The same process follows in the analysis.
Sorry, nope.
As soon as you have two different capacitors in the bottom branches, voltage and current waveforms no longer follow a simple exponential rise or fall. You need a different approach to analyzing the behavior of the circuit.
One of the ways to do it would be to write out some differential equations:
v
g = v
C1 + v
C2 + R
1*C
2*dv
C2/dt
v
g = v
C1 + v
C3 + R
2*C
3*dv
C3/dt
C
1*dv
C1/dt = C
2*dv
C2/dt + C
3*dv
C3/dt
These are actually the same three equations I wrote in my post above, just written with differential equations.
KVL1: Vg = VC1 + VC2 + VR1
KVL2: Vg = VC1 + VC3 + VR2
KCL: i = iC2 + iC3
You would then have to mash the three equations together and solve for v
C2, for example. But solving for a Heaviside (step) input voltage function would be quite difficult in the time-domain, so a better solution would be to switch into the s-domain (Laplace transform).
The same three equations then become:
v
g = v
C1 + v
C2 + R
1*v
C2*C
2*s
v
g = v
C1 + v
C3 + R
2*v
C3*C
3*s
v
C1*C
1*s = v
C2*C
2*s + v
C3*C
3*s
The Heaviside function is quite simple to write in the s-domain:
v
g = 1/s
Solve for desired quantity and perform an inverse Laplace transform, so you get the output function in the time domain.
This is how one would solve the circuit analytically. The process is rather tricky and time consuming, but (contrary to what some in this thread claim) it
can be done. Kirchoff's laws apply. Always.
It is a good idea to help yourself solve problems like these with programs like Matlab.
______________________________________________
We could simplify our life and approach the problem by slapping everything into SPICE and running the simulation. This is what you get, if you choose C
1=2µF, C
2=3µF and C
3=1µF:
You can see something interesting happening, which may seem somewhat counterintuitive at first glance. The current through C
3 goes
negative at one point (V(n3) is voltage across R
2, therefore directly proportional to current in that branch).
Just after switching, the capacitors are all at 0V and the currents are only dictated by the resistors. Both resistor values are equal and therefore the currents are equal. The function of voltage on a capacitor is 1/C * integral(i
C dt). Because C
3 is only a third of C
2, the voltage on the former starts increasing three times as fast as on the latter. However, at the end of the transient, they both need to settle at the same voltage (as they are connected in parallel), so the current in C
3 has to go negative in order for the voltage to drop and match the other capacitor. Interesting, huh?