That looks reasonable. It might simplify down some, but it might not. I didn't look at the nitty-gritty details, so it might not be correct. But the approach you are using is fine.



That's actually going further than you need to. Because R3 is part of an open circuit, no current can flow through it. As a consequence, there can't be any voltage across it and therefore V1 = V24.



No, you can't use the fact that for an ideal opamp V+ = V-. You are not working with an ideal opamp!

And the problem says nothing about how big A is. I've designed circuits that have used opamps for which A was less than 10.

Simply replace the opamp with a voltage-controlled voltage source in which the control voltage is the difference between V+ and V- and the output voltage is A times this difference.

@WBahn
For #4, if at t=0, V1 = Vdd, wouldn't that make the voltage across R1 = 0 and then we will only have the capacitor C1 and R2 in series at t = 0?

 

@WBahn
For #4, if at t=0, V1 = Vdd, wouldn't that make the voltage across R1 = 0 and then we will only have the capacitor C1 and R2 in series at t = 0?

Momentarily. But ONLY momentarily. At t=0.0000000000000000001 s, R1 is having an effect.

 

Momentarily. But ONLY momentarily. At t=0.0000000000000000001 s, R1 is having an effect.

Thank you @WBahn ,
Can we then say that at the very moment that t = 0 we only have C1 and and R1 alone in parallel. And C1 is not charged ( no current flowing) and VC1(t=0) = 0 and nothing is happening.

At t>0, C1 starts charging. And VC1(t) = VR1(t).

 

Thank you @WBahn ,
Can we then say that at the very moment that t = 0 we only have C1 and and R1 alone in parallel. And C1 is not charged ( no current flowing) and VC1(t=0) = 0 and nothing is happening.

@WBahn ,
And therefore,
At t>0, C1 starts charging. And VC1(t) = VR1(t).

And V1(t) = Vdd- VC1(t)

 

Momentarily. But ONLY momentarily. At t=0.0000000000000000001 s, R1 is having an effect.

For #4, I have attached a picture of my attempt to solve the problem.
Thank you very much for your help.

 

Thank you @WBahn ,
Can we then say that at the very moment that t = 0 we only have C1 and and R1 alone in parallel. And C1 is not charged ( no current flowing) and VC1(t=0) = 0 and nothing is happening.

At t>0, C1 starts charging. And VC1(t) = VR1(t).

But something IS happening. What is the current in R2 at t=0?

Also, be careful about equating "not charged" with "no current flowing". These are two different things and one does not tell you anything about the other. In an AC circuit in steady state, the current in a capacitor is a maximum when it is the capacitor isn't charged and vice-versa.

 

For #4, I have attached a picture of my attempt to solve the problem.
Thank you very much for your help.

You're pretty close. The biggest issue is the time constant. You have the effective resistance in the RC time constant being the series combination of the two resistors. Ask yourself if that makes sense. Let's say that R1 wasn't there. What would the time constant be? Now, let R1 in your equation for the time constant go to infinity. Does it reduce to the same thing?