# Conceptual question about capacitors in DC circuits After a long time when the circuit above has reached steady state and the capacitor can therefore be treated as an open circuit, the voltage at the node directly above the 20 ohm resistor is found by voltage division and equals 20/30(30V) or 20 V. Similarly the current is 30V/30 ohms or 1A since the current through the capacitor is zero. However, although in these calculations the resistors are treated as though they were in series (and by definition if they have the same current they must be in series) the equivalent Thevenin resistance is not 30 ohms but 20/3 ohms as the resistors are in parallel. Can someone explain this inconsistency?

• Why do you say that they are in parallel? They obviously do not have the same potential difference across their ends. Mar 27, 2020 at 3:27
• @Krishna, they're in parallel when you zero the 30 V to solve the problem by superposition. Mar 27, 2020 at 5:12
• As The Photon says, with a short circuit at that location, the two resistors are in parallel relative to the load, that is, relative to the two terminals of the capacitor. Mar 27, 2020 at 5:22

First, this has nothing to do with the capacitor. Regardless of what load you connected there (capacitor, inductor, some complicated transistor circuit, or whatever), you'd find the Thevenin equivalent of the source the same way.

Second, when you ask for the Thevenin equivalent resistance of the circuit, you're asking what's $$\frac{dV_o}{dI_o}$$, where $$V_o$$ is the output voltage and $$I_o$$ is the current going in to the output port.

So we can consider it as a superposition problem. If we attach a current source on the output, then, because of the superposition property of linear circuits, the voltage source inside the given circuit won't affect the $$\frac{dV}{dI}$$ due to the source we just attached to its output.

We can analyze the effect of the external source by setting the internal source to 0. Now if the internal source is 0 V, that's the same as a short circuit. And with a short circuit at that location, the two resistors are in parallel.

Just a suggestion do away with series or parallel combination completely and use Kirchoff's Rule

Se that in loop on right

Applying Voltage Law $$30 V -i(10)-i(20)=0$$ solving gives $$i=1A$$

Now you can see effective resistance $$R={V\over I}$$ or $$R=30$$ ohms

P.S -

(and by definition if they have the same current they must be in series) the equivalent Thevenin resistance is not 30 ohms but 20/3 ohms as the resistors are in parallel.

Well for series you do have $$R_{eff} = (R_1) + (R_2)$$ so $$R_{eff}$$ is indeed 20+10=30 ohms

• Current is same in steady state in both resistors in case you are wondering how i assumed it to be same in that equation Mar 27, 2020 at 6:05