capacitor discharging

hi all. nice site you have here

i was wondering if any of you can help me with this problem.

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

basically from my understanding, in the upper position, the capacitor gets charged by the voltage source to (20/3) V (i got this from the thevenin equivalent cct). So at t=0, it goes into the lower position, then what happens? Will any current flow? I dont think so... cuz theres no difference between the above and below circuits. lol. thats my theory - but i cant find any theory regarding its correctness - so forget that. the capacitor will discharge, but through where? Is it given by the textbook formula I = [Vc*e^(-t/RC)]/R? If so, what's R -> will the current flow to the mesh on the left with the voltage source?

textbooks always deal with the case where capacitor discharging circuit just consists of the capacitor and a resistor in series, but this isnt

 

During the time between switch contacts the capacitor will discharge through the 20K resistor in parallel with it.

 

Do you mean, BETWEEN the time the switch is in the upper and lower position, as in, in mid-air? or do you mean t>= 0

 

vanush,

You have noticed that the two power sources are reversed from one another, right?

hgmjr

 

im not sure what you mean?

 

What I mean is the 10V dc voltage source associated with the switch in the upper position is oriented in a direction to charge the cap with a negative voltage with respect to the reference point which I take to be the bottom end of the capacitor and the 10V dc voltage source associated with the switch in the bottom position is oriented in a direction to charge the capacitor with a positive voltage with respect to the same reference point.

hgmjr

 

at t=0, the voltage is -6.66v.
when the switch touches the lower pole, the voltage source become +6.66v, with a 6.66 k ohm resistor in series. the voltage at c is -6.66, so the total voltage difference is 13.33v, hence i(t)=13.33/6.66k.
what happens is the capacitor descharging it's -6.66v, all the way through 0v until it reaches 6.66v.

 

but is i(t) the same as the current derived from the thevenin equivalent cct as u have just posted

 

yes it is. just simplify it.