I am performing an analysis of an output signal in the interval $t<0$ from the operational amplifier given in the following imageenter image description here

where $v_{C_1}(0^-)$ and $v_{C_2}(0^-)$ are the initial conditions to be computed of two capacitors. As you can see, the solution is just below the image.

What I can't understand is why all the measured voltages are $1V$ without knowing the values of the resistances and just knowing that $V_-=V_+$. For sure I am missing something that I don't know. Can someone give me a hint? Thank you.

  • $\begingroup$ That's not really a signal proc question, more of a "basics of the three golden opamp rules" question... (Hint: three golden rules. Apply them. $\endgroup$ Jun 20 '18 at 17:32
  • $\begingroup$ @MarcusMüller I would post it in electronics stack exchange but you don't know how mean the people is there. Ok, with that said my problem is solved: The inputs draw no current. Thank you good sir! $\endgroup$
    – Martín
    Jun 20 '18 at 17:36
  • $\begingroup$ I'm one of the people over there. We try hard to not be mean :( $\endgroup$ Jun 20 '18 at 20:57
  • 4
    $\begingroup$ I'm voting to close this question as off-topic because this is a circuit analysis question, not a SP question. $\endgroup$
    – MBaz
    Jun 20 '18 at 21:37

You should know charaterstics of an opamp it has infinite input resistance and zero output resistance hence opamp doesn't draw any current in the input side hence it is an open circuit and current through is zero apply kvl(Kirchoff voltage law) you will know


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