Capacitor voltage at time = infinity

AI Thread Summary
The discussion centers on determining the voltages Va and Vc in a circuit with a capacitor, where it is assumed the circuit has been in a steady state for a long time. The user initially believes that Va should be 80V and Vc 0V based on voltage calculations, but realizes this contradicts Kirchhoff's voltage law (KVL). It is clarified that the capacitor acts as an open circuit for DC, leading to no current flow and a constant voltage across it. The user concludes that Vc remains at 40V as time approaches infinity, due to the capacitor's inability to change voltage under steady conditions. The conversation emphasizes the importance of considering the circuit's initial conditions and the role of the switch in determining voltage values.
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Homework Statement



I am trying to understand how to get the voltages Va and Vc in the following circuit.
It is assumed the circuit has been like this for a very long time.

Yp8mJ.jpg


Homework Equations



Kirchoff's voltage law

The Attempt at a Solution



So I know that the capacitor acts like an open circuit to DC. The current through the resistor is obviously 0, so the voltage across the resistor is also 0.

I redraw it like this ( easier to visualize for me ).

oYHtz.jpg


Now, I have the answer from my book to be that Va is 0 and Vc is 80. This allows KVL to work.

How do you know that Va is not 80 and Vc is 0? To find the voltage between nodes "a" and "b" I usually do Voltage @ a - Voltage @ b (using correct signs). Doing this between the top of the 80V source, and the top of the resistor gives: Va = 80 - 0 = 80. If I do this at the bottom, I get Vc = 0 - (-80) = 80. Obviously this is wrong because KVL won't hold.

Thanks for any help
 
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new_id_7 said:

Homework Statement



I am trying to understand how to get the voltages Va and Vc in the following circuit.
It is assumed the circuit has been like this for a very long time.

Yp8mJ.jpg


Homework Equations



Kirchoff's voltage law

The Attempt at a Solution



So I know that the capacitor acts like an open circuit to DC. The current through the resistor is obviously 0, so the voltage across the resistor is also 0.

I redraw it like this ( easier to visualize for me ).

oYHtz.jpg


Now, I have the answer from my book to be that Va is 0 and Vc is 80. This allows KVL to work.

How do you know that Va is not 80 and Vc is 0? To find the voltage between nodes "a" and "b" I usually do Voltage @ a - Voltage @ b (using correct signs). Doing this between the top of the 80V source, and the top of the resistor gives: Va = 80 - 0 = 80. If I do this at the bottom, I get Vc = 0 - (-80) = 80. Obviously this is wrong because KVL won't hold.

Thanks for any help

There must be a switch opening or closing in this problem, or it makes no sense. Was a switch closed at time t=0 across that gap or something?
 
Ya, thanks I think I figured it out. I tried to simplify the problem by leaving the switch out, but that let's you figure out the voltage.

What I didn't tell you was that the switch causes Vc to be 40V before and after 0.

Then, I just think that no current can flow through the capacitor, so 0=I=C dv/dt means dv/dt = 0. Therefore, Vc does not change as t goes to infinity. Thus it remains at 40V.

If there is another way of thinking about it please let me know.

Thanks
 
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