capacitor

[electricked]

Hi all,

If I have the following:
A B
----o o-----
| |
--- 10V --- +
- ---
| |
--------------

Then the voltage between the two plates of the cap is 0V right? So when the
switch is closed, the positive side inverses its voltage gradully, until the
difference in potential between the cap's plate becomes 10V and the
direction of current is opposite to the direction of current of the battery.

Now, if however I have the following:

+
----| |-------
| |
--- 10V o A
- o B
| |
--------------

Then the voltage between the two plates is 10V when the switch is open,
right? In this case though, the positive side of the cap decreases its
voltage gradually until it reaches 0V and the negative side is 10V. Is this
right?

That's it. Just want to make sure I understand how cap charges and
discharges.

Thanks!

--Viktor

 

[John Popelish]

electricked wrote:

Hi all,

If I have the following:
A B
----o o-----
| |
--- 10V --- +
- ---
| |
--------------

Then the voltage between the two plates of the cap is 0V right?

It might be. Or it can be any arbitrary voltage left from some
previous experience. The important property of capacitors is that at
zero current, they hold their voltage (whatever that is). Current
causes the voltage across a capacitor to change.

So when the
switch is closed, the positive side inverses its voltage gradully,

Gradually is probably not the best adjective. Initially, if the cap
has zero volts across it, it looks like a short circuit across the
battery and a big current flows, limited only by the internal
resistance of the battery. That current causes the voltage across the
capacitor to change (top end getting more positive than the bottom
end. As this happens, the battery resistance drops whatever
difference there is between the battery voltage and the capacitor
voltage. So this resistive voltage drops as the capacitor charges and
the current through the resistance goes down as its voltage drop goes
down (ohm's law).

until the
difference in potential between the cap's plate becomes 10V and the
direction of current is opposite to the direction of current of the > battery.

At the moment when the cap reaches 10 volts the voltage remaining
across the battery's internal resistance approaches zero, so the
battery current (which is also the capacitor current) approaches
zero. At zero current, the capacitor voltage stops changing.

Now, if however I have the following:

+
----| |-------
| |
--- 10V o A
- o B
| |
--------------

Then the voltage between the two plates is 10V when the switch is open,
right?

Assuming the switch had been closed long enough for the current to
approach zero, then yes.

In this case though, the positive side of the cap decreases its
voltage gradually until it reaches 0V and the negative side is 10V. Is this
right?

When the switch was closed, the left end of the capacitor reached a
potential 10 volts more positive than the right end.

When the switch opens, there is no path for current so that voltage
remains on the capacitor, so there remains no voltage across the
switch contacts.

It is as if you had two 10 volt batteries connected in series, but one
of them is turned around backwards. Their voltages add up to zero.

--
John Popelish

 

[electricked]

"John Popelish" <jpopelish@rica.net> wrote in message
news:40623637.B395F6CA@rica.net...

electricked wrote:

Hi all,

If I have the following:
A B
----o o-----
| |
--- 10V --- +
- ---
| |
--------------

Then the voltage between the two plates of the cap is 0V right?

It might be. Or it can be any arbitrary voltage left from some
previous experience. The important property of capacitors is that at
zero current, they hold their voltage (whatever that is). Current
causes the voltage across a capacitor to change.

Right, I was making the assumption that the voltage was allowed to be
balanced out so ground is at the usual 0V. I know this is not true "most" of
the time but it was an assumption to let me understand the main concept,
which was how the cap charges/discharges.

So when the
switch is closed, the positive side inverses its voltage gradully,

Gradually is probably not the best adjective. Initially, if the cap
has zero volts across it, it looks like a short circuit across the
battery and a big current flows, limited only by the internal
resistance of the battery. That current causes the voltage across the
capacitor to change (top end getting more positive than the bottom
end. As this happens, the battery resistance drops whatever
difference there is between the battery voltage and the capacitor
voltage. So this resistive voltage drops as the capacitor charges and
the current through the resistance goes down as its voltage drop goes
down (ohm's law).

Right, I was generalizing here. It's about 63% of voltage gain per time
constant, right?

until the
difference in potential between the cap's plate becomes 10V and the
direction of current is opposite to the direction of current of the
battery.

At the moment when the cap reaches 10 volts the voltage remaining
across the battery's internal resistance approaches zero, so the
battery current (which is also the capacitor current) approaches
zero. At zero current, the capacitor voltage stops changing.

Now, if however I have the following:

+
----| |-------
| |
--- 10V o A
- o B
| |
--------------

Then the voltage between the two plates is 10V when the switch is open,
right?

Assuming the switch had been closed long enough for the current to
approach zero, then yes.

In this case though, the positive side of the cap decreases its
voltage gradually until it reaches 0V and the negative side is 10V. Is
this
right?

When the switch was closed, the left end of the capacitor reached a
potential 10 volts more positive than the right end.

When the switch opens, there is no path for current so that voltage
remains on the capacitor, so there remains no voltage across the
switch contacts.

It is as if you had two 10 volt batteries connected in series, but one
of them is turned around backwards. Their voltages add up to zero.

So we have 10V by battery and -10V from cap, sweet.

