# Open circuit with capacitor

The problem asks to find the steady state voltage across the capacitor. i attached the circuit (sorry for the poorly drawn and yes, that open part of the branch is supposed to be there) This is a multiple choice question where the answers are:
a) 0V
b) V_dc/2
c) V_dc/3
d) V_dc
e) 2*V_dc

In steady state, a capacitor behaves as an open.
If the capacitor was placed before the resistors, I think its voltage would be V_dc.
However, this capacitor is placed after the resistor.
In steady state, no current will flow through these 2 branches due to the open circuit, thus the voltage drop across the 2 resistors are 0.
This means that the ends of the capacitor are at the same voltage (V_dc). Thus the voltage difference is V_dc-V_dc=0.
So the answer is a. Am I right?

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• Circuit.png
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NascentOxygen
Staff Emeritus
The problem asks to find the steady state voltage across the capacitor. i attached the circuit (sorry for the poorly drawn and yes, that open part of the branch is supposed to be there)
I expect that gap is supposed to be bridged by a switch, otherwise the circuit is rather pointless.

I expect that gap is supposed to be bridged by a switch, otherwise the circuit is rather pointless.

The diode used to be where the gap is. But the orientation of the voltage source implies that the diode will become a gap because it will be reverse biased.

NascentOxygen
Staff Emeritus
The capacitor (if it carries any charge) will discharge via the resistor path, until the capacitor voltage reaches zero.

The capacitor (if it carries any charge) will discharge via the resistor path, until the capacitor voltage reaches zero.

Yes. So I think my proof in my first post was correct.

Just for practice, if this was not steady state and the capacitor just began to charge, would its voltage be 0 still? I think that because initially when (time is just greater than 0) a capacitor behaves as a short branch.

NascentOxygen
Staff Emeritus