64

tl;dr- The maximum data rate you're looking for would be called the maximum entropy flux. Realistically speaking, we don't know nearly enough about physics yet to meaningfully predict such a thing. But since it's fun to talk about a data transfer cord that's basically a $1\mathrm{mm}$-tube containing a stream of black holes being fired near the speed of ...


53

It is less of a "sending electrons," and more of jiggling them. Think of a wave, done by the crowd at a sporting event. One person raises their hands high, and then sits back down. The next person raises their hands, and sits back down. So on and so forth. When we get to the end, its not that someone's hands moved from one side of the ...


34

The electrons themselves don't move all that fast. The wave energy is the part that moves quickly. Picture it this way. You have 500 meters of pipe, with a small hole at the other end. The pipe is full of water and you increase the pressure at your end. Water will flow out the other end immediately. This is the electrical energy (pressure) and the copper(...


33

The Shannon-Hartley theorem tells you what the maximum data rate of a communications channel is, given the bandwidth. $$ C = B \log_2\left(1+\frac{S}{N}\right) $$ Where $C$ is the data rate in bits per second, $S$ is the signal power and $N$ is the noise power. Pure thermal noise power in a given bandwidth at temperature $T$ is given by: $$ N = k_BTB $$ ...


31

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element. To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit. The voltage across the combination is $$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$ The ...


28

An capacitor has one intuitive property: Its voltage can't change instantly since its voltage is dependent on the charge it has stored, and charge doesn't move at infinite speeds (there is always resistance somewhere), therefore you can't instantly charge up a capacitor without infinite current. More capacitance means less voltage for the same amount of ...


27

I have drawn a diagram to illustrate the mechanisms of how telecommunications work. This is a highly simplified cartoon of what happens. What I’ll be describing is in part often described in Telecom as Layer 0. All other layers have more to do with software than physics. That being said lets go through he parts. Blue is for Microwaves sent from Cell Phone ...


15

All I really have to add to the other answers is to provide some context to complete the picture. You probably know that when you send something (like a text, or a web request) over a network, it is packaged in some data structure, then passed through several layers of software (various network libraries and so on), that manipulate it in various ways, break ...


14

In fact, electron's speed is not so fast that light bulb glows up immediately. It is the electromagnetic field which travels in the circuit at near the speed of light that is resposible for it. After turn on the light, electron only acquires a little speed in addition its thermal speed. The thermal speed of electron can be estimated by $mv^2/2\approx k_BT/...


13

Anyone who's ever set up a public address system knows that we do have this issue. It's generally called feedback, and tends to result in a high-pitched screaming sort of sound. It can be kept under control by careful use of EQ (graphic equaliser) and correct positioning of the microphone and speakers. (Pro audio people probably have lots of other tricks for ...


12

Calling it a built-in voltage is something of a misnomer. People usually think of "voltage" as "what you measure with a voltmeter". So "voltage" is normally synonymous with "electrochemical potential of electrons" (in stat mech terminology) and with "difference in fermi level" (in semiconductor terminology). Under this definition, the built-in "voltage" is ...


10

There is a physical meaning behind the imaginary component of the impedance. You can re-cast the complex impedance $Z = R + jX$ (using engineering's notation $j$ for the imaginary unit) in polar form to get $Z = |Z|\exp(j\phi)$. $|Z|$ is the magnitude of the impedance, and scales the amplitude of the current to get the amptlitude of the voltage. $\phi = \...


10

What is the physical behaviour which allows a capacitor to act as a high or low pass filter? A capacitor alone cannot act as either. To create a filter you need a combination of resistance and capacitance or inductance and capacitance (or RL). You need two immittances, at least one of which is reactive. Let's take a practical example, an RC circuit. This ...


9

Imaginary components in physics often mean phase shifts. In this case the impedance is like a resistance, but it kicks in when the current is changing by messing with its phase.


9

Strictly speaking, a phase coherent electron device is an electronic device whose dimensions is smaller than the phase coherence length of the electrons. This definition is the one adopted in mesoscopic physics. So, what is a phase coherence length? To each electron, one associates a wave-function $\Psi=\Psi_{0}e^{i\varphi}$, with $\varphi$ the phase of the ...


