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Let's say I have an open source like below, then, at the moment the wires are connected to the source, do the electrons in the left wire still move to the positive terminal, and electrons in the right wire move away from negative terminal?

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I ask it like that because I have this exercise, they told me to determine the R1 and R2 when the output voltage (which is CD voltage) is 15V

enter image description here

When I use KCL at node B, I realize that there is no current flow in the BC wire.

Why does this happen?

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  • \$\begingroup\$ Voltage is just potential to move electrons. Current is the rate of charge flow limited by the resistance I=V/(R1+R2). Thus any two equal R's will give the same "potential" of V/2 \$\endgroup\$ Mar 16, 2022 at 4:37

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Your analysis is using steady state conditions for a DC circuit which is for continual current flows. No continuous current can flow in BC.

However, there is a transient (transmission-line effect) effect similar to pumping water into an empty dead end pipe. The water flows momentarily until the pipe is full and then stops. For example, if the voltage supply is off then the entire circuit is at zero potential. However, when you connect the circuit, there will be a momentary flow of charge into BC to bring it up to the potential that it should be at (in this case 15V) after which current will stop. Similar to charging up a capacitor. It could be modelled as a very small capacitor between node C and the negative terminal of the battery. In reality, there's parasitic capacitance everywhere which means there's a small capacitance between every point of the circuit to every other point of the circuit.

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  • \$\begingroup\$ Does that mean if I touch either BC or AD, I only get electric shocked at a moment, after that I will have positive or negative charge but still fine? \$\endgroup\$
    – Becker
    Mar 16, 2022 at 4:52
  • \$\begingroup\$ @Becker No. There are more variables when that comes into play. Your body is a resistor, not a capacitor. If the circuit was very high voltage and the negative battery terminal was connected to earth and you touched C, your body would be a resistor connected between C and the negative terminal of the battery and you would get electrocuted. If your feet were very well insulated from earth then that would be like a capacitor due to the space between your body and earth. You then get a momentary jolt which may or may not harm you as the charge flows into you and brings your body up to potential \$\endgroup\$
    – DKNguyen
    Mar 16, 2022 at 5:02
  • \$\begingroup\$ @Becker if the negative terminal of the battery was not connected to earth then your body would also be like a capacitor since there's a parasitic capacitance between the negative battery and your body (or between the negative battery and the earth, and then between the earth and your body). The so-called "free-space capacitance" which is also the capacitance used to model ESD. \$\endgroup\$
    – DKNguyen
    Mar 16, 2022 at 5:07
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No current can "flow" in BC, because as far as nodal circuit analysis is concerned, B and C are the same node, and should be denoted with a single letter: there's no "C", just B. Same with A and D: there's just A.

But suppose you wanted to have separate B and C: you could insert a zero-ohm resistor between those nodes. No matter what current would flow though it, there would be no voltage drop, and thus the potential at C and B must be the same. So such a resistor can be removed, and the C is just B.

As for whether the electrons move: nodal circuit analysis doesn't particularly care, because you're not simulating what electrons do, but instead you're dealing with abstract concepts of nodes, currents, and potentials. The electrons are out of this picture. You don't even need to know they are there.

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