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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 ...

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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 ... 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 ... 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 … 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 ... 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 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 It's called a diode because the device has two terminals. Devices that have three terminals are called triodes, and those with five pentodes. Words of that type have fallen by the wayside except in the realm of vacuum tube electronics ... except for the word diode, which has hung on. Note that the Wikipedia article that you cite refers specifically to ... 3 Yes, there is a fundamental reason why electricity is so universal. It is because matter is made of electric charges bound together (protons and electrons). When you think of non-electric technologies such as the wheel, realize that the wheel relies on the rigidity of matter which depends on the bonds between atoms which are electric in nature. So even ... 3 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 ... 2 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. 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 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 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 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 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 ... 1 Radios work with a form of radiation called non-ionizing radiation. This means the EM waves contain enough energy to move the atoms (charges) around but not enough energy to break particles loose. Ionizing radiation removes particles because they carry a lot more energy and can break atomic bonds. These travel as UV-rays, x-rays or gamma-rays. In ... 1 Generally in a ferromagnet each atom has$Z-1$paired electrons, with zero net spin, and one unpaired electron. The unpaired electron spins on neighboring atoms like to be parallel, which is why the material has a bulk magnetization. If you try to inject another electron into the ferromagnetic material, that other electron has to go somewhere — into an ... 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 ... 1 As radio amateurs we've all learned the various relationships of power, voltage, current and resistance as expressed in Ohm's Law Ohm's law is: $$E = IR \tag{1}$$ This doesn't directly say anything about power. There is the related Joule's first law, which relates to electrical power converted to heat in resistive materials: $$P = I^2 R \tag{2}$$ ... 1 As I recall some models of Samsung phones display the same message but many phones don't. My Nexus 5 does not and my LG tablet does not. Modern phones use lithium batteries. It is important that these are not overcharged because this degrades the batteries by forming deposits of metallic lithium. However all modern phones will automatically stop charging as ... 1 I am not an expert in thermodynamics but I think the following is reasonable: When heat flows from A to B (temperature$T_a$and$T_b\$) then you could theoretically do work - efficiency given by the ratio of temperatures. A Peltier is an inefficient heat engine running in reverse (a heat pump), and I thought the efficiency is the ratio between the work ...

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You may be aware that both torque and angular momentum can be represented as a vector - and that such vectors follow the normal rules of vector addition. Thus, if you have equal rotation about both the X and the Y axis, what you really have is rotation about the XY axis; and in general, rotation about an arbitrary axis can be projected onto the X, Y and Z ...

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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 ...

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