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Find the voltage at node c.

  1. Dec 31, 2013 #1
    Hello, I'm stuck on this question, it asks to find the voltage at node c (on the circuit below) then use this value to work out other values of the op amp circuit.

    I tried using kirchoff's laws but that didn't work, then I used thevenins. Basically I took out the 6k and 8k resistors and calculated the voltage difference between the two 3k resistors and got 10.5V. Then I took away the two 3k resistors and put back the 8k and 6k then using the 10.5v found the voltage drop across the 6k resistor i.e. (6k x 10.5v)/(6k + 9k) and got 4.5v. I then took 4.5 from 10.5 and got 6 volts. However I don't have answers to this question so I'm unsure if I've got the right answer and/or my method is correct, which I doubt.

    Any help on how to approach this question would be much appreciated.
     

    Attached Files:

  2. jcsd
  3. Dec 31, 2013 #2
    The correct answer is 6V. Why you don't use nodal analysis? Also are you aware that thanks to negative feedback action the voltage at point B must be equal the voltage at point A? If so VA = 0V so the VB also must be equal to 0V.
     
    Last edited: Dec 31, 2013
  4. Dec 31, 2013 #3
    Mainly because I'm an idiot.
     
  5. Dec 31, 2013 #4
    Don't be so harsh on yourself, new year is coming.

    I think that you should try again but this time use KCL.

    For this diagram

    attachment.php?attachmentid=65257&stc=1&d=1388503829.jpg

    We can write for node C the KCL

    I1 = I2 + I3 + I4 and additional

    I1 = (21V - Vc)/3K

    I2 = Vc/6K

    I3 = (Vc - Vd)/8K

    I4 = ??

    Next do the same thing for B node and next instead VB put VB = 0V and solve this simultaneous equations.
     

    Attached Files:

  6. Dec 31, 2013 #5

    I'm missing something here I'm putting I4 as (Vc-Vb)/3k

    and for node B

    (Vc-Vb)/3k = (Vb-Vd)/5k.

    But this doesn't seem to be right.
     
  7. Dec 31, 2013 #6
    Very good

    Why? I don't see any mistake. The ideal opamp has infinite input impedance. That means its inputs don't draw any current at all. And this is why I4 = I5
    So you have all information needed to solve this problem.
     
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