# Capacitor polarity confusion in a circuit

I was reading about opamp integrator concept in a text as such: I marked the terminals of caps as a and b. Above formula assumes the b terminal as being the positive end I guess.

But regarding the direction of I2, can't we say that the positive terminal of the capacitor is a?

But then if it the positive terminal is a, then it is connected to the ground. What is the + and - ends of the capacitors in the above figure?

I know it is basic circuit theory but somehow I'm confused at this point.

Isn't there only one option of the polarity of a cap at a moment for a given current direction across it?

EDIT:

Besides these, I also do not get why in sim for the inverting input (point X) the voltage is not zero for 10 secs of time duration. According to op amp theory Vx must be around zero but it is around 1V here: • The simulation shows you the steady-state response. Op-amp is in negative saturation. In transient options, you need to check this box "Skip initial operating point solution". And for R1 = 1k and Vin = 1V the capacitor (1uF) will charge to -12V in $T= 1\mu F *\frac{12V}{1mA} = 12ms = 0.012s$,
– G36
Jun 19 '17 at 17:59
• But whatever the response type is, isn't it violating one of the opamp golden rules which is Vx = V- = V+ = GND in this case? Jun 19 '17 at 18:18
• Could you write this to your answer in a detailed way? i thought "Skip initial operating point solution" would skipping something but in this case it reveals something. Jun 19 '17 at 18:20
• No, when the opamp output is in saturation (positive or negative) this "golden rule" do not hold anymore. Because the opamp output voltage cannot be large than supply voltage.
– G36
Jun 19 '17 at 18:20
• – G36
Jun 19 '17 at 18:22 