# Open circuit with capacitor

1. Feb 24, 2013

### pyroknife

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?

#### Attached Files:

• ###### Circuit.png
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2. Feb 24, 2013

### Staff: Mentor

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

3. Feb 24, 2013

### pyroknife

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.

4. Feb 24, 2013

### Staff: Mentor

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

5. Feb 24, 2013

### pyroknife

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.

6. Feb 24, 2013

### Staff: Mentor

Yes, but not for that reason. To change the voltage on a capacitor you have to add charge to it. Adding charge takes time. Immediately after t=0 there has been no time to add much charge, so the voltage across the capacitor plates hasn't changed.

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