# Node voltages in the first-order op-amp circuit

Here is a first-order op-amp circuit:

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.

• 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. – Javi Aug 2 '19 at 7:07
• What you say doesn't make sense. – Andy aka Aug 2 '19 at 7:10
• In the grey equation, what's V? – Chu Aug 2 '19 at 7:50
• @Javi, where does this implication come from? – tenghiz Aug 2 '19 at 8:34
• @Chu, voltage across the capacitor. – tenghiz Aug 2 '19 at 8:35