AC-DC voltage sources-capacitor-diode

1. The problem statement, all variables and given/known data
For the following circuit assume that the diode is ideal and that the time constant RC is very big compared to the period of the generator.Find the v(out).

2. Relevant information
Thought to start this with the superposition theorem.(I wrote it wrong in the pdf btw)
And when i had reached the stage with only the AC source,even though it seemed very alike to the capacitor-input filter,i couldn't do anything,as the diode was after the capacitor,so i improvised and i am not proud of it.


3. The attempt at a solution
Here is the pdf.I am almost certain it is not right and i have many questions that i will pose after the first instructions.Not all together.Don't want to scare you off.

 

Firstly let me thank you gentlemen for your time.Now let's cut to the chase.

I wasn't aware of that.However now in this particular exercise,in which it happens the diode to be ideal,can't i substitute it with a switch and use superposition?

I totally lost you.I must be missing something...How we got from<<there is no series resistance>> to <<So C charges to max etc etc>>?Can you explain a little more?

About the time constant i read that,being that bigger than the period,the capacitor doesn't have the time to either fully charge or fully discharge.So because of this difference between them,we can assume that the voltage in the capacitor is steady.Right?

Well no,because of difference between the time constant and the period?

About that.If the time constant is theoritically infinite times bigger than the period,can i assume that the voltage across the capacitor is 0(in case it had 0 charge in that point)?So that is one of the cases why it is said,that in ac circuits the capacitor is like a short circuit?

Well i cannot understand the relation between the large time constant and the negligible current.It seems to me as the resistance has similar behavior with the ac source.


2 extra questions from the ones i had before your replies,so as to reduce the amount a bit.

1)When the current flows through the capacitor,doesn't it <<prefer>> to go through the diode(ideal) and the dc voltage source(zero resistance in that branch),instead going through the resistance?If that is the case can't i remove the resistance?
2)I can place a ground wherever i want,and do the calculations in relation with it,right?

 

Still confused.Let me ask you some more.

1)I read that if we have in a circuit both DC and AC voltage sources,then we analyze them seperately.For DC analysis i open all the capacitors.For AC analysis
i substitute all DC voltage sources with AC grounds and i substitute the capacitors(coupling-bypass capacitors) with short circuits.So there are capacitors i cannot substitute with short circuits right?

2)If i do what i told earlier then in DC analysis(opening the capacitor) i get that VR=Vout=10V.
If i do in AC analysis what i said then i have the AC source connected across with both the resistance and the diode(resistance and diode in parallel).Now the current will
<<prefer>>(for the reasons you well explained) to pass through the diode.So current through resistance almost zero.So ac voltage through resistance almost zero?
And basically because the diode is ideal,the AC source cannot have any impact on R(=Vout) the way i see it.But earlier we reached to the point that it has the same behavior with the AC source,so it does have an impact on Vout.So we reach to the point that in this case i cannot substitute it with a short circuit?

3)If in a circuit there is absolutely no reference to steady states,switches(open-closed) etc can i substitute them with either short or open circuits(depends on source)?Because it is a little confusing.In first and second order circuit analysis,it is very different.We don't just open the capacitor,we find relations.But in other books,without any reference to even tell us that it was in steady state,it just opens the capacitors and move on with the calculations.I am not sure if made myself clear...

Because it cannot discharge once charged?

You said that it charges to a voltage of the peak voltage on the left minus the 10 battery voltage,so i assume it starts from -10,rises up to +10 and stays there.

But i have a question
4)How can it reach the peak,if the difference between time constant and period is that big.I mean it needs 4-5 time constants to get fully charged,but during just one time constant the AC source would probably have done many cycles.How does it allow the capacitor to reach the peak?

 

Ah for these as well huh?Well it is a lot clearer now.

I believe i forgot the 5 DC voltage source.Wouldn't it be 5-10=-5 volt.Meaning C charges to a voltage of the peak voltage on its left side minus the 5V battery voltage on its right
What is the peak voltage?It is 20 now.Or peak is 25 and it would go minus 10 instead of minus 5.Did you mean that or something else?

Well i made a wrong assumption.I believed that the removal of the resistance only had practical meaning for the calculations(current,voltage) and that it wouldn't affect the time constant.

Now a question about this .Well i read that the resistance plays no role,as far as the peak of the charge of the capacitor is concerned.However it does play a role on the speed eh,i mean on how fast or slow the charge or the discharge will take place.Also i read that the charge or discharge don't happen in an instance.So if we assume that R=0 and we omit it,will the capacitor charge to its peak following entirely the AC voltage's rythme?

 

Well thanks you to you sir and to mrGneill ofc, many questions were answered and now i can see the problem a lot better than before.I am grateful.

 

Yes you are right,they seem harsh but i don't get easily discouraged.I downloaded LTspice and i will work with it.Moreover,i will keep posting more exercises and questions as i move.So for now,goodnight :)