In power supply, each house connected to the grid consumes a certain amount of power which leaves less power to be consumed for the rest of the houses.

Similar to that, when a radio station broadcasts a radio signal and the receiver receives it, does it decrease the power of the broadcasted signal? Or in other words, does the signal get weaker as more and more receivers are picking it up? How can 1000s of receivers pick up the same radio signal?

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    \$\begingroup\$ The recieved power is much smaller than the transmitted power. The vast majority of the transmitted power is absorbed by the earth, buildings, atmosphere, or radiates off into space. \$\endgroup\$
    – pjc50
    Commented Jan 23, 2022 at 15:18
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    \$\begingroup\$ Tesla famously thought that if he transmitted power from an antenna on the Earth's surface, only other antennas on the Earth's surface would be able to capture it. Thus, he thought that he could convey power from a generator to customers with efficiency comparable to a wire transmission line. So, you are in company with a genius asking this question. Tesla was, however, wrong here. \$\endgroup\$
    – John Doty
    Commented Jan 23, 2022 at 22:18
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    \$\begingroup\$ Does a lighthouse appear dimmer just because there are a bunch of other ships also seeing its light? \$\endgroup\$ Commented Jan 23, 2022 at 22:55
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    \$\begingroup\$ @JohnDoty Tesla was not under any such misconception. His idea wasn't just to broadcast radio waves. One of the approaches he tried was near field coupling, which is used for wireless charging today. Another was to use the resonant cavity formed by the ground and ionosphere. This was a bad idea for many reasons, but not the one you cite. \$\endgroup\$ Commented Jan 24, 2022 at 0:37
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    \$\begingroup\$ @jasonharper to be fair, it does appear a fair bit dimmer if you are behind one of those ships... \$\endgroup\$ Commented Jan 24, 2022 at 14:56

8 Answers 8


... without any loss in it ...

First, physics says there must be some loss, because energy is received at the antenna and is available to be consumed by the receiver.

... be received by 100s of receivers ...

Because the amount of power received by a typical receiving antenna is minuscule compared to the amount that is transmitted.

Let's consider a not-atypical exchange over amateur radio. I send someone a message in Morse Code, using a 100W transmitter. Assume that someone out there needs to receive at least -100dBm to understand my signal*. That's \$10^{-13}\mathrm{W}\$ -- less than a picowatt, and \$10^{-15}\$ times smaller than the signal that I transmitted.

There are not that many people in the whole world, and perhaps not in the whole history of the world -- and that's with a very modest 100W of transmitting power. Change out my 100W amateur radio transmission for an AM or FM station that's transmitting tens or hundreds of kilowatts, and I think that you can see that while the energy efficiency of broadcast radio may be poor, you can still reach a whole lot of people with it.

* I'm pulling numbers out of my head, but that implies a fairly poor receiver.

  • 3
    \$\begingroup\$ And lots of people use 5W or less to communicate between continents... \$\endgroup\$
    – Jon Custer
    Commented Jan 24, 2022 at 20:22
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    \$\begingroup\$ Yup. On a good day, at the right frequency, 5W will get your signal around the world. \$\endgroup\$
    – TimWescott
    Commented Jan 24, 2022 at 21:14
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    \$\begingroup\$ I think the main thing missing from this answer is to explain that the receiver generally has to amplify the tiny signal it receives, and to do so, needs it's own power source, that power doesn't come from the transmitter. \$\endgroup\$
    – crobar
    Commented Jan 25, 2022 at 0:18
  • \$\begingroup\$ @crobar crystal radios don't. Nfc doesn't. \$\endgroup\$
    – Passerby
    Commented Jan 25, 2022 at 3:31
  • \$\begingroup\$ Tim, so why do we need 100W or kilowatts if the receiver only needs a fraction of a fraction of it? \$\endgroup\$
    – Passerby
    Commented Jan 25, 2022 at 3:33

You're right ... but

Let's put some numbers to this. 3 mV (70dBuV) is a good strong signal into an FM receiver, on the 300 ohm (dipole antenna) input. That is a current of 10uA or a power of 30nW, so a perfectly efficient 1W transmitter could power 33 million such radios.

