0
\$\begingroup\$

As far as I know there should be a current when a high potential point is connected to a low potential point, but apparently that is not true. I understand that a circuit must be closed in order for it to flow. I just don't get why a return path to source is essential for creating current.

Why do electrons have to go back in a loop to the source? Shouldn't they just want to move to a point with a different potential, regardless if it's the source or not?

\$\endgroup\$
  • 1
    \$\begingroup\$ I get the impression you are regarding a "point with a different potential" as something that needn't have an electrical connection to the original point. \$\endgroup\$ – Andy aka Aug 23 '18 at 11:38
  • \$\begingroup\$ Yes exactly. Shouldn't that create a voltage, potential difference, and therefore create a current? \$\endgroup\$ – poh Aug 23 '18 at 11:45
  • \$\begingroup\$ Have a look at electronics.stackexchange.com/q/72875/152903 \$\endgroup\$ – Solar Mike Aug 23 '18 at 11:51
  • 2
    \$\begingroup\$ Possible duplicate of Am I insane to question that only with a closed path can electrons move? \$\endgroup\$ – Dmitry Grigoryev Aug 23 '18 at 13:03
  • \$\begingroup\$ The same current flows into a source as flows out of it. But electrons are not homing pigeons, so it doesn't need to be the same ones flying in as flew out. \$\endgroup\$ – Chu Aug 23 '18 at 14:18
2
\$\begingroup\$

If there is no "return" galvanic connection between two independent circuits, those circuits will rapidly charge to the same voltage and then current will stop flowing. The speed at which they charge is down to their relative capacitance to ground and their initial potential difference. Imagine two buckets of water (connected via a pipe at their bases and a tap that is turned off preventing flow). The two buckets have different heights of water but, after the tap is opened, the levels equalize and no more water flows. Water-height = voltage, water-flow = current.

If there is a galvanic link i.e. the two circuits are electrically connected (even by high value resistors), current will continue to flow because there will always be a potential difference generated by the current flowing through the resistors maintaining a static voltage difference.

If they are galvanically linked and have the same potential, current will not flow.

\$\endgroup\$
  • \$\begingroup\$ So you mean that the source continues to create potential difference, hence creating a constant current flow? \$\endgroup\$ – poh Aug 23 '18 at 12:02
  • \$\begingroup\$ A source that can produce a constant flow of charge (aka electrons) will create a rising voltage that can only be curtailed by having a resistive galvanic connection back to the source return terminal. Without that resistive connection the rate of voltage increase is determined partially by the capacitance between the two galvanically isolated circuits. It's more complex than I suspect you think it might be AND your question only partially covers the different scenarios that I tried to deliver in my answer. \$\endgroup\$ – Andy aka Aug 23 '18 at 12:06
  • \$\begingroup\$ But how does for instance a battery maintain a potential difference? Doesn't charge balance itself out going from - to +? What is the battery doing to create potential difference? \$\endgroup\$ – poh Aug 23 '18 at 12:24
  • 1
    \$\begingroup\$ I'm interested in this too. What if we theoretically could touch an extremely large, well conducting object with one terminal of a battery? Would the object still rapidly charge to the same voltage regardless how big it is compared to the battery? \$\endgroup\$ – S. Rotos Aug 23 '18 at 12:38
  • \$\begingroup\$ @poh Batteries create a potential difference due to a chemical reaction. \$\endgroup\$ – Andy aka Aug 23 '18 at 15:17
2
\$\begingroup\$

Electricity is nothing more than the flow of electrons.

Atoms are made of protons and neutrons in the center, and are surrounded by an electron "cloud" orbiting this. The number of electrons usually balances out the positive charges of the protons, but the exact count is variable; some materials are willing to take on an extra electron temporarily, and give it back later. These materials are called "conductors" because they facilitate adding and removing electrons. Add or remove electrons across the surface of a material, and it can be considered a flow of electrons. Materials which are not willing to accept electrons temporarily are termed "insulators", as they resist the flow of electrons from one atom to another, so cannot conduct electrical current.

"Current" is the flow of these electrons. Zero current = zero electron flow. To cause a current, a source of excess electrons is needed. Imagine you had a steel plate with all of it's electrons in equilibrium. And you have a pin-sized source of excess electrons. When the pin is touched to the plate, the excess electrons flow quickly from the pin to the plate (because equilibrium must be maintained) and scatter over the surface. During this time, we can say that a current is flowing. After a very short amount of time, the excess electrons are scattered over the surface uniformly, and the whole steel plate now has a slightly different electrical potential than when it started - we can say it has gained charge. Interestingly, if you touched this plate to another (uncharged) plate, current would also flow to that, and both would have some charge.

This is how "static" electricity works. Static electricity is an impractical way to do much real work however, so to move more electrons around more quickly, a way to "sink" or "return" those excess electrons (and more quickly balance the charge) is needed. This is exactly what a battery does. One terminal supplies excess electrons, and the other terminal eagerly wants electrons back. Connect a wire between these two, and a lot of current flows. How much current flows, depends on how eagerly the battery supplies and returns electrons (voltage) and how good the wire is at conducting the flow of electrons (the wire resistance.)

If you take a 9v battery and a 6v battery and connect + to + and - to -, the 9v battery is going to try to force electrons into the 6v battery (in reverse) because the 9v can supply more electrons (at a faster rate) than the 6v.

\$\endgroup\$
  • \$\begingroup\$ Would it then be possible to connect the - of a battery to a point to a metal plate and still gain normal current flow? \$\endgroup\$ – poh Aug 23 '18 at 12:45
  • \$\begingroup\$ The instant one conductor is touched to another conductor, any latent charge present on those conductors seeks to find equilibrium. So if a terminal of a battery was connected to a pin (which had a surplus of electrons on it) and this was touched to a metal plate - current would flow only to create equilibrium, then current would stop. This all happens on the order of millionths of a second. To get current flowing again, you must connect the other battery terminal to the plate somehow, to provide a path for electrons to return back to the battery. \$\endgroup\$ – rdtsc Aug 23 '18 at 20:54
0
\$\begingroup\$

If your voltage source is AC, then the changing electric fields, continually changing, allow a flow of current to the surrounding environmental conductors such as the floor if concrete or dirt, to the metal grids behind tile walls, and to the ever-present wall wiring of the 50/60Hz power system.

Lets model your feet, above concrete/grounded floor as 0.1meter * 0.1 meter, and assume the insulation is 1cm thick, with polarizability being 5X that of air or vacuum, thus 5X energy can be stored. How much capacity is this?

C = Eo * Er * Area/Distance

C = 8.98e-12 Farad/meter * 5 * 0.1m*0.1m / 0.01m

C ~~ 50pF * 0.01/0.01 = 50pF

What will be the (displacement, or field charging) current?

Q = C*V , now differentiate to get dQ/dT = dC/dT * V + C * dV/dT

and proclaim the C to be constant so dC/dT is zero. Result is

I = dQ/dT = C * dV/dT

For 117vac rms 60Hz, where the 60Hz is 377 radians/second, the slewrate the dV/dT is 117 * peak/rmsconvert * 377

dV/dT = 117 * 1.414 * 377 or approx. 70,000 volts per second. The current thru 50pF is

I = 50pF * 70,000 volt/sec = 5e-11 * 7e+4 = 35e-7 or 3.5 microAmperes at 60Hz.

On the other hand, if your voltage source is DC, once the electric-fields become stable, no more current/charge/electronics will move.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.