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I know that current flows from higher potential towards the lower. I was wondering why it does not flow towards point "a" and why isn't it divided into "i_y" and "i_x"? And if it happened, it would stuck inside the open wire at point "a", and current "i_y" would head back towards the source. Here, I was thinking, what is the potential of an open circuit in this picture and is it lower enough for the current i to flow in it?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ You're answering this question yourself. i_x == 0, because every rule says that as much charge flows into that branch needs to flow back at the same time. So, your question "why does it not flow" makes no sense – the question you should be asking yourself would be "why should it flow?" \$\endgroup\$ – Marcus Müller Dec 9 '17 at 13:59
  • \$\begingroup\$ @MarcusMüller it should because "maybe" the open wire has less potential than wire at the back of point "a". plus, does it go back?! why would it go back? \$\endgroup\$ – parvin Dec 9 '17 at 14:03
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    \$\begingroup\$ @user16307 Current does not "choose the easiest way". Current "chooses" all paths inversely proportional to the resistance of that path. \$\endgroup\$ – Matt Dec 9 '17 at 21:45
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    \$\begingroup\$ This is one case where the water-in-pipe analogy doesn't work. If you cut a wire that is carrying an electric current, the electrons do not tumble out onto the floor; they stop flowing. \$\endgroup\$ – Chu Dec 10 '17 at 0:38
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    \$\begingroup\$ @IwillnotexistIdonotexist, If I open a valve, water flows; if I open a switch, current does not flow. Analogies can be confusing and should be used with caution - remember, analogies are usually used to help students, not experts \$\endgroup\$ – Chu Dec 10 '17 at 9:54
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Under steady state conditions there is no place for the current in that branch to flow as others have said. The potential at the end of that wire is equal to the potential at the start of the branch so there's no push. At time zero though there is no potential anywhere except for at the source. When the supply is activated for the first time or a switch closed charge carriers begin to move down the line (we'll use the positive convention for simplicity). At this time the supply has no idea the end of the line is open so it will continue to pump.

As our friendly charge carriers slam into the open circuit they will begin to build up until they are they are strong enough to push back against the flow. Then the system will fall back into equilibrium at which point no more carriers will flow down the open end of the line. The potential will be the same. You could do an experiment with a pulse generator and an oscilloscope and some coax and look at the voltage waveform that appears at the end of the line.

Now depending on the physical structure of the lines and speed of the input signal more interesting things could happen. For instance a high frequency signal might find a coupling path and radiate like an antenna.

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I have no idea whether this will help you or anybody else but your question, "I was wondering why it does not flow towards point "a" and why isn't it divided into "i_y" and "i_x"?" makes me think that you imagine that the 'a' branch can accept current.

If I can use a road and traffic analogy it might look like this:

enter image description here

Figure 1. Top: open circuit. Bottom: closed circuit.

As Figure 1 attempts to show, the whole road network is full of cars. When the switch is closed and traffic is allowed to flow all the vehicles in the closed loop align and begin to circulate. None branch down the dead-ends as there is no room and no way out for the vehicles. (The main circuit is already full and can't take vehicle from one of the dead-ends.)

You could make similar analogies with water pumps, etc., but the important thing to realise is that the system is already primed. There are no voids for "current to flow into".

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    \$\begingroup\$ I wish I had graphic skills. I would make a gif! \$\endgroup\$ – mkeith Dec 9 '17 at 17:56
  • \$\begingroup\$ Hmm how you draw these ? O_O \$\endgroup\$ – Mitu Raj Dec 9 '17 at 18:58
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    \$\begingroup\$ Redraw the schematic using CircuitLab, screen-grab the schematic, paste into Inkscape on one layer and lock it. Image search for a car icon, screen grab it very tightly to save time making it have a transparent background. Paste into Inkscape on another layer. Copy'n'paste as many as required. Use 'H' and 'V' to flip some of them. Screen-grab the result, save and upload. Probably ten minutes work. greenshot.org for the screengrabber. \$\endgroup\$ – Transistor Dec 9 '17 at 19:10
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There is no potential there! it is like throwing a ball when there is no gravity, what would happen? a loop must be closed in order to flow current.

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  • \$\begingroup\$ I was just thinking, maybe it has less potential than a charges wire (the wire at the back of point "a") so current would flow in it \$\endgroup\$ – parvin Dec 9 '17 at 14:04
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    \$\begingroup\$ "maybe it has". It has not. How should it? Physics isn't a "maybe I can imagine a reason" game – you need a cause for an effect. \$\endgroup\$ – Marcus Müller Dec 9 '17 at 14:05
  • \$\begingroup\$ told you the cause! it's just, i'm not so sure about it, havent's studied this enough to be sure enough! An uncharged wire has less potential than a charged wire, an uncharged wire is on the way of current flowing from a charged wire, so it goes to the uncharged wire due to its less potential. some friend just told me that it "does" go inside, it's just too little and flows only for a short while, and in that short while it has a very little energy loss due to self resistance of the wire. \$\endgroup\$ – parvin Dec 9 '17 at 14:34
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There is a small amount of biasing current at point a. I have been point a. I felt it. It taught me a lot about electrical safety.

The effect is similar to a hydraulic line that is unpressurized but entirely full of oil. When you open the valve to connect it to a 2000psi source, the oil compresses very, very slightly, and there is this momentary "blip" of flow at a. It then balances out and flow stops.

Essentially it is placing an electrostatic charge on the wire past a. That can be measured or calculated, but it is inconsequential unless the physical material past a is quite large, or the voltage is quite high. I can tell you biasing a 20,000 square foot metal roof, to 600V, is noticeable.

You can think of it as a small capacitor.

AC is a different deal. That tiny biasing current will flow every half cycle, or rather, almost all the time as it changes the electrostatic charge.

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  • \$\begingroup\$ But there is a short-circuit between the two branches. There will be no oil pressure differential. \$\endgroup\$ – Transistor Dec 10 '17 at 8:28
  • \$\begingroup\$ @Transistor what I mean is, like on your drawing, at "a" the cars will push themselves a tiny bit closer. No more than the actual capacitance of the wire, too little to be worth modeling for sane wire lengthe ot low voltages. \$\endgroup\$ – Harper Dec 10 '17 at 10:06
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In an open circuit condition , state that voltages is present between them at across " a "of the source , that is zero current , there is zero power dissipoation . P=I E , i is zero and any multiplier by zero is zero .

In an open circuit , the end wire between two point are infinite resistance I= V/R v/infinite. Is I = zero .

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