The FM transmitter covering London, at Wrotham in Kent transmits 125kW on each main BBC station, covering a service area between 50 and 100 miles in radius.

If there were 33 million radios within that service area, their antennae would consume 1W of that power, leaving 124,999W to dissipate across the countryside and into space.

Alternatively, if you could cram 4.125 trillion radios into SE England, close enough to see a 70dBuV signal, they would consume the entire 125kW transmitter power on that frequency. But that's more than 500 radio receivers for every person on Earth, assuming they all want to listen to the same station.

  • \$\begingroup\$ Thanks for giving concrete numbers. I had no idea that good old radio transmits such amounts of power! \$\endgroup\$
    – AnoE
    Commented Jan 25, 2022 at 16:31
  • \$\begingroup\$ There are bigger AM transmitters, like the 500kW ones at Droitwich. \$\endgroup\$
    – user16324
    Commented Jan 25, 2022 at 17:47
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    \$\begingroup\$ This needs more votes. Liked the humour. Might be worth mentioning to those who ask themselves this question that the rest is absorbed everywhere else as heat (eventually, after many partial reflections and diffusions). \$\endgroup\$ Commented Feb 4, 2022 at 6:48
  • \$\begingroup\$ @MisterMystère Actually not all does dissipate as heat ... as Carl Sagan pointed out in "Contact", the televised 1936 Olympics from Berlin are about 86 light years away by now (and OK, perhaps impossibly weak to receive by any receiver of our devising) \$\endgroup\$
    – user16324
    Commented Feb 4, 2022 at 13:38
  • \$\begingroup\$ They will, eventually. Every wave's destiny is to get reflected, diffracted, or absorbed (as heat). If it's not absorbed yet, it's just still going - "alive". \$\endgroup\$ Commented Feb 4, 2022 at 15:13

If you imagine the radio signal as the sun and each person as a radio. Only those indoors or behind someone's shadow on the ground is attenuated.(yet not completely in the dark)

Why not millions of radios? Same results.

The difference is that the grid distribution loss is by heat of {Pd= resistance * current²} in a loop back to source.

In AM/FM radio signals, it is one way and radiated thru insulation. Air is a dielectric insulator not much different that space vacuum except for moisture and reflection losses off atmosphere. So from a vertical line source, it is horizontally broadcast with inverse distance² loss in the air.

The receiving antenna does not contribute any loss to the rest of the air. (only in that tiny space it occupies and an attenuated shadow under it)

It must capture the modulated RF electric E-field created by this transmitter. This E field is measured in uV/m and has the impedance of free space. The antenna must try to match that to capture this tiny signal then filter and then amplify and filter 3 times to reject all the other signals.

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    \$\begingroup\$ "The receiving antenna does not contribute any loss to the rest of the air." Conservation of energy says it does. While the power that's picked up in an antenna is minuscule, it's still a finite amount of power that could have continued to radiate past the antenna had the antenna not been there. So by one of the most fundamental laws of physics, yes, the receiving antenna does contribute some loss of signal. \$\endgroup\$
    – TimWescott
    Commented Jan 23, 2022 at 5:30
  • \$\begingroup\$ @TimWescott yes that like comparing the energy of a flea to an elephant or saying that each tower bushing also absorbs losses on the grid which is also true but irrelevant \$\endgroup\$ Commented Jan 23, 2022 at 5:32
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    \$\begingroup\$ I certainly wouldn't argue with saying that the receiving antenna contributes an entirely negligible amount of loss -- but \$\lim_{\mathrm{loss}\to 0}\frac{\mathrm{negligible}}{\mathrm{loss}}\$ is a pretty significant non-number. \$\endgroup\$
    – TimWescott
    Commented Jan 23, 2022 at 5:44
  • \$\begingroup\$ There have been court cases where people operated flourescent lamps near strong radio towers from antennas .... \$\endgroup\$ Commented Jan 24, 2022 at 13:40
  • \$\begingroup\$ @rackandboneman I can respect that but I would expect that the issue is not the absorption of power to light but the reflection of distorted EMI near-by to the tower affecting subscriber mobile quality. You can do that all you want to a HVAC grid and they won't care. The more likely factor is lawsuits from ignorant unsafe users , thus litigation. If my assumptions are wrong . Do links. youtube.com/watch?v=L50x35df2nI&t=85s \$\endgroup\$ Commented Jan 24, 2022 at 13:55

In both cases (electrical grid and radio), energy is transported by electromagnetic fields.

