 20 MΩ.
     
  9. Nov 12, 2014 #8
    ok here are my updated current equations, I was hoping somebody could let me know if I'm on track here:
    [itex] i_{6} = \frac{10-v_{3}}{20M \Omega} [/itex]
    [itex] i_{4} = \frac{v_{2}-v_{3}}{1m \Omega} [/itex]
    [itex] i_{1} = \frac{v_{1}-v_{2}}{10 \Omega} [/itex]
    [itex] i_{3} = \frac{v_{2}-0}{20M \Omega} [/itex]
    [itex] i_{5} = \frac{v_{3}-0}{10 \Omega} [/itex]
    And for [itex] i_{2} [/itex] I would not need an equation because I calculated the resultant of resistors [itex] R1 \hspace{2 mm} {and} \hspace{2 mm} R2 [/itex] right?
    [itex] \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [/itex].
    I will post my updated Node equations momentarily
     
  10. Nov 12, 2014 #9
    Here are my node equations:
    Node 3: [itex] \frac{10-v_{3}}{20M \Omega} + \frac{v_{2}-v_{3}}{1m \Omega} - \frac{v_{3}-0}{10 \Omega} = 0 [/itex]
    Node 2: [itex] \frac{v_{1}-v_{2}}{10 \Omega} - \frac{v_{2}-v_{3}}{1m \Omega} - \frac{v_{2}-0}{20M \Omega} = 0 [/itex]
    Node 1: [itex] 1 - \frac{v_{1}-v_{2}}{10 \Omega} = 0 [/itex]
    Now I can formulate my matrix, no? Also, the problem asks what the voltage would be at the right hand end of R3, it would just be 10V right?
     
  11. Nov 12, 2014 #10
    Also, for the above equations I assumed that current going into a node is positive and current leaving a node is negative.
     
  12. Nov 12, 2014 #11
    I'm a little confused about setting up this matrix, is there a specific formula I should follow?
     
  13. Nov 12, 2014 #12

    ehild

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    You know that v1-v2= 10 V from the third equation.
    You are left with 2 unknowns, v2 and v3. Do you say that you can not solve a simple two variable system of equations? :D

    Collect the terms with v2 and also the terms with v3. You get two equations of form av2+bv3=Q; cv2+dv3=R.
     
  14. Nov 12, 2014 #13
    Thanks for the response again, I'm sure I could solve by hand but I was having trouble figuring out how to formulate a matrix to put in matlab...I'll try by hand first and probably that will help...ps I can solve systems of equations with the best of them :D
     
  15. Nov 12, 2014 #14
    Also, just to be sure, what I did(ignoring the extended portion of the circuit) is a legal move because I found the resultant resistance, right?
     
  16. Nov 12, 2014 #15

    ehild

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    I do not understand what you did. Show your calculation and result.

    I am not good in Latex.
    You have system of linear equation for v1, v2 and v3 in the form

    a11 v1 + a22 v2+a13 v3= Q1
    a21 v1 +a22 v2 +a33 v3 = Q2
    a31 v1 + a32 v2 +a33 v3 = Q3
    Arrange the aik-s in matrix form, and the right side as a vector Q. v1, v2, v3 are also components of the vector v.
    Then you have a matrix equation A v = Q
     
  17. Nov 12, 2014 #16
    Ok, I was confused about how to deal with the extended portion of the circuit, I was told that you could do this:
    [itex] \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} [/itex]
    I performed that calculation and found that the resultant was 10 ohms, I was also told that after doing this I could ignore the extended portion of the circuit as long as I wrote that [itex] R_{1} = 10 \Omega [/itex]. Somehow, this doesn't seem correct to me though, so I am confused about how to account for the extended portion...I feel like [itex] i_{2} [/itex] must impact the circuit somehow or else it would not have been included in the problem..
     
  18. Nov 12, 2014 #17
    Basically I am wondering if [itex] i_{2} [/itex] should appear in my node equations for nodes 1 and 2...
     
  19. Nov 12, 2014 #18

    ehild

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    You can solve for the voltages v1, v2, v3, without including i1 and i2. Determine them from the equation i1= (v2-v1)/16.67 and i2=(v2-v1)/25.
    Or you can write the equations without simplifying, then include both i1 and i2 in the equation for node 1. 1=(v2-v1)/25 + (v2-v1)/16.67.
    So you either include both i1 and i2 or only their sum, which is 1 A.
     
  20. Nov 12, 2014 #19
    ok, so then my above equation was correct, thanks for the help!
     
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