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After all these years I am back to the start position of how capacitors work. I don't know to laugh or to cry. Okay even before that. When I hook up two wires, one on each terminal of a 12V DC battery and I measure the voltage difference at the end of the wires, it reads 12V. That goes without saying.

Now we learned, at least I seem to have that no current of charges moves at all in an open circuit. So what moved from the battery to the end of the wires? If I place an ammeter in any of the two wires also before I test the voltage at the end of the two wires, I must not read any current in the ammeter but I assume I would read volts in the voltmeter.

Since I certainly read volts, it must imply that energy ( I had to throw in that word) moves without any charge movement, unless charges actually do move in an open circuit.

Capacitors by actual design is an open circuit no matter how close are the plates. To my understanding, that is the whole idea, not to allow charges to move through.

So you can see what my mind is looking at here. And capacitors are not like a battery where the plates are connected by the fluid/electrolyte in some respect.

( I realize also that Antennas are basically open circuits....)

Furthermore current of charges is at a maximum in a capacitor and it is described as a short circuit assuming no resistors are there.

Can anyone explain this without using formulas? Thanks in advance.

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  • \$\begingroup\$ I realize also that Antennas are basically open circuits.... Not really, an antenna is still an impedance that depends on frequency. How else would antennas be able to transmit and receive radiowaves. There are plenty of sources that explain the details of capacitors, explain why you need another explanation, which will be a repetition of all those other sources anyway. \$\endgroup\$ Mar 7 at 20:37
  • \$\begingroup\$ So you want to know how capacitors work, but only because you want to know why connecting wires to 12V supply will read 12V at the other end of the wires even if no current flows? But yet you don't find it strange why it reads 12V at the battery terminals even if current does not flow, and the terminals are also conductors like wire is? I suggest reading Wikipedia how capacitors store energy in electric field though. \$\endgroup\$
    – Justme
    Mar 7 at 20:39
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Since I certainly read volts, it must imply that energy ( I had to throw in that word) moves without any charge movement, unless charges actually do move in an open circuit.

I don't see how the reading of volts implies that energy has moved. (Unless you take the non-finite impedance of the voltmeter into account, but then you must acknowledge that charge has moved)

Capacitors by actual design is an open circuit no matter how close are the plates. To my understanding, that is the whole idea, not to allow charges to move through.

In a static situation, capacitors are indeed an open circuit. No doubt about that.

( I realize also that Antennas are basically open circuits....)

Like capacitors, they are open circuits in a static situation. Antennas do not transmit DC.

I think you are confusing static and dynamic behaviour. Consider a capacitor as a reservoir of charge. Each conductor has this property, but capacitors are designed to have more of this property. Likewise, inductors are a reservoir of current (much like a moving object has momentum), but in a static situation an inductor behaves just as a wire.

To put it into different words: a capacitor is something that hinders change of voltage (over the capacitor), and an inductor is something that hinders change of current (through the conductor).

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We know that Current is defined as the rate of flow of charges. I=dQ/dt.

Also, we know the current in any dielectric or insulator or capacitor is Ic=C dV/dt.

Thus dQ/dt = C dV/dt or Capacitance is the ability to store charges with a change in voltage. C = dQ/dV and once charged and flow stops we can compute the charge stored by Q = CV.

Now try to figure out the Q and C of a battery starting at 3V and resting at a full charge 3.8V rated for 3600 mAh.

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Metal has incredible many charged particles. The positive ones - metal atom nuclei - are heavy and quite strongly bound to their places in the metal molecular structure. A part of the electrons fly here and there randomly between the atoms and can move as electric current along the metal if there's a battery or other device which generates an electric field.

If you connect 2 otherwise unconnected metal wires to the poles of a battery the electric field of the battery forces electrons in the wires to move. The plus pole of the battery sucks electrons and the minus pole injects them to the wire. The movement stops when the movable electrons in the wires have found a new distribution so that the electric field caused by non-uniformly distributed positive metal atomic nuclei and the moved electrons together create an electric field which exactly compensates the field caused by the battery. You have already seen that a 12V battery redistributes the electrons in the wires so much that the voltage of the electric field is between the wires 12V also at the free ends of the wires.

