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15

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 ...


6

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 = ...


5

Nowadays, the answer is negligibly so. Video cameras now digitise the image as pixels in parallel using charge coupled device technology. Former technologies, however, would emit appreciable bremstrahlung from decelerating electron beams, as I now describe. Before the coming of CCD arrays, the main video technology was the scanned photocathode, also called ...


5

This is a part answer - my (published) research has been in detecting UVA and UVB using smartphone cameras, but there has been research and a successful app made to detect gamma radiation using a smartphone camera by the Australian Nuclear Science and Technology Organisation, as reported on their webpage Smartphone radiation detector app tests positive. ...


5

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 ...


4

From "The Transistor, A Semi-Conductor Triode", by J. Bardeen and W. H. Brattain, Phys Rev. 74(2), 230-231 (1948): "The device consists of three electrodes placed on a block of germanium as shown schematically in Fig. 1. Two, called the emitter and collector, are of the point-contact rectifier type and are placed in close proximity (separation ~0.005 to ...


4

… an ideal power source capable of providing infinite current with no drop in the voltage it supplies. … Let's ignore the effects of current density on superconductors for now. … In these phrases is the explanation for the contradictory possibilities you have computed: you have supposed an impossible circuit. As a mathematical model, the behavior of ...


4

If we believe this measurement shown by Omen, smartphone cameras are basically useless below Dose rates of 10uSv/h. The max. exposure limit for a human who is not a radiation worker is 1mSv/year, which translates into roughly 0.11uSv/h. In other words, the camera chip in a phone would have to be 100 times more sensitive to pick up relevant amounts of ...


4

Your eye has a lens in it. Without a lens, the light is all spread out and overlapping, just like you say. The light from any given pixel goes out in all directions, but a lens can make it re-converge back to a point. Hold up a sheet of white paper. Is there an image on it? No, of course not. It has light on it---light coming from each object in the ...


3

A capacitor is often used for "decoupling". The wires into any electrical appliance have inductance (because they are long and thin). This means that if there is a sudden increased demand in current, there will be a significant voltage drop. A capacitor can act as a "tiny battery" that briefly supplies this current while the main supply catches up. A fan ...


3

As Kevin Reid aptly explains, the circuit you have drawn is not realizable. But, let's take the closest physical thing you could build, assuming: your voltage source can supply enough energy that we don't hit its limits like all physical things, this apparatus has non-zero size Then, the circuit you actually built is this: simulate this circuit ...


3

The electrons need to get from the top to the bottom without any interference from any gas molecules that might be in the channels. If nothing else, collisions with gas molecules will degrade performance. At atmospheric pressure, I don't think the device would work at all. You can blow a hole through an MCP with over-voltage, but I'm not sure how this ...


3

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 ...


2

The light output of a LED is pretty linear with the current through it, over its normal operating range. Light does usually drop off from linear with current at the high end. Sometimes that high end is not included in the normal operating range, so the graph you see in the datasheet will be linear. Common T1-3/4 20 mA indicator LEDs are usually linear ...


2

First of all note, that ions do not move in the semiconductor. Only electrons and holes move there. The depletion results from diffusion of the free electrons from the n-region into the p-region and the diffusion of holes from the p-region into the n-region. In either case it is a thermodynamical process. The electrons form something like a dense gas in ...


2

The analogies may be built in various ways – similar simple mathematical relationships like $U=RI$ are among many of them – but I would choose the analogy consistent with the Czech language where "napětí" [nuh-pyeh-tyea] means both "voltage" and "tension". I guess that even English speakers must sometimes say "electric tension" instead of "voltage". In the ...


2

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


2

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 ...


2

Is it possible to produce gamma radiaton using radio emitter? Unlikely. A 'radio emitter' consists of, at least, some type of antenna and a transmitter to drive that antenna. The size of the antenna is related to the wavelength of the transmitted radio wave, e.g., half-wave dipole, quarter-wave monopole. But the wavelength of gamma rays is less than ...


2

One form of evidence is the ionization energies of silicon. Nth ionization energy is the energy needed to remove the nth electron. There is a big jump going from the 4th ionization energy (~4000 kJ/mol) to the 5th ionization energy (~16000 kJ/mol). Another form of evidence is the compounds silicon makes. Silicon forms $\mathrm{SiH}_4$, $\mathrm{SiF}_4$, ...


2

For good doping you need two things: (1) get enough dopant in to be useful in changing carrier concentrations, and (2) having an energy level close to a band edge to generate electrons (holes) in the band, rather than making a mid-level recombination center. The below is assuming you are trying to dope Silicon. Data is generally from Sze's excellent ...


2

You are massively overthinking the problem. The collector current is given (by the diagram) to be 150x the base current. The sum of base and collector current has to flow through the emitter... That's all you need to solve this. In particular, a current source will look to a circuit like "whatever resistance" it needs to be in order for the correct current ...


2

As the producer of one of these apps (GammaPix, available for Android and iOS, if you'll forgive the plug), allow me to weigh in here. Yes, smartphone, and other CMOS and CCD cameras, can detect radiation. While cameras are less sensitive then Geiger-Muller counters, specialized solid state detectors, and scintillators, they are sensitive enough for quite a ...


2

A couple of suggestions: (1) the EE stackexchange site a better home for this question (2) simply solve for the voltage across the capacitor and the current through the inductor. Once you have those, the energies stored, as a function of time are just $$W_L(t) = \frac{L}{2}i^2_L$$ and $$W_C(t) = \frac{C}{2}v^2_C$$ Since this is evidently a DC circuit ...


1

The inherent idea is, from that equation $$I = \exp (\frac{eU}{k_B T})$$ if you plot $(\ln I)$ versus $U$, that would be a straight line (of the form $y=mx$), with a slope $$\alpha = \frac{e}{k_B T}$$ That's all you have to do in the experiment, use least square fitting to find an accurate value of $\alpha$ and then find the Boltzmann constant using the ...


1

Typically it is the ferrite cores in inductors/transformers that resonate mechanically, or through magnetostrictive effects that produce a high pitched whine. Switching PSUs are the main culprit. It can also occur when the EM fields interact with steel components in the PSU.


1

What's so hard to believe about this? You have an awful low frequency phase noise of -40dBc/Hz at 1Hz. Lock it to a rubidium standard like the SRS FS725 and that problem should go away. The phase noise of the FS725 is given as <-100dBc/Hz at 1Hz. In other words... there is a VERY good reason why the N5173B has a 10MHz reference input... its internal ...


1

If motor A does the same job as motor B, but with a 10x greater load, and the mechanical advantage (gearing etc) is the same, then I would expect that the torque that A supplies is ten times greater as well. But that is not quite how you phrased the question. It necessarily follows that a higher HP motor can supply greater torque - at least, with the right ...


1

take only first 2 resistors and rest as $x$ now the vertical resistor and your $x$ will be in parallel, effective resistance would be $Rx/R+x$ with series in horizontal resistor. Now equivalent resistance would be $$Req. = R+ (Rx/R+x). $$ take Req. as $x$ again Form quadratic equation and solve for $x$. This will be the answer.


1

Would you post an answer for me to accept? It is stipulated that the 18V voltage delivers 8A which I interpret to mean that the 8A leaves the positive terminal of the source to enter the top node thus, the currents entering the top node sum to $$8A + 13A$$ The currents leaving the top node sum to $$\frac{18V}{R_A} + 3A$$ Setting both sums equal ...



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