Here is a first-order op-amp circuit:

enter image description here

I have to find V0 given that V(0) = 4 V (i.e. the capacitor is charged). If the capacitor is charged, then the voltage at the node 1 is V1 = 4V and KCL must look something like this: $$ \frac{0 - V_1}{R_1} = C\frac{dV}{dT} + \frac{V_1 - V_0}{R_f} $$ However, the solution postulates that since the node 2 has a zero voltage, then V1 = 0. It seems that there is some contradiction. Does the voltage on the capacitor influences the voltage of the node 1 or not? I have studied op-amp theory, however, I feel perplexed because university curriculum and textbooks takes onto consideration only two cases: either the voltage source is between the ground and the inverting input or it is between the ground and the non-inverting input. This is the first time when I see such a circuit that's why I have posted this question.

  • 1
    \$\begingroup\$ why do you say V1 starts at 4V? It looks like V1 = 0 all the time. The initial state of the capacitor implies that 4V are initially at the output of the opamp. \$\endgroup\$
    – Javi
    Commented Aug 2, 2019 at 7:07
  • \$\begingroup\$ What you say doesn't make sense. \$\endgroup\$
    – Andy aka
    Commented Aug 2, 2019 at 7:10
  • \$\begingroup\$ In the grey equation, what's V? \$\endgroup\$
    – Chu
    Commented Aug 2, 2019 at 7:50
  • \$\begingroup\$ @Javi, where does this implication come from? \$\endgroup\$
    – tenghiz
    Commented Aug 2, 2019 at 8:34
  • \$\begingroup\$ @Chu, voltage across the capacitor. \$\endgroup\$
    – tenghiz
    Commented Aug 2, 2019 at 8:35

2 Answers 2


Your question is ambiguous. First, you haven't placed a sign convention on the capacitor, so it's not clear which side is 4V with respect to the other side.

Next, in initial value problems, one should really understand EXACTLY what happens at time t=0, and that isn't clear at all here. This is often handled by including a switch that closes at t=0 in the circuit.

If this is some sort of HW, a clearer question would be better.

  • \$\begingroup\$ If you mention the sign convention, then the capacitor charge does influence the voltage on the op-amp terminals, right? Can you explain, please, how? As for the signs themselves, I use the symbol for the electrolytic capacitor, convex line is "-". \$\endgroup\$
    – tenghiz
    Commented Aug 4, 2019 at 12:58
  • \$\begingroup\$ The original problem as it is in the textbook: "For the op amp circuit find V0 for t>0 if V(0) = 4V. Assume that Rf = 50 kOhm, R1 = 10 kOhm, C = 10 uF". No switch. \$\endgroup\$
    – tenghiz
    Commented Aug 4, 2019 at 13:03

Since the + input of the opamp is at 0V and the opamp has negative feedback then the output is also at 0V and the circuit does not work. The circuit needs the + input of the opamp to be biased at about half the supply voltage or there should be an additional negative supply.

  • \$\begingroup\$ We don't know that the opamp doesn't have a negative supply. \$\endgroup\$ Commented Jul 17, 2020 at 13:03

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