If you move the wires closer each other (but not let them touch each other) or insert bigger metal chunks to the free ends of the wires more electrons need to change place before the new balance is found with the same 12V battery connected to the other ends. This can be also calculated with vector field math, but I guess you are not going to read any math proofs.

Capacitors are made to have the same effect - let the current flow until a new charge balance is found when the capacitor is connected to a voltage source or more complex circuit. The capacitance is the measure of that function. 1 farad means that voltage =1 volt moves worth of 1 coulomb of electrons from the positive side plate through the battery to the negative side plate when the capacitor is connected to a voltage source.

1 Coulomb as an amount of electric charge is defined "If we had somewhere current =1 ampere it would move one coulomb in 1 second".

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Two words: Displacement current

It's such an important topic, but it is almost completely neglected by electronics textbooks. My original electronics books (although almost two decades old by now, but it's not like this is a new concept) doesn't mention it at all. All there was, was a small paragraph in the phyisics book.

Displacement current is important in

  • Antennas
  • Cable and trace impedance
  • Capacitors
  • Signal Propagation
  • EMI
  • ..and more

I am convinced you will several eureka moments when you read up on displacement current.

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Current does not flow through a capacitor, apart from a tiny leakage current. A capacitor is almost an open circuit.

Capacitor charge/discharge circuit.

To start with SW1 & SW2 are both open and C is discharged and so both both plates of the capacitor are at 0V.

Now SW1 closes. Immediatly the whole voltage of the battery is developed across R1 which forces a current to flow through R1 and positive charge flows from the positive terminal of the battery and accumulates on the top plate of the capacitor increasing its voltage. Simultaneously a current flows from the lower plate of the capacitor to the negative terminal of the battery removing positive charge from the bottom plate of the capacitor.

The currents in a capacitor's two leads must always be equal to each other and flowing in the same direction, one in and one out, making the current appear to be flowing through the capacitor even though the capacitor is actually an open circuit.

As positive charge accumulates on the top plate and is removed from the bottom plate, the voltage across the capacitor rises reducing the voltage across R1, reducing the current flow, rate of charge transfer and rate of voltage rise. After a time period of 5RC the voltage across the capacitor will be almost equal to the battery voltage.

Now SW1 opens and SW2 closes. The full voltage of the capacitor is now developed across the resistor R2. Positive charge flows from the top plate of the capacitor through R2 and onto the bottom plate of the capacitor. Again, at all times the current in the capacitor's two leads is equal and in the same direction (one in and one out) giving the illusion of the current flowing through the capacitor.

As positive charge is removed from the capacitor's top plate and accumulates on the bottom plate, carried by the current flowing through R2, the voltage across the capacitor reduces and after 5RC the voltage across the capacitor will be almost equal to 0V.

Now replace the two resistors with wires and repeat the experiment. Because the resistance of the two wires is very small, when the voltage is developed across them (at switch closure) the forced current will be very high (rate of transfer of charge very high) and so the charge and discharge of the capacitor will happen very quickly, still taking 5RC to almost fully charge and discharge but now R1 & R2 have very low values.

I've talked about positive charge flow rather than electron flow because I feel the concept of positive charge flow is more intuitive. I've also assumed that the output resistance of the battery and that the capacitor's ESR are both equal to zero Ohms in order to simplify an explanation.

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    \$\begingroup\$ Your first paragraph is misleading; current does flow through a capacitor, but only AC. \$\endgroup\$
    – Hearth
    Mar 7 at 22:38
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    \$\begingroup\$ @Hearth Only a tiny leakage flows through a capacitor. Not even ac current flows through a capacitor. When an ac voltage is applied to a capacitor the two plates will charge and discharge repeatedly. The current in the capacitor's two leads will always be equal and in the same direction as each other (one in and one out) the direction alternating with the applied voltage. That is what is meant when it is said that ac current flows through a capacitor, the plates are charging and discharging repetitively giving the illusion of current flowing through the capacitor. \$\endgroup\$
    – James
    Mar 7 at 22:58
  • \$\begingroup\$ I would consider the displacement current to be a real current, but I suppose it's valid to not think of it that way. \$\endgroup\$
    – Hearth
    Mar 7 at 23:14

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