Wires in the electrical grid act as waveguides. Electrical field (voltage) exists mostly between the Live and Neutral wires and current flows in the wires. Think of waveguides as pipes that guide an electromagnetic wave where we want it to go. They're a bit leaky, some of the electromagnetic field gets out, but most of the energy goes where it is supposed to. And the important part regarding your question is that if there is no load plugged in at the end of the line, the energy is not wasted.

On the other hand, a transmitting radio antenna is more like a firehose spraying electromagnetic energy around. The shape and orientation of the antenna will control the pattern and the direction of the spray, but once the waves leave the antenna and propagate in free space, there is no more waveguide. They will bounce around and diffract around obstacles, and most of them will end up absorbed by the ground, buildings, clouds, or shooting up in space. The receiving antennas will only pick up what fraction of the electromagnetic field they can pick up, wherever they are.

So the important distinction is that, with a transmitting antenna, all the transmitted power is already spent once the wave leaves the antenna. Unless there is a large reflector in front of it to send the energy back into the antenna, it never comes back. It just propagates away. So if you put a receiving antenna somewhere, it will receive some of it, whether the receiving radio is turned on or off, that doesn't matter. If you remove the antenna, the waves that it would have picked up will instead keep going and end up being absorbed somewhere else. The transmitter cannot "know" if its signal ended up in your antenna or in a tree.

The only way to "steal" someone else's signal would be to have conditions conducing to straight line propagation, and to place an object significantly larger than the wavelength between the transmitter and the other guy. Whether this object is a receiving antenna or anything else is irrelevant: if it absorbs the radio waves or reflects them in another direction, it will create a "shadow" and a receiver placed in this shadow will get less signal strength.

To keep the water analogy, if it rains and you put a bucket outside, you'll get some water. But you're not influencing the amount of water that someone else's bucket receives, unless you put yours on top of theirs.

  • \$\begingroup\$ "Wires in the electrical grid act as waveguides. " This is not correct. If this were the case, then there would be no way for DC current to flow. \$\endgroup\$
    – SteveSh
    Commented Jan 24, 2022 at 16:45
  • \$\begingroup\$ "In both cases (electrical grid and radio), energy is transported by electromagnetic waves." This statement is not true in so far as the electrical grid goes. \$\endgroup\$
    – SteveSh
    Commented Jan 24, 2022 at 16:55
  • \$\begingroup\$ I edited to replace "waves" by "fields". Would you like to explain how DC power is transported if it's not by electromagnetic fields? \$\endgroup\$
    – bobflux
    Commented Jan 24, 2022 at 17:13
  • \$\begingroup\$ So we have wire carrying DC current from a battery to a load - a resistor. Current flows through the wire, right? But if you measure the voltage difference between the battery and load side of the wire - along the length of the wire - what do you measure? Zero, right? So no electromagnet field in that direction, along the length of the wire. \$\endgroup\$
    – SteveSh
    Commented Jan 25, 2022 at 1:15

An interesting part about antennas is that if two or more antennas (elements) are closer together than a fraction of a wavelength, they influence each other through mutual coupling effects.

The result of this interaction is that having two antennas close together does not result in twice the power collected. Looking at this in a qualitative manner, the antenna (elements) "share" the RF energy impinging on them. To some extent, one antenna (element) is stealing power from the other.

In order to gather more power, antennas should be separated by approximately one-half wavelength or greater, thus increasing the total aperture.

This is different than light, because almost always when we are dealing with light we use devices (mirrors, lenses, photocells, etc) that are many many times greater in size than the wavelength of light.

