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One should really be careful when mixing circuit theory and electromagnetic theory... Fact of the matter