No, the circuit is broken, the power supply stops supplying current to the circuit and the only remaining current in the circuit is that which is stored in the capacitor.

Yes this is correct.

Is there now any voltage across R1 or any current through it?

P.S. does it makes a difference weather the switch is on the positive side of the power supply or the negative? (in my task it is between the power source and R1) .

No difference.
I placed it where I put it because of the mix up naming R1 and R2.

Oh, and I have to find the current function through the resistor R2 !

Ok we are getting there.

So at t=0 the voltage across R1 is zero, the voltage across R2 is the voltage across AB with A more positive than B. This is equal to 130 volts.

We can now say that all the current from the capacitor passes through R2 only (This would not be true for a different circuit so be careful in your reasoning). Thus the capacitor discharge current = the current in R2 at all times t>0

I have drawn the direction of my I2 arrow consistent with this discharge current flowing from positive to negative from the capacitor through R2. this direction is called conventional current.

At this point you will need to help me because I do not know what you have been taught about the equations for the discharge of a capacitor through a resistor so tell us about this and we can continue to crack the problem.

Thus we can

 

@WBahn, are we discussing AC circuits here or are you needlessly complicating matters for a beginner?

Please note that if there is not alternating voltage across the capacitor ie the alternating voltage is non existent then the conditions I stated hold good.

 

Yes this is correct.
Is there now any voltage across R1 or any current through it?

No, there is not!

At this point you will need to help me because I do not know what you have been taught about the equations for the discharge of a capacitor through a resistor so tell us about this and we can continue to crack the problem.


V = V(initial) * e^(-(t/RC)) , where RC is the time constant and V(initial) = V(C(t=0)) = 130 V . Am I right?
So the answer to my problem could be: i( R2 ( t > 0 ) ) = ( 130 / R2 ) * e^( - ( t / ( 6.5 * 10^-3 ) = e^( - ( t / 6.5m ) ) / 10 A (?)

 

@WBahn, are we discussing AC circuits here or are you needlessly complicating matters for a beginner?

Please note that if there is not alternating voltage across the capacitor ie the alternating voltage is non existent then the conditions I stated hold good.

Fine. Then let him believe that when the voltage across a capacitor is zero that the current through the capacitor is zero. After all, that is one of the two conditions that you've stated in no uncertain, absolute terms. I'm sure that getting that firmly set in his mind now will serve him oh so well down the road.

Furthermore, the conditions you state do not "hold good" even when there is no alternating voltage. I can take just the components in this circuit and make a minor change to the topology and end up with a circuit in which, after the switch is thrown, the voltage across the capacitor starts out with one polarity and ends up with the other polarity with the result being that the capacitor has current flowing in it even at the point in time that it has zero voltage across it, something you told him can't happen.

(Actually, to make the change I had in mind I need a SPDT switch instead of the SPST switch in the schematic, or a third resistor, but the example is still valid.)

 

Looks good.

Note that, as WBahn said, things are a bit different for AC circuits.

If the source was 330 volts alternating, then there would not be 200 volts across R1 and 130 volts across R2 and the capacitor with the switch closed.

Nor could we assume any particular instantaneous voltage across the capacitor when we open it.

But that is another story for another thread.

Now you know how to develop your questions here.

Go well in your future studies.

 

Thank you very much for your help, Studiot !!!
Will be returning here
Have a good one, you all!