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.