# Charging a capacitor parallel to a resistor

This case is different to bog-standard capacitor-resistor circuits in that one resistor is also in parallel with the capacitor, which I'm unable to handle. How does that affect the charging of the capacitor over time?

• Maybe doing a steady state analysis (DC) will help you, what is the voltage when the capacitor is fully charged? Nov 6, 2015 at 14:54
• or maybe you have to calculate Thevenin equivalent? Nov 6, 2015 at 14:54
• @Arsenal The fully charged voltage will be V_{in}, if I'm not mistaken. Nov 6, 2015 at 14:58
• @Columbo I don't think so, as with the capacitor fully charged you'll still have a current through both resistors. Nov 6, 2015 at 15:00
• You are mistaken - the capacitor is parallel to one of the 10k resistors and not parallel to the input voltage. (The schematic is a bit mean in that regard maybe) Nov 6, 2015 at 15:01

At least to me, it seems likely that a tiny bit of redrawing makes the circuit more recognizable:

simulate this circuit – Schematic created using CircuitLab

If we ignore the capacitor for a moment, and look only at the resistors, it's now pretty obvious: a textbook example of a simple voltage divider.

• ...... oh, that makes a lot of sense, since we covered this.... I feel retarded now. +1 (I'll still use Thevenin since that makes me look more expertized, but your solution is certainly the one to go for!) Nov 6, 2015 at 16:33

This is a pretty straightforward case using Thévenin's theorem to model the source and resistors as seen from the cap.

With the values given, the circuit will behave as though both the voltage and serial resistance is half as large.

• Thanks for your answer, Thevenin is certainly useful! Accepted Jerry's since that's the one intended by the creator of that question. Nov 6, 2015 at 16:36