solving a deceptively simple-looking switched RC circuit (singular circuit ?)

It is very difficult for me to figure out exactly what you are talking about. From looking at your circuit diagram, which I also do not fully understand, you are falling into a common logical trap when analyzing capacitor charging other capacitor problems. The schematic does not present the entire circuit.

Parasitic inductance and resistance are neglected. Such circuits tend to ring at high frequency using parasitic inductance to form a resonator. Wire resistance also limits the current flow. Even without such resistance, there would still be radiation resistance that dissipates the stored resonant energy.

Bill

-- Ferme le Bush

As Bill says, your circuit is not a practical one.

Without any additional components, at the moment of switch an infinite current will flow, redistributing charge between your theoretically perfect capacitors, instantaneously.

You can simply ignore the presence of the resistor that is present, at the moment of switch.

Up to the moment of switch, the calculations are pretty trivial. After the switch and redistribution of charge, the decay calculations are pretty trivial.

All you have to allow for is that the charge will re-distribute in zero time, at the moment of switch.

Looking at the formula and drawing of your first figure, it seems you have swapped V1 and V2. Consider the steady-state situation of this circuit. As t->infinity, the second term of the formula for V1 vanishes and you have V1 = Vdc. But looking at the circuit, this is clearly wrong since as t-> infinity, the capacitor C2 voltage = Vdc and the current through R1 is zero (C1 voltage = 0). Similarly, as t-> infinity V2 (by your formula) approaches zero, yet your schematic shows that the capacitor C2 would charged to Vdc.

Hope this helps,

daestrom

You and Bill have essentially pointed out a limitation of circuit theory. However, one can assume the same sum of c*v at t=0+ as at t=0- (conservation of charge similar to the situation with flux linkages). That is - redistribution of charge as you indicated. Energy balances before and after are meaningless as the model doesn't account for this because of the points that Bill raised but for calculations of circuit behaviour after t=0 there is no problem in finding the post switching behaviour. This is related to the conundrum of connecting an ideal charged capacitor to an uncharged ideal capacitor -energy disappears according to circuit analysis -hard on sophomore students.

Use ATP. Downloadable free with subscription. This kind of problem illustrates zero resistance problems extremely well and you'll have to put some very low value of resistance in for the program to solve it. It's quite illustrative as you put different values of low resistance in to see how the circuit simulation varies. You can of course do it just using basic old fashioned diff equations/laplace transform methods aswell...but again Maple can be of great help to show you graphically what's happening.

Danny

------------- You can use Laplace transforms but the situation doesn't require that. By inspection, when the battery is connected, the transient voltage across the RC1 parallel branch is of the form Ae^-t/R(c1+c2) and the steady state voltage is 0. Initially there will be a delta function to allow instantaneous charging of the capacitors so that the initial voltage across C1 and R will be EC2/(C1+C2) so the voltage across the resistor (V1) is EC2/(C1+C2)e^-t/T where T=R(C1 +C2). This will decay to 0 and the voltage across C2 which is in series with the source will rise to E. At any intermediate time the voltage V2=E-V1

When the source is removed and the third capacitor is switched in, you will have (C2 and C3 in series) in parallel with C2 giving a total capacitance Ct=C1C3/(C1+C3) +C2 and a time constant CtR. The total charge immediately after switching is CtV(0+) =C1V1(0-) +C2V2(0-) so the initial voltage V(0+) across the resistor can be found. so the final transient is V(0+)e^-t/CtR.

Of course, this circuit analysis doesn't actually deal with what happens at t=0 so we skip over the discontinuity by assuming conservation of charge for the whole system(or flux linkages for inductive cases) rather than for each capacitor individually.

Another "of course" is that you should not automatically take my solution as correct.