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How does current flow through antennas to transmit radio signals? From what I know, in order for current to flow, there must be a load across the circuit.

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There IS a load. Energy radiates from an antenna, so the equivalent circuit for that antenna has an energy-losing element, a resistance. It doesn't represent an Ohm's Law compliant resistor with heat generation, in this case. You have to consider that an antenna is a kind of coupling with a waveguide, and then (because it's in open space) remove the 'guide' bits. A cable, or a waveguide, has a characteristic impedance, and so does an antenna in air.

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  • \$\begingroup\$ How is the energy lost in the antenna quantified? Is it a constant amount? Does it rely on the frequency of the wave? The antenna? \$\endgroup\$ – Carrot M Aug 12 '16 at 7:46
  • \$\begingroup\$ Energy isn't 'lost in' an antenna, it's 'radiated by' an antenna. Exactly matching a wavelength to an antenna is possible, but usually there's an LC network, a 'transmatch', with a simple bandpass region (a dipole has multiple passbands, a kind of 'comb'). \$\endgroup\$ – Whit3rd Aug 13 '16 at 6:58
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You're right that there won't be a DC current flowing, but RF is all about high frequency AC. You can think of an antenna as a capacitor. You can shove some current in there for a little while and charge it up. Then you can pull current out and discharge it. The alternating current in the antenna will generate EM radiation.

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How does the source know whether there is a load at the end of the wire or not? It doesn't.

It takes time for that information to travel down the wire, reflect off the open-circuit end of the wire, and travel back to the source. I think for coaxial cable, the information travels at about 67% the speed of light. What this means is the energy source can start sending a wave down the wire but it won't now whether the wire is an open-circuit until the signal bounces back. What ends up happening is the sent wave interferes with the bounced wave and creates a standing wave in the wire. The frequency must be fast enough and the wire the right length so that the source gets through a cycle of the signal before the bounced wave returns and they constructively interfere. Wikipedia has a nice GIF of this here. Imagine the blue signal is the sent wave, the red signal is the reflected wave, and the right end of the GIF is the end of the wire. The black wave is the sum of the red and blue.

This is why both frequency and length are important. If you try to transmit a slow signal, lets say 1 kHz, the wire will need to be really long. The period is 1 ms, so you need the wave to travel to the end of the wire, reflect back, and arrive at the source in 1 ms. That means the wire would have to be:

(2/3)*c*0.001/2 = 100 km

Yes! 100 kilometers long. So, we stick mostly to waves with higher frequencies because the antennas are practical lengths and we can send more information.

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