  • 1
    \$\begingroup\$ Antenna element shaping is basically black magic. \$\endgroup\$
    – Passerby
    Commented Jan 25, 2022 at 3:52

First of all, remember that you're comparing photons (transmitted EM) to electrons (the power grid). This is important.

Allow me to use a couple of metaphors

When drawing power from an electrical grid, it's like drawing water out of a swimming pool. The individual pipe may draw very little water, and the result may be hard to detect, but the volume of water in the pool is decreased by the pipe no matter where the pipe is.

But that doesn't easily reflect how EM transmission works. A radio receiver is not drawing energy out of a pool (kinda, I'll get to that). Imagine, instead, a balloon being inflated. The "receiver" is a point on the balloon that's enjoying the benefit of the portion of energy that one point receives as the whole balloon expands. The fact that the single point enjoys that bit of the whole doesn't impact any other point on the surface of the balloon enjoying their portion of the energy needed to inflate the balloon.

Setting aside the metaphors

When a transmitter sends out energy, the energy delivered to the antenna is the rating (e.g., 50 kw). That energy expands away from the antenna (simplifying things a lot) in all directions - and each little point along the sphere of that expansion carries a portion of the 50 kw energy. The receiver captures that little amount — which has no impact on your next door neighbor because they're getting their little amount at a different point along the surface of the expansion.

Thus, it doesn't matter (again, really simplistically) how many receivers there are. They all capture their little portion.

Where you can get into trouble is when two receivers are too close together, interfering with the ability to capture the energy from just one point along the expansion. This is because receivers are not infinitely small — but that's another story.

Yeah, but when I think about those metaphors, they don't really work, do they?

Of course not. That's the problem with metaphors. What you're not realizing is that you can't compare photons to electrons. You can, e.g., draw an electron away by providing a lower-resistance path, thereby robbing your neighbor of electricity. But you can't draw a photon away like that. It's basically going in a straight line from the transmitting antenna until it hits something it can't move through — like a receiving antenna.


Same way you can speak in a room and everyone in the room will hear you regardless of how many people are in the room. Does your professor sound quieter when attendance is higher?

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    \$\begingroup\$ I think this might be a bad answer, because the professor actually would sound quieter - both because of the irrelevant part (the non-ear part of peoples bodies absorbing sound that would otherwise have been reflected) and the relevant part (the eardrums absorbing sound) \$\endgroup\$
    – BeB00
    Commented Jan 23, 2022 at 6:48
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    \$\begingroup\$ Its a simplified analogy, youre purposely missing the point. \$\endgroup\$
    – Shredder
    Commented Jan 23, 2022 at 7:43
  • \$\begingroup\$ It's a valid comparison for wave propagation. \$\endgroup\$ Commented Jan 23, 2022 at 7:48
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    \$\begingroup\$ @BeB00 - A better RF analogy for that might be a room with a bunch of RF absorber in it. \$\endgroup\$
    – SteveSh
    Commented Jan 23, 2022 at 17:21
  • \$\begingroup\$ @BeB00 That is also true with antennas and radios. But the effect is so small it's negligible. \$\endgroup\$ Commented Jan 24, 2022 at 0:10

Ultimately ALL of the electromagnetic radiation is absorbed by something or exits into space. The transmitter does not care or know what is absorbing the power it is sending out.

The receivers absorb only a tiny tiny fraction of the source power, and this tiny signal is boosted by an amplifier in the radio. The amplifier has to supply power to do this, usually sourced from the electrical grid or batteries.

This is why radio works without wires, but transmitting a reasonable amount of power needs wires. Where a large proportion of the energy can be delivered to you without it being absorbed or deflected by other things along the way.

Another way to think of it is, imagine you used more and more receivers, until one effectively enclosed the transmitter in a hollow metal sphere made of receivers. This will absorb all of the power, and any receivers outside the ball will receive nothing. You basically reach a limit where you can't fit any more receivers around the source, and they absorb all the power, splitting it up between them.


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