This won't be that difficult and I do not know what "ig" is.

So for example you may charge C1 up with 1.5v, then use that to charge C2, then charge C1 again and again use that to charge C2 more, etc., etc.
Something like that?
I just want to be sure what is allowed here.

What was the book answer?

ig is instagram.

the book(pathfinder for olympiad and jee advanced) is one of the hardest books here in india so u can expect they are asking some crazy hard stuff not just simple things , so yes a lot of things are allowed, and this question is from the hardest section of the book .

yes exactly, this was my method : u first charge both c1 ,c2 to Vo , then u use c1+ battery to charge c2 , then charge c1 with battery again , now use c2+ battery to charge c1 qnd keep doing this until the final voltage on the capacitor being charged converges to a maxima.

this method isnt the one that leads to the answer though , this was proved to me by @WBahn , his method was slightly different than this , you could check his most recent reply to see what he did .

the answer in the book is 2Vo * (c1 +3c2)/(c1+c2)

i have reached the answer though with the help of @WBahn , but ur still very welcome to give any more thoughts regarding this , thanks a lot for trying to help me .

 

ig is instagram.

the book(pathfinder for olympiad and jee advanced) is one of the hardest books here in india so u can expect they are asking some crazy hard stuff not just simple things , so yes a lot of things are allowed, and this question is from the hardest section of the book .

yes exactly, this was my method : u first charge both c1 ,c2 to Vo , then u use c1+ battery to charge c2 , then charge c1 with battery again , now use c2+ battery to charge c1 qnd keep doing this until the final voltage on the capacitor being charged converges to a maxima.

this method isnt the one that leads to the answer though , this was proved to me by @WBahn , his method was slightly different than this , you could check his most recent reply to see what he did .

the answer in the book is 2Vo * (c1 +3c2)/(c1+c2)

i have reached the answer though with the help of @WBahn , but ur still very welcome to give any more thoughts regarding this , thanks a lot for trying to help me .

Hello again,

Ok.

Starting with two capacitors in series C1 and C2 and battery E3, and initial voltages for C1 is E1 and for C2 is E2, the infinite time equations are:
v1=(C2*E3-C2*E2+C1*E1)/(C2+C1)
v2=(C1*E3+C2*E2-C1*E1)/(C2+C1)
where
v1 is the final voltage for C1, and v2 is the final voltage for C2.
However, this is arranged so that we can charge C1 from C2 after we have pumped C1 up many times and then kept charging C2 up with that so that the voltage across C2 reaches 2*E3. It is assumed that it is easy to figure out that part. The harder part is to figure out what happens when we use the C2 voltage plus the battery voltage to charge C1. Those equations provide us with that. They come from the time equations for two caps in series and letting time 't' go toward infinity. We could do the same thing for charging C1 first if needed, doing that multiple times and charging C2 also.

USING THOSE EQUATIONS
First, we get rid of C1 and C2 because these values are not needed. We only need to know the ratio of the two. If the ratio is n (where n=3 for this problem) then we end up with these two:
v1=(n*E3-n*E2+E1)/(n+1)
v2=(E3+n*E2-E1)/(n+1)
and these two are both functions of E1, E2, and E3, and the ratio n which is the ratio of C2 to C1 (which here is 3/1=3).
Now due to the assumption above, we know that E2=2*E3 and E1=E3, but since E2 will be used to charge C1, we have to set E2=-2*E3.
Setting E2=-2*E3 and E1=E3 we end up with:
v1=((3*n+1)*E3)/(n+1)
v2=-(2*n*E3)/(n+1)
Since the caps are in series and the final topology change is to put them in series and in series with the battery E3, that leads us to:
v1-v2+E3=(2*(3*n+1)*E3)/(n+1)
and note we made v2 positive by making the above solution for v2 negative because we flip it around at the last step and that changes the polarity again.
So the final solution is:
Vout=2*E3*(3*n+1)/(n+1)
where E3 is the battery voltage and 'n' is the ratio of C2/C1.
This should match the previous solution by WBahn.

Taking the limit as n goes to 0 we get 2*E3, and taking the limit as n goes to infinity we get 6*E3. So the smallest voltage is 2 and the largest is 6 using other cap values, and for E3=1v. For n=65.6666 we get within 1 percent of 6 volts when E3=1v. To get within 0.1 percent n=665.6666, to get within 0.01 percent n=6665.6666, etc.

I used the time equations because that takes us back to first principles in circuit analysis, and it also allows me to check the final result using a program that uses the actual time domain equations to arrive at the result. Without the time domain solutions I would not be able to check it that way.
We could also use a simulator to check the last step.