antenna-theory.com

[tensor20]

Dear Forum,

I found this pdf article that some of you may find useful:

http://www.hottconsultants.com/pdf_files/dipoles-1.pdf

I am wondering the following:
If we take a DC battery (constant voltage source) and connect only one terminal to ground (either the + or the -) and the other to a light bulb, nothing will happen because

1) The circuit is open and DC current needs a closer metallic path to flow.
2) The terminal that gets connected to ground assumes the same potential as ground.
3) To make current flow we need a potential difference.
4) Once we connect the battery terminal to ground, a fast current will take plance that will transfer electric charge around and make the battery terminal+ wire+ ground at the same potential. So there is a transient situation that leads to an electrostatic situation eventually.
It is true that initially there is a potential difference between the battery terminal and ground but it cannot be sustained.

AC case:
AC current does not need a close path to flow. Circuits involving antennas are an example: they are open circuits where AC current flows back and forth.

Now, if we connect one terminal of an AC battery to a light bulb and we don't connect the other terminal to anything (we actually cover it up) will current flow through the light bulb or not? I don't think so. but why?

If seems we would have a monopole antenna....

I assume the right frequency for the AC source.

Thanks,
tensor 20

[bigSteve]

You wrote:

>> "1) The circuit is open and DC current needs a closer metallic path to flow."

Yes.

>> "2) The terminal that gets connected to ground assumes the same potential as ground. "

Yes.

>>" 3) To make current flow we need a potential difference. "

Current will flow if you have voltage and a non-infinite resistance (V=IR)

>> "4) Once we connect the battery terminal to ground, a fast current will take plance that will transfer electric charge around and make the battery terminal+ wire+ ground at the same potential. So there is a transient situation that leads to an electrostatic situation eventually.
It is true that initially there is a potential difference between the battery terminal and ground but it cannot be sustained."

I don't know what you're talking about here. The battery always maintains the potential difference between its terminals. Then you move to the AC case:

>>"AC current does not need a close path to flow. Circuits involving antennas are an example: they are open circuits where AC current flows back and forth. "

AC current does need a path to flow. Antennas are a rare example, and this occurs only when the paths are large relative to a wavelength (about one half wavelength). Then if you have a structure that will radiate (i.e. not a transmission line, then it is possible to consider the unconnected antenna mode).


>>"Now, if we connect one terminal of an AC battery to a light bulb and we don't connect the other terminal to anything (we actually cover it up) will current flow through the light bulb or not? I don't think so. but why? "

No current will flow. There is no closed loop, there is no antenna. AC current requires a closed path unless you have a radiating structure. You don't have it here.

>>"If seems we would have a monopole antenna....

I assume the right frequency for the AC source. "

No, a monopole uses a ground plane as the negative (ground) side of the atnenna. This is not "connected to nothing".