I don't understand a particular feature of voltage division. Consider the circuit below (we are trying to find Vo):


simulate this circuit – Schematic created using CircuitLab

Now, if the 10-KOhm resistor was not there, it would be obvious that the voltage across the capacitor would simply be the Source Voltage multiplied by the voltage divisor

Vo = 30 x (40/(40+20))

However, we have a 10-KOhm resistor here in the same branch where the capcitor is. I always understood voltage as "pressure", and whenever voltage meets a resistor, some of that pressure is lost forever (i.e. until the current flows back into the voltage source).

Thus, in this case I would be inclined to think that the 10-Ohm resistor "eats up" some of that voltage and thus the Voltage across the capacitor would not be found using the classic voltage divider.

Well, I am wrong apparently, since the solution to this circuit is indeed given by the voltage divider.

So how come the 10-KOhm resistor does not affect the voltage across the capacitor?

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    \$\begingroup\$ Uhm, there's no capacitor nor a 10 ohm resistor in that schematic. \$\endgroup\$ – Bimpelrekkie Oct 19 '17 at 7:32
  • \$\begingroup\$ My bad, I meant 10KOhm resistor. Also, The capacitor is not shown in the schematic because this problem is from the RC/RL circuit chapter of the textbook. The capacitor is just being shown as an open circuit to reflect the fact that the circuit is currently in steady-state. I am going to reupload the image and fix the typo. \$\endgroup\$ – iOShit I Screwed Up Oct 19 '17 at 7:34
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    \$\begingroup\$ I recommend using the build-in schematic editor, Edit your question and draw the schematic \$\endgroup\$ – Bimpelrekkie Oct 19 '17 at 7:42
  • \$\begingroup\$ If your capacitor is ideal (no leakage) and your source DC, then after enough time has elapsed you can analyze the circuit by ignoring the capacitor. However, in the short term after a change in the applied voltage, that RC time constant is substantial. \$\endgroup\$ – Chris Stratton Oct 19 '17 at 7:49
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    \$\begingroup\$ Current flowing through 10k Resistor would be 0A. Since no current would be flowing through the resistor, the voltage difference between the two terminals of the resistor must be 0. Therefore the voltage drop happens entirely across the capacitor. Wow, I completely get it now. Thank you!! Edit: If you want to post this as the answer to the question, I will mark it as solved and you can get kudos for it. Cheers. \$\endgroup\$ – iOShit I Screwed Up Oct 19 '17 at 7:58

In the steady state analysis you treat the capacitor as an open circuit.

As you now realize, this means there is no current flowing through the 10K resistor, and as a result Ohm's law indicates that there is no voltage difference across the resistor.

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