I'm studying $RC$ circuits and I'm in doubt on how to deal with one kind of situation. Well, first when we have simply a circuit with just one resistor and one capacitor in series, or a circuit that is equivalent to one of those simplest ones I know how to deal with it: we just use Kircchoff's loop rule and then it leads to a first order linear differential equation which can be easily solved: from here I obtain many results, like the final charge when charging is $Q\varepsilon$ and so on.

However, there are some cases in which it's not possible to reduce to this simple case, and so I'm unsure on how to proceed. The example I've trying to tackle for instance is the same I've asked here:

enter image description here

Now, as I've been answered on that question I made it's not possible to reduce what's on the right to an equivalent resistor. I cannot work with impedance, just resistance and capacitance really. I've tried using Kircchoff's loop rule again, however the currents are different and I don't know how to tackle this.

Can someone give a little help on how do we tackle this kind of problem?


For this particular circuit, the voltage across the R1/C1 branch #1 is fixed by V1, and that across R2 (branch #2) is also fixed, again by V1. That is, the fixed V1 decouples the two branches, so they can be solved separately (circuit #1 = voltage source V1 across branch #1, and circuit #2 = V1 across branch #2). Once these circuits are solved, the current flowing from V1 in the original circuit is the sum of the two branch currents.

  • $\begingroup$ Thanks for your answer @ArtBrown! So, when working with $RC$ circuits like that we just split the circuits in simpler one using this reasoning, solve them one by one and then put everything together in the end? Thanks very much again. $\endgroup$ – user1620696 May 22 '13 at 3:31
  • $\begingroup$ @user1620696: You're welcome. Unfortunately, life is usually not so simple. Most circuits do not split up nicely like this one, and other techniques are required. $\endgroup$ – Art Brown May 22 '13 at 3:34

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