Analysis of Electrical Networks

We need to find a wawe of voltage output of an ideal operational amplifier after the switch gets turned on in the time t=0.
apologies for bad english

I am not even sure if this is a commutation problem or not.
If it is, I know that the network is well defined.
We can have linear and non linear work regime.
I have never seen something like this and i really have no idea what so ever what to do.
Help is much appreciated

Current on the amp is probably 0 on both clamps.
Voltage between them is probably also 0.
I have only seen these with resistors only.

so nobody?
nothing?

go to stupid questions thread.
You just fucked up the board a little more by creating this thread.

Don't reply, don't bump

i don't understand
is it that easy?
i mean, you have to prove it, not just say that all the elements are linear so no new wawes

Stupid question threads are actually:

"Questions that don't deserve their own thread"

Yours obviously doesn't, as you could wait for a classmate to tell you or ask your professor.

yeah, the thing is, classmates know nothing about it, just like me, and the prof is an idiot.
these two are obviously related
so i went to the one place where people always helped me, even with the retarded questions.
but now you come here with your /b/ attitude.
if it's so easy, why didn't you solve it?

Eeq=E*R2/(R1+R2)
Ic=Eeq/(R1||R2)
U_R3=-t*Ic/C
or something like that I guess

so you skipped like a dozen steps or something cuz i'm not seeing any of that
sorry but i'm really not good with amps

You've got an inverting integrator hooked up to a constant input voltage, meaning the output voltage will start at 0 and then decrease.

The op-amp saturates at 0 since its negative rail is tied to ground.

So the output will be 0 forever.

The end.

actually the task says that the op-amp saturates at Ez

the positiv rail is tied to ground

I'm not good either, I just look at a circuit and feel what's probably going on. I didn't skip anything, that's just the story I felt could be suitable. Maybe some expert can have a look at it.

...

Looks like a high pass filter, OP. It's set up to invert the output too, so you'll have full attenuation at DC, but full output voltage past the cutoff frequency. I'm not sure what the gain is, but you can calculate that too.

Way too much reality. This is academia: the positive supply is grounded and the negative is not shown. You could even short R2 and still expect the circuit to work. In this realm infinite idiocy is the normal case.

yeah. in the picture of the "solution" they shorte R2. thats the only way E can equal R1*i1

Current must constantly flow through the cap so the output voltage will continue to fall until it hits the rails (Ez / V- / Vs-) were it will clamp there.

>the positive supply is grounded and the negative is not shown

I'm guessing that's to show the Op-amp can sink current to ground and not a rail.

yeah it's shown on the graph that Ui will be -Ez after the specific time

If there is negative feedback (wich is the case) the voltage in the + and - pins of the op amp are the same, therefore R2 doesn't matter and you have a typical integrator

how can you tell there is negative feedback?
i just assumed that the voltage there is 0, so i shorted R2, bcz i didn't know any better

The best way to check the feedback is to imagine what would happen if you increase a bit the voltage on one of the op amp inputs, if one follows the other then there is negative feedback, if they separate then there is positive feedback