Potential of a capacitor at each terminal

[Hugo]

Hi,

I've come across a question concerning the following circuit:

- Voltage Source: Node 0, Node 1. (Voltage Vmax)
- Resistor 1: Node 1, Node 2. (Resistance R1)
- Resistor 2: Node 0, Node 3. (Resistance R2)
- Capacitor: Node 2, Node 3. (Capacitance C)
- Ground: i. e., Node 0.

The voltage across the capacitor, assuming that it was discharged at
the beginning of the process, is given by:

Vcapacitor=Vmax(1-exp(-t/(C*(R1+R2))), representing t the time

The potential at Node 2 (in relation to ground) is given by:

Vnode2=Vmax-i*R1

And at Node 3:

Vnode3=i*R2

being i the current across the circuit.

Vnode2 and Vnode3 will be equal at the beginning of the process (since
the capacitor is discharged).

However, the initial value of Vnode2 (or Vnode3) will be closer to 0V
or Vmax depending on the values of the resistors. It is given by:
Vnode3,initial=Vnode2,initial=Vmax/(R1+R2)*R2.

Since it is a series circuit, the current will be the same across the
whole circuit. This also means the the coulombs transferred to each
"plate" of the capacitor will be the same.

However, the variation of the voltage at Node3 and Node2 will not be
the same, for it will depend on the values of the resistors.

And this is where my question arises: Why does the same amount of
carriers create a different variation of voltage on each "plate"? Is
it because adding or removing an electron from the "plate" connected
to the lower resistor, for example, requires less work than doing it
on the other "plate" and therefor its addition of removal only implies
a smaller change in the electric field?

The voltage difference between two nodes (V=J/C) tells us how much
work we have to exert in order to bring a charge from one node to the
other. How much does each electron added or removed from each "plate"
changes the potential at that node?

Thank you very much for your answers.

 

[Tim Wescott]

On Sat, 12 Sep 2009 05:59:10 -0700, Hugo wrote:

Hi,

I've come across a question concerning the following circuit:

- Voltage Source: Node 0, Node 1. (Voltage Vmax) - Resistor 1: Node 1,
Node 2. (Resistance R1) - Resistor 2: Node 0, Node 3. (Resistance R2) -
Capacitor: Node 2, Node 3. (Capacitance C) - Ground: i. e., Node 0.

The voltage across the capacitor, assuming that it was discharged at the
beginning of the process, is given by:

Vcapacitor=Vmax(1-exp(-t/(C*(R1+R2))), representing t the time

The potential at Node 2 (in relation to ground) is given by:

Vnode2=Vmax-i*R1

And at Node 3:

Vnode3=i*R2

being i the current across the circuit.

Vnode2 and Vnode3 will be equal at the beginning of the process (since
the capacitor is discharged).

However, the initial value of Vnode2 (or Vnode3) will be closer to 0V or
Vmax depending on the values of the resistors. It is given by:
Vnode3,initial=Vnode2,initial=Vmax/(R1+R2)*R2.

Since it is a series circuit, the current will be the same across the
whole circuit. This also means the the coulombs transferred to each
"plate" of the capacitor will be the same.

However, the variation of the voltage at Node3 and Node2 will not be the
same, for it will depend on the values of the resistors.

And this is where my question arises: Why does the same amount of
carriers create a different variation of voltage on each "plate"? Is it
because adding or removing an electron from the "plate" connected to the
lower resistor, for example, requires less work than doing it on the
other "plate" and therefor its addition of removal only implies a
smaller change in the electric field?

The voltage difference between two nodes (V=J/C) tells us how much work
we have to exert in order to bring a charge from one node to the other.
How much does each electron added or removed from each "plate" changes
the potential at that node?

Thank you very much for your answers.

Trying to view the action of each plate independently will only get you
confused. Because it's a series circuit putting one electron into one
end of the capacitor means that you need to take one out the other
side*. The reason that the voltage on each end of the capacitor is
different is because the current through the circuit determines the
voltage drop in each resistor; the reason that the current flows is
because the voltage across each resistor determines the current; your job
is to wrap your brain around those two facts and make sense of it.

* Quantum mechanics, tunneling, capacitance to the rest of the world, yes
yes yes blah blah blah.

--
http://www.wescottdesign.com

 

[Hugo]

On 12 sep, 23:53, Tim Wescott <t...@seemywebsite.com> wrote:

On Sat, 12 Sep 2009 05:59:10 -0700, Hugo wrote:
Hi,

I've come across a question concerning the following circuit:

- Voltage Source: Node 0, Node 1. (Voltage Vmax) - Resistor 1: Node 1,
Node 2. (Resistance R1) - Resistor 2: Node 0, Node 3. (Resistance R2) -
Capacitor: Node 2, Node 3. (Capacitance C) - Ground: i. e., Node 0.

The voltage across the capacitor, assuming that it was discharged at the
beginning of the process, is given by:

Vcapacitor=Vmax(1-exp(-t/(C*(R1+R2))), representing t the time

The potential at Node 2 (in relation to ground) is given by:

Vnode2=Vmax-i*R1

And at Node 3:

Vnode3=i*R2

being i the current across the circuit.

Vnode2 and Vnode3 will be equal at the beginning of the process (since
the capacitor is discharged).

However, the initial value of Vnode2 (or Vnode3) will be closer to 0V or
Vmax depending on the values of the resistors. It is given by:
Vnode3,initial=Vnode2,initial=Vmax/(R1+R2)*R2.

Since it is a series circuit, the current will be the same across the
whole circuit. This also means the the coulombs transferred to each
"plate" of the capacitor will be the same.

However, the variation of the voltage at Node3 and Node2 will not be the
same, for it will depend on the values of the resistors.

And this is where my question arises: Why does the same amount of
carriers create a different variation of voltage on each "plate"? Is it
because adding or removing an electron from the "plate" connected to the
lower resistor, for example, requires less work than doing it on the
other "plate" and therefor its addition of removal only implies a
smaller change in the electric field?

The voltage difference between two nodes (V=J/C) tells us how much work
we have to exert in order to bring a charge from one node to the other.
How much does each electron added or removed from each "plate" changes
the potential at that node?

Thank you very much for your answers.

Trying to view the action of each plate independently will only get you
confused.  Because it's a series circuit putting one electron into one
end of the capacitor means that you need to take one out the other
side*.  The reason that the voltage on each end of the capacitor is
different is because the current through the circuit determines the
voltage drop in each resistor; the reason that the current flows is
because the voltage across each resistor determines the current; your job
is to wrap your brain around those two facts and make sense of it.

* Quantum mechanics, tunneling, capacitance to the rest of the world, yes
yes yes blah blah blah.

--http://www.wescottdesign.com

Thank you for your answer. I think I've got it. The electric field
across the capacitor will set the potential of each plate according to
the resistors, so that the potential difference across each of them
produce the same current through each resistor. Thinking it the other
way around is much more complicated =)