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Solving a DC Circuit with OpAmp

  1. Mar 13, 2012 #1
    Hey everyone,

    I am trying to solve for the circuit that I have attached.

    So I know that for a opamp, it does not draw any currents into its input side and the nodes b and c are the same potential. To figure out the potential at node b, i need to know I_6, but that 6k resistor is messing everything up.

    If I assume that the 6k resistor is parallel to other 6k resistors then I can solve for I_6=1 mA
    However, if that is the case then I_12=1 mA as well. But that means that the resulting voltage at node d is 0...which does not make sense..
    Furthermore, from node c, I can use that to solve for the potential at node d.

    I cannot say all three 6k resistors are in series since from that diagram it does not seem to be unless the ground has to do something with it.

    Am I missing something here?

    Here is my math:

    I_6=12V/12k = 1mA
    therefore: V_b=V_c=6V
    I_3=(V_c-3V)/3k = 1mA

    this does not make sense since from V_a, I_12=1mA and that would result in V_d=0V...

    P.S sorry the diagram is so small!

    Attached Files:

  2. jcsd
  3. Mar 14, 2012 #2


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    Staff: Mentor

    This can't be right. Why ignore the current I to ground through the 6kΩ?
    V_a=12 – I*6k

    The potential divider with a pair of 6kΩ resistors gives
    V_b=V_a / 2
    This potential divider is unloaded because the current into the op-amp + input is so tiny.
  4. Mar 14, 2012 #3
    Ok, that kinda makes sense to me.

    So if I care my calculations with V_a=12 – I*6k,
    V_c = V_b = 6-I*3k
    V_d = V_c -3 +I_3*9k = 12-12I
    V_a-I_12 = V_d --> I_12=0.5I
    V_b = V_a -I_6*6k --> I_6 = 1-0.5I
    Therefore I = I_12 + I_6 = 1 mA

    This would mean that the potential at node d is zero. Can opamps do that?
  5. Mar 14, 2012 #4


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    Staff: Mentor

    12-12I https://www.physicsforums.com/images/icons/icon5.gif [Broken]

    I can't see it.
    Last edited by a moderator: May 5, 2017
  6. Mar 14, 2012 #5
    well i calculated I_3 as (V_c-3V)/3k = (6-3I-3)/3 = 1-I

    so subbing that into V_d = V_c -3 +I_3*9k = 6-3I-3+9(1-I) = 12-12I

    unless i calculated I_3 wrong?
  7. Mar 15, 2012 #6


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    Staff: Mentor

    Check this.
  8. Mar 15, 2012 #7
    I see, I could exclude the 3V source from the calculation of V_d which would make it 15-12I
    So now,
    V_a-I_12*12k = V_d = 15-12I
    12-6I - I_12*12k = 15-12I
    I_12 = 0.5I-0.25

    Since V_b = V_a -I_6*6k --> I_6 = 1-0.5I
    then I= I_12 +I_6 = 0.75

    As for I_2,
    It should simply just be I_2 = V_d/2k
    and I could just find the value of every other current and voltage by subbing in I.
  9. Mar 17, 2012 #8


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    Staff: Mentor

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