9

Imagine electricity as water in a pipe. The current can flow in either direction (direct current, DC) or one way then the other way (alternating current, AC). Now put a rubber membrane in the pipe. This is the capacitor. Now it will slow and then stop DC, but AC can still keep wobbling back and forth. In this way, capacitors block DC but enable AC. ...


8

In this case, the magnitude is telling you how to scale your input signal, and the argument is telling you how to phase shift it. Complex numbers usually represent 'amplification' and 'twist'. So, say, 1 means 'leave it the same', 2 means 'double it', 0.5 means 'halve it', i means 'one quarter turn', -1 means 'one half turn', -3i means 'triple it and give ...


8

Well, it does. The Earth's magnetic field is about half a gauss, or $0.5\times10^{-4}\rm\,T$. So if you have a meter of wire carrying one ampere of current from east to west, it'll feel a magnetic force of $0.5\times10^{-4}\rm\,N$ in some mixture of upwards and the north-south direction that depends on the tilt of Earth's field at your location. (I'm in ...


8

The short answer is that you need a complete circuit for a battery to work. However, I find a longer answer to be very useful to help with understanding. If you have a circuit, you are solving a problem in what we call "electrodynamics." Electrodynamics is studying what happens when electrons are moving. In a circuit, electrons are constantly circling ...


8

A matter of thresholds The reality of spread-spectrum is complicated but let's imagine that the WiFi router and cell phone tower have both allocated 1 Watt to transmit to your phone, and it in turn can transmit 1 Watt back in both cases. If the WiFi router is 10 meters away and the cell phone tower is 1 km away, then it's possible to imagine a four order ...


7

Using imaginary numbers for current in reactive components just happens to make the maths a lot simpler. In AC circuits there is typically some phase difference between the voltage and the current. Manipulating these quantities without the use of complex numbers, but instead just keeping track of the phase differences (such as the power factor), is a right ...


7

Sometimes for physical intuition, it's nice to think about the extreme cases. For instance, a zero frequency signal is just a DC voltage. If we send it through the RC high pass filter, the capacitor is just like a break in the circuit, and prevents any current from flowing. Slightly more quantitatively, the capacitor equation $Q = CV$ implies that if we ...


7

It is unlikely that the electronics of the device is responsible for that. The times of high static voltage in TVs are long gone. The only electronic thing I think could be responsible for that is improper grounding. But I think even a manufacturer of cheap TVs cares about proper grounding, unless he wants to risk expensive lawsuits filed by the relatives of ...


6

The key difference between a Zener diode and a normal diode is that the Zener diode has a low breakdown voltage - typically in the few volts range. The breakdown voltage is low because the heavy doping means the depletion layer is very thin, and even at a low voltage the field strength over this thin depletion layer is very high. With a conventional diode ...


6

But how can there be current without electron potential (voltage)? In the case where there is no resistance, current (once flowing) does not require any voltage to continue flowing. If you start a current flowing in a superconductor, then even with no applied voltage, it continues to flow. It doesn't take any force to keep a ball rolling if there is no ...


6

The lumped elements approximation of electrical circuits uses the quasi-stationary approximation for the solution of Maxwell's equations. This means that the speed of electromagnetic field propagation c can be neglected (can be assumed to be infinite). Roughly this means that the dimensions $l$ of the circuit are much smaller than the vacuum wave length $l≪\...


6

To take your question at face value: what if we somehow connected the two ends of a charged capacitor with an ideal wire, that somehow avoids having self-inductance or radiation losses or anything? Well, we can write a differential equation using Kirchoff's voltage law: $$ \frac{Q}{C}=0 $$ Whoops! That's nonsense- our initial conditions are not $C=\infty $ ...


6

here is why you can kill yourself with the knife-and-toaster trick. the AC power line into which the toaster is plugged consists of a hot line (black wire) that has 120VAC on it and a ground return line (white wire) which is at zero VAC or very close to it. The toaster has a switch inside which feeds power to the bare resistance wire inside the slots when ...


6

It can, and does. However, the absorption of a photon with wavevector $\mathbf k$ causes an electron transition from $(\mathbf{k}_i,E_i)$ to $(\mathbf{k}_f,E_f)$ subject to the constraints that $E_f-E_i = \hbar c|\mathbf k|$ and $\mathbf{k}_f-\mathbf{k}_i = \mathbf k$. Consider a transition corresponding to an energy difference of $2$ eV. This would ...


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