# capacitor discharging

Discussion in 'Homework Help' started by Vanush, Apr 19, 2008.

1. ### Vanush Thread Starter Active Member

Apr 19, 2008
46
0

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

2. ### beenthere Retired Moderator

Apr 20, 2004
15,808
295
During the time between switch contacts the capacitor will discharge through the 20K resistor in parallel with it.

3. ### Vanush Thread Starter Active Member

Apr 19, 2008
46
0
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

4. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
vanush,

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

hgmjr

5. ### Vanush Thread Starter Active Member

Apr 19, 2008
46
0
im not sure what you mean?

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
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

7. ### rwmoekoe Active Member

Mar 1, 2007
172
0
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.

8. ### Vanush Thread Starter Active Member

Apr 19, 2008
46
0
but is i(t) the same as the current derived from the thevenin equivalent cct as u have just posted

9. ### rwmoekoe Active Member

Mar 1, 2007
172
0
yes it is. just simplify it.