Remember, if the voltage across a capacitor is steady (DC), then no current is flowing into or out of the capacitor.
It requires a change in voltage.
A capacitor does not act like a resistor.
If it did, we wouldn't need capacitors

So even here :



I shouldn't think of it as a voltage difference and thus the current flow ? Like for resistors ?

Because here :



It was not flowing. But as I know I should look at the voltage difference. But should I look at this the same ? That the flow of current from capacitor is not because of voltage difference ? :

 

If I had voltage source, resistor and capacitor connected in series then this rule high voltage flow to lower voltage works.
Because for example capacitor has 5V and voltage source has 4V so 5V goes to 4V.

But I don't know when I can see it like that and when not.

You have the exact same issue. If you have a voltage source, resistor, and capacitor in series, then how much current would flow in the circuit if the capacitor were removed? Zero, since you now have an open circuit. What is the voltage drop across a resistor with no current flowing through it? Zero. So the voltage across the capacitor would be equal to the voltage of the source and there would be no current flowing.

Also, voltage does not flow -- current flows. Voltage is a description of the difference in potential energy between two points in a circuit.

 

You cannot treat a capacitor as a resistor.
A resistor is a linear device. It obeys Ohm’s Law.
A capacitor is a non-linear device. It does not obey Ohm’s Law.
The equation that governs the current-voltage behaviour of a capacitor is a differential equation.

Resistors, capacitors, and inductors are, indeed, linear devices.

The constitutive relationship between voltage and current for all three are linear differential equations.

Ohm's Law is merely the name we give to the particular constitutive relationship for a resistor.

 

That the flow of current from capacitor is not because of voltage difference ? :

As I stated, capacitor current does not flow when there is a steady DC voltage across the capacitor.
It only flows when during a change in the voltage, given by the formula I = ΔV/Δt * C, where ΔV/Δt is the change in voltage with time.

Is that not clear?

 

So even here :

View attachment 273754

I shouldn't think of it as a voltage difference and thus the current flow ? Like for resistors ?

At the moment you are showing, namely with 4 V on the left side of the resistor and 5 V on the right, there will be a current flowing, namely 1 mA. But that means that 1 mA of current is flowing out of the top of the capacitor, which means that the voltage on the capacitor is decreasing. In this case, the voltage is decreasing at a rate of about 0.1 V per us. So in a few microseconds, the voltage on the capacitor will be 4.5 V and the current in the resistor will have dropped to 0.5 mA and the rate of discharge of the capacitor will have dropped to about 0.5 V/us. But it will still be discharging. Eventually the voltage on the capacitor will reach 4 V, at which point the voltage across the resistor will be zero and there will be no current flowing in it, meaning that the rate of discharge on the capacitor will be zero and it will sit there with a voltage of 4 V across it and no current flowing in it from that point on (until something in the circuit changes).

Because here :

View attachment 273755

It was not flowing.

That is because the voltages you have shown there correspond to a time after everything has settled, whereas the voltages you showed in the prior circuit correspond to a time when things are still in the process of settling.[/QUOTE]

 

Also, voltage does not flow -- current flows. Voltage is a description of the difference in potential energy between two points in a circuit.

Sorry my bad I wanted to say current flow

But what I wanted to say is that I shouldn't look at the capacitor there as a voltage differential that causes the current to flow when the capacitor is there.
I shouldn't look at it like that when I had capacitor parallel with resistor where there was voltage differential 5V and 2,5V.
But I shouldn't also look at that moment where current is flowing from high voltage to low in series circuit.

As I stated, capacitor current does not flow when there is a steady DC voltage across the capacitor.
It only flows when during a change in the voltage, given by the formula I = ΔV/Δt * C, where ΔV/Δt is the change in voltage with time.

Is that not clear?

It is.
I just wanted to confront two circuits one with parallel and one with series.

Which in one when there was voltage difference capacitor wasn't discharging.
For series there was also voltage difference but capacitor was discharging.

What do I mean here is that I shouldn't look in capacitor a voltage difference that causes the current flow.

 

What do I mean here is that I shouldn't look in capacitor a voltage difference that causes the current flow.

As has been stated several times, current flow in a capacitor is related to CHANGES in the voltage across it. If the voltage across a capacitor is not changing, then there is no current flowing through it. Equivalently, if there is current flowing through a capacitor, then the voltage across it must be changing.

In DC circuits such as the ones you have been showing, there is a transient response when the circuit starts up and during this time the voltages and currents within the circuit are changing. Eventually, they reach their final, steady-state values at which point they are no longer changing. When they are in steady state, the voltage across any capacitors is not changing, meaning that there is no current flow in the capacitor in steady state.

Whether or not current was flowing in the capacitors in your examples had absolutely nothing to do with whether they were series, parallel, or a combination of the two. It had to do with the fact that one of them was in steady state while the other wasn't.

