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