--
John Popelish

Thanks John!

--Viktor

 

[John Popelish]

electricked wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:40623637.B395F6CA@rica.net...
of them is turned around backwards. Their voltages add up to zero.

So we have 10V by battery and -10V from cap, sweet.

--
John Popelish

Thanks John!

My pleasure.

--
John Popelish

 

[petrus bitbyter]

"electricked" <no_emails_please> schreef in bericht
news:L5OdnaJ23oZisf_dRVn-gg@comcast.com...

Hi all,

If I have the following:
A B
----o o-----
| |
--- 10V --- +
- ---
| |
--------------

Then the voltage between the two plates of the cap is 0V right? So when
the
switch is closed, the positive side inverses its voltage gradully, until
the
difference in potential between the cap's plate becomes 10V and the
direction of current is opposite to the direction of current of the
battery.

Now, if however I have the following:

+
----| |-------
| |
--- 10V o A
- o B
| |
--------------

Then the voltage between the two plates is 10V when the switch is open,
right? In this case though, the positive side of the cap decreases its
voltage gradually until it reaches 0V and the negative side is 10V. Is
this
right?

That's it. Just want to make sure I understand how cap charges and
discharges.

Thanks!

--Viktor

Viktor,

You can't understand how caps charges without a resistor in the circuit.
Otherwise the current will - theoretically - become infinite when the switch
closes. Further, both circuits are essentially equal. The difference you
make is only the result of the points you take for your measurements.

When the cap is initially uncharged you will find 0V for the cap, 10V for AB
and 10V for the battery. When you make a trip in the circuit according to
Kirchhoffs law, one of that 10V values become -10V depending on the
direction you are traveling. * In both circuits *. When you close the
switch, the cap will be charged by a current through the resistor mentioned
until its voltage has raised to 10V. Then the current stops. Doing
Kirchhoffs travel again (cw) you will find 0V for AB, 10V for the battery
and -10V for the cap. When you open the switch nothing changes. *Again in
both circuits *.

So what makes the difference? I repeat it is the choice of the points you
use for your measurements. If you keep the bottom line for a common, you
cannot measure the AB-voltage in the upper circuit and you cannot measure
the caps voltage in the lower circuit. But you can calculate it from the
results of measuring the other two elements using Kirchhoffs law.

To tell the same story with other words: As it seems you consider the bottem
line a common and the upper right corner a kind of an output you use the
Ucap for it in the upper circuit and Uab in the lower circuit.

Hope it's clear.

petrus


---
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[electricked]

"petrus bitbyter" <p.kralt@reducespamforchello.nl> wrote in message
news:SdA8c.35035$s9.32350@amsnews02.chello.com...

"electricked" <no_emails_please> schreef in bericht
news:L5OdnaJ23oZisf_dRVn-gg@comcast.com...
Hi all,

If I have the following:
A B
----o o-----
| |
--- 10V --- +
- ---
| |
--------------

Then the voltage between the two plates of the cap is 0V right? So when
the
switch is closed, the positive side inverses its voltage gradully, until
the
difference in potential between the cap's plate becomes 10V and the
direction of current is opposite to the direction of current of the
battery.

Now, if however I have the following:

+
----| |-------
| |
--- 10V o A
- o B
| |
--------------

Then the voltage between the two plates is 10V when the switch is open,
right? In this case though, the positive side of the cap decreases its
voltage gradually until it reaches 0V and the negative side is 10V. Is
this
right?

That's it. Just want to make sure I understand how cap charges and
discharges.

Thanks!

--Viktor



Viktor,

You can't understand how caps charges without a resistor in the circuit.
Otherwise the current will - theoretically - become infinite when the
switch
closes. Further, both circuits are essentially equal. The difference you
make is only the result of the points you take for your measurements.

I'm not concerned with the timing of the circuit at this point and I do
understand the practical considerations that must be taken to limit the
current flow. I'm concerned with the concept of charging/discharging and the
voltage changes within the circuit.

When the cap is initially uncharged you will find 0V for the cap, 10V for
AB
and 10V for the battery. When you make a trip in the circuit according to
Kirchhoffs law, one of that 10V values become -10V depending on the
direction you are traveling. * In both circuits *. When you close the
switch, the cap will be charged by a current through the resistor
mentioned
until its voltage has raised to 10V. Then the current stops. Doing
Kirchhoffs travel again (cw) you will find 0V for AB, 10V for the battery
and -10V for the cap. When you open the switch nothing changes. *Again in
both circuits *.

Isn't this what John pointed out in his last paragraph? Battery voltage is
10V and the cap's voltage when fully charged is -10V. If that's not it
please explain further.

So what makes the difference? I repeat it is the choice of the points you
use for your measurements. If you keep the bottom line for a common, you
cannot measure the AB-voltage in the upper circuit and you cannot measure
the caps voltage in the lower circuit. But you can calculate it from the
results of measuring the other two elements using Kirchhoffs law.

To tell the same story with other words: As it seems you consider the
bottem
line a common and the upper right corner a kind of an output you use the
Ucap for it in the upper circuit and Uab in the lower circuit.

Hope it's clear.

petrus

--Viktor