 

As has been stated several times, current flow in a capacitor is related to CHANGES in the voltage across it. If the voltage across a capacitor is not changing, then there is no current flowing through it. Equivalently, if there is current flowing through a capacitor, then the voltage across it must be changing.

In DC circuits such as the ones you have been showing, there is a transient response when the circuit starts up and during this time the voltages and currents within the circuit are changing. Eventually, they reach their final, steady-state values at which point they are no longer changing. When they are in steady state, the voltage across any capacitors is not changing, meaning that there is no current flow in the capacitor in steady state.

Whether or not current was flowing in the capacitors in your examples had absolutely nothing to do with whether they were series, parallel, or a combination of the two. It had to do with the fact that one of them was in steady state while the other wasn't.

I was only expecting the answer yes or no though

I was just saying that whether the capacitor is series like in the second picture or parallel like in the first picture.
Voltage difference doesn't work there. Even If it looks like it works for series. Because in one was 5V and in second 4V but not the voltage difference says here anything for capacitor.

 

I was only expecting the answer yes or no though

I was just saying that whether the capacitor is series like in the second picture or parallel like in the first picture.
Voltage difference doesn't work there. Even If it looks like it works for series. Because in one was 5V and in second 4V but not the voltage difference says here anything for capacitor.

I don't know if you under stood my simulation and what I have said, and since what you are saying is unclear, my answer is yes or no.

 

I don't know if you under stood my simulation and what I have said, and since what you are saying is unclear, my answer is yes or no.

I'll use pictures Because my english is pretty bad.



In both cases just because current flows is not because of voltage difference. In first it flows in second it doesn't.

 

Just get it out of your head that a voltage across a capacitor causes current to flow in or out of the capacitor. All of your misunderstanding comes back to that incorrect assumption.

Here is the rule for a capacitor:

dV / dt = I / C

Just knowing the voltage across the capacitor says nothing about how much current is flowing. You must know how the voltage is changing. If the change (dV / dt) is 0, then the current is zero. And it is the current that changes the voltage, not the other way around.

 

You really need to start over again with Ohm's Law and grow from there in order to out-think your wrong thinking.

 

You really need to start over again with Ohm's Law and grow from there in order to out-think your wrong thinking.

I know the equation from Ohm's Law.

 

That is not enough to analyze circuits with capacitors!

 

That's good. So you know what happens when you have two resistors in series. That is essential. (Be thankful that we can skip kirkoff's circuit law! --at least for now
)
Now take a look at this: https://en.wikipedia.org/wiki/RC_time_constant

 

I'll use pictures Because my english is pretty bad.

View attachment 273757

In both cases just because current flows is not because of voltage difference. In first it flows in second it doesn't.

Again, in the circuit on the left current is flowing in the capacitor because current is flowing in the resistor. Because current is flowing in the capacitor, the voltage across the capacitor is changing, thus this circuit is NOT in DC steady state. which requires that the voltage across all capacitors be constant (i.e., steady).

The circuit on the right IS in DC steady state because the current in the capacitor is zero and thus the voltage on the capacitor is not changing.

There is still current flowing in both resistors, but none of it is flowing through the capacitor.

Take those same two pictures and, in the left hand one, just change the 5 V on the top left node to 4 V. That will place it in DC steady state with no current flowing in the capacitor. Then change the 2.5 V in the right hand picture to, say 4 V, and now that circuit is not in DC steady state and current will be flowing in that capacitor. Series or parallel does not matter when it comes to whether or not current is flowing in the capacitor. What matters is whether or not the circuit is in steady state.

 

I was supposed to post this earlier today but got distracted.
Bob filled in on post #31.

The current in the capacitor leg is given as
I = C x dV/dt

Have you covered differential equations in calculus as yet?

 

I think I see now where your assumption is flawed.
A capacitor with a resistor in series is charging.
A capacitor with a resistor in parallel is discharging.
These are both wrong.

You need to analyze the initial condition of the circuit.
Then you need to determine the final or steady-state condition.

Begin by analyzing the circuit with the capacitor removed.
This is only part of the initial condition.
Now assign an initial condition of the capacitor. This is critical.
We normally assume that the initial condition of the capacitor is 0V. But this does not have to be the case. If we were to make the initial condition of the capacitor to be charged at 10V for example, this changes the behaviour of the circuit. The capacitor could be charging or discharging.

Regardless of the initial condition of the capacitor, the final condition remains the same. The final condition of the circuit will always be the same as that of the circuit without the capacitor. There will be no changes in voltage. dV/dt = 0 and therefore the current through the capacitor will be zero. The capacitor is neither charging or discharging. This is the steady-state condition.

 

I've taken this example :

https://tinyurl.com/2fansnsf



It's just so wierd for me that it doesn't go from High potential to low potential. And it is the same with voltage source.
It works only for resistors yes ?

 

I do not understand your question.

You have shown three new circuits with voltage sources.
You need to analyze each circuit separately using Kirchhoff’s current and voltage laws and Ohm’s Law.