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KCL/Nodal Analysis Problem

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    solve for voltage V with polarity as shown.
    1ethjm.jpg


    2. Relevant equations
    KCL - current into a node = current out


    3. The attempt at a solution
    I tried to do nodal analysis using the top center node as v.

    v/240 + (v-9)/100 - 20mA - v/100 = 0

    => v=26.4V

    Obviously, I'm doing something wrong. Any tips to get me back on the right track?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 8, 2009 #2
    I think current leaving the node is positive (+20mA) and the current through the middle resistor must be the same as that current source since it's the same wire
     
  4. Feb 8, 2009 #3

    LowlyPion

    User Avatar
    Homework Helper

    The top node is not at V.

    The current through the middle leg is constrained to 20 ma however, and you know that the currents coming in the top must sum to 20 ma.

    I1 + I2 = 20 ma.

    Hence construct your Voltage loops.

    I think that will resolve into the current I2 that you need to determine the voltage across R3.
     
  5. Feb 8, 2009 #4
    I'm confused again. How do I solve for the i1 and i2?

    sum of the voltages in a loop = 0, so for loop 1 with current i1 on the eft:

    v1 + v2 = 0
    i1*r1 + i1*r2 = 0
    i1*240 + i1*100 = 0

    loop 2 with current i2 on right:

    v2 + v3 = 0
    i2*r2 + i2*r3 = 0
    i2*100 + i2*100 = 0


    I know I'm not calculating the loops correctly, what am I doing wrong?
     
  6. Feb 8, 2009 #5

    LowlyPion

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    Homework Helper

    240Ω*I1 = 20ma*100Ω

    9v + 100Ω*I2 = 20ma*100Ω

    and as before

    I1 + I2 = 20ma
     
  7. Feb 8, 2009 #6
    What's going wrong now?

    240Ω*I1 = 20ma*100Ω

    ==> I1 = (20mA*100Ω)/240Ω
    I1 = 8.33mA

    9v + 100Ω*I2 = 20ma*100Ω

    ==> I2 = [(20mA*100Ω)-9]/100Ω
    I2 = -70mA


    I1 + I2 =/= 20
     
  8. Feb 8, 2009 #7
    Here we go, I think I got it.

    the two equations:
    I1 - I2 = .02A (KCL top center node)
    240I1 + 100I2 = 9V (KVL outer loop)

    => I1 = 20.59mA
    => I2 = 40.59mA

    => VR3 = I2 * R3 = 40.59 * 100 = 4.06V
     
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