Capacitor voltage at time = infinity

In summary: I figured it out on my own. In summary, the switch closes and causes Vc to be 40V before and after 0.
  • #1
new_id_7
3
0

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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  • #2
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?
 
  • #3
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
 

What is the significance of capacitor voltage at time = infinity?

The capacitor voltage at time = infinity is the final voltage that the capacitor will reach after being continuously charged. It represents the maximum amount of charge that the capacitor can hold.

Does the capacitor voltage at time = infinity depend on the initial voltage?

Yes, the capacitor voltage at time = infinity is directly proportional to the initial voltage. This means that a higher initial voltage will result in a higher final voltage at infinity.

Why is the capacitor voltage at time = infinity often referred to as the "steady state" voltage?

The capacitor voltage at time = infinity is referred to as the "steady state" voltage because at this point, the capacitor is fully charged and the voltage remains constant. There is no more change in voltage over time.

Can the capacitor voltage at time = infinity ever be exceeded?

No, the capacitor voltage at time = infinity is the maximum voltage that the capacitor can hold. Any attempt to exceed this voltage will result in the capacitor breaking down and potentially causing damage.

Does the capacitor voltage at time = infinity have any practical applications?

Yes, the capacitor voltage at time = infinity is an important factor in the design and functionality of electronic circuits. It is used to determine the maximum voltage that a capacitor can handle and helps in selecting the appropriate capacitors for a given application.

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