Analyzing Discharge Circuit with Capacitor and Voltage Source: Finding i(t)

[Vanush]

hi all, wonderign whether you can help with this problem with capacitor discharging.
take a look at the pic

the question is

In the circuit shown above, the switch has been in the upper position for a long time and moves to the lower position at t = 0. Find i(t).

Basically, first i wanted to find out wat the voltage between the capacitor terminals would be after the long time it had been in the upper position. I found this as 20/3 V using thevenins theorem. now, i(t) is given in the textbooks as [-V*e^(-t/RC)]/R where V is the voltage across the OC capacitor. however, the discharge circuit the capacitor is connected is not just a simple circuit which contains only resistors - it contains a voltage source. how can i get an expression for i(t) in the discharge circuit? or am i going about this the wrong way.
thank you.

EDIT: I am not supposed to post this here, but i can find no way to delete the thread?

 

[varunag]

This looks like a homework problem, still I'm giving a few hints.

Just use the KVL for the switch placed in the lower position. (I've not tried but I suppose this should help)

moreover, look carefully, is this circuit(switch in lower position) really discharging circuit?

hope this helps.

 

[Vanush]

So, in between the upper and lower contacts, the capacitor discharges into the resistorr parallel with it, but when it reaches the lower contact, there is a voltage source charging the capacitor again - so i(t) would be a sum of the currents frm these sources?

 

[varunag]

as you say that the switch goes to the lower position at t=0, the capacitor will really not get the time to discharge.
notice one thing that the polarity of the voltage across capacitor changes after the switch remaiins in the lower position for a long time.

and yes, you are right, "there is a voltage source charging the capacitor again". but I didn't understand what did you mean by, "so i(t) would be a sum of the currents frm these sources". Please explain.