# Can current flow in an open circuit?

• I
• Salmone
In summary, there is a short time current in the wire when a battery is connected, as the electrons from the battery push the free electrons on the wire towards the end until an equilibrium is reached. This is similar to how a capacitor charges up briefly when connected to a battery. The system represented in the GIF is a receiving dipole antenna, which would also show a brief current when connected to a resistor. The wire in the GIF is actually one plate of a capacitor, with the other plate being everything else in the Universe. This results in a very tiny value of capacitance.

#### Salmone

I have two related but separated questions:

1. If I have an open circuit connected to a battery, let's say an alkaline battery connected to a straight wire and nothing more, once I connect the battery to this wire there will be for a short time a current in the wire? Since in the battery due to redox a certain number of free electrons are "produced" and accumulated on the negative terminal, when I connect the battery to the wire, does the force excercised by these electrons push the free electrons on the wire toward the end of the wire until an equilibrium is reached and then the current will be zero? I have this doubt since I know that a capacitor in DC current will charge for a little amount of time with a certain current I flowing in the circuit, is this the same? I think that even a capacitor with only one plate connected to a battery will charge;

2. Is the system represented in this GIF physically realizable? Let's imagine that the "white tube" in the image is a wire with a resistor R and it is immersed in a variable electric field like the one in the image, will there be a current in the circuit with an accumulation of alternating positive and negative charges at the ends of the wire? Like if we put a multimeter on the resistance, will we always measure a passage of current?

vanhees71 and Delta2
-1- You are indeed charging the parasitic capacitance, so that circuit is a closed circuit just like it would be with an explicit capacitor that you were charging up.

-2- is just a receiving dipole antenna, no? So yes you measure the RX AC voltage across the resistor.

Salmone and vanhees71
berkeman said:
-1- You are indeed charging the parasitic capacitance, so that circuit is a closed circuit just like it would be with an explicit capacitor that you were charging up.

-2- is just a receiving dipole antenna, no? So yes you measure the RX AC voltage across the resistor.
Sorry, as always I don't understand answers at first sight:

1. I'm changing the question a bit: if we connect the battery to a wire with a resistor and a multimeter to the resistor, in the moment we connect the battery do we see current with the multimeter (current that will disappear after a certain time)? Like in the image below I've made with a highly refined graphic program(Paint).

2. I don't know what object is, I've used the GIF because I needed an example like that, a wire with a resistor in that kind of time-varying electric field will feel that movement of electrons, like in the GIF, and that current?

-1- You've labeled your meter with "A", but an ammeter would be in series with the resistor to measure the series current. As drawn in parallel, you would be using a voltmeter to measure the voltage drop across the resistor to infer the current via V = I*R. And yes, if you used a fast and sensitive enough measurement of the voltage across the resistor, you would see some current flow briefly as the parasitic capacitance from the piece of wire to the right of the resistor back to the battery + end was charged up.

-2- Well, what you have shown is just a dipole receiving antenna. In fact, you didn't say where you got the GIF (you should always provide a link for attribution), but it appears to be on the Wikipedia page for Dipole Antenna:

https://en.wikipedia.org/wiki/Dipole_antenna

vanhees71 and Salmone
Salmone said:
If I have an open circuit connected to a battery, let's say an alkaline battery connected to a straight wire and nothing more, once I connect the battery to this wire there will be for a short time a current in the wire?
Yes.
Salmone said:
I have this doubt since I know that a capacitor in DC current will charge for a little amount of time with a certain current I flowing in the circuit, is this the same? I think that even a capacitor with only one plate connected to a battery will charge;
That short piece of wire is actually one plate of a capacitor, but of a very tiny value.

The other plate is everything else in the Universe, and the value of a capacitor is proportional to the size of its smallest plate and inversely to their separation (Area/Distance).

The "very tiny value" arises because the surface area of the wire is small and most of the Universe is so far away.

Hope this helps!

Cheers,
Tom

[edit:]
p.s. Strictly speaking, there is also some capacitance between the wire and the other battery terminal too.

Salmone, berkeman and vanhees71