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KVL and node voltages

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img163.imageshack.us/img163/408/circuit2w.jpg [Broken]
    We are asked to solve for I. I found I = -4 Amps.
    I want to verify my answer using KVL in the bottom left mesh

    2. Relevant equations

    3. The attempt at a solution
    I follow the convention: voltage drops are positive, which means that the first sign I hit is the sign of the voltage. Starting at the 8 ohm resistor:

    - 8*(-4) + ( ????? ) - ( ?????) = 0

    I am confused because we usually apply KVL using the voltage drops/rise between elements. For the current source and the 2 ohm resistor, I have the node voltages between these elements. Therefore, I don't know if the current source is + (v2 - v1) or - (v2 - v1). Same thing for the 2 ohm resistor, I hit the - sign first, but is it - (v2 -0) or - (0-v2) ????

    I am a little bit confused.. thank you very much
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 14, 2011 #2


    User Avatar

    Staff: Mentor

    Since your circuit has only current sources in it, did you consider using KCL node equations instead of KVL loops? If you solved the node equations for the node V1, then the current you're looking for is just V1/8.

    If you insist on using KVL loops, then one way to approach current sources is to assign variables to the unknown voltage drops across them (choose whichever polarity you want, the math will take care of the details) and then use the current sources to place constraints on the relationships between the loop currents. This gives you the extra simultaneous equations you need to solve for the variables you introduced.

    So suppose that the top loop is Loop2 with current I2 flowing around it. Loop1 is your loop with current I as shown. Both currents are assumed to be running counterclockwise around their loops. Then a suitable constraint equation as imposed by the 4A current source would be I2 - I = 4A.
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