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Nodal Analysis of a Circuit

Discussion in 'General Electronics Discussion' started by rainbow9, Mar 18, 2016.

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  1. rainbow9

    rainbow9

    3
    0
    Dec 15, 2012
    Im trying to learn nodal analysis.I have the following circuit

    [​IMG]

    I have selected V1 as the reference node.
    I assume that since current flow from + to - terminal.The direction of current is from the positive terminal of the battery (12V) to the negative terminal.Current Splits from the point V1 and goes to the branches.

    So i have

    (v4-v1)/2=(v1-v2)/3+(V1-Vref)/2

    I'm i correct?
     

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  2. dorke

    dorke

    2,342
    665
    Jun 20, 2015
    The equation is correct.
    The directions of currents are chosen arbitrarily,
    but it must be kept consistently through all node equations.

    What do you mean by "V1 as the reference node"?

    I would choose the "ref" point to be the junction between the power supplies.
    This way you have 3 equations with 3 unknowns for the 3 nodes V1-3 to solve.
    In this case,The equation for node V1 is:
    (12-V1)/2=(V1-V3)/2 + (V1-V2)/3


    ref-v.jpg
     
    Last edited: Mar 18, 2016
    rainbow9 likes this.
  3. Ratch

    Ratch

    1,094
    334
    Mar 10, 2013
    Then why do you mark "ref" where the 1,4,2 ohm resistors come together?

    Charge may or may not flow. Current is already flowing. "Current flow" means "charge flow flow", which is wrong and confusing. You should not be wondering which direction the current has. Do the algebra and let the values tell you.

    No, You have one supernode and two regular nodes if your reference is at "ref". That means you need 3 equations to solve the problem. The equations are shown below. You only listed one equation. Ask if you have any questions on how the supernode is calculated. The answers are v1 = 10.4262, v2 = 7.23934, v4 = 22.977

    Rainbow9.JPG

    Ratch
     
    rainbow9 likes this.
  4. rainbow9

    rainbow9

    3
    0
    Dec 15, 2012
    Really sorry for the late reply.Thanks a lot for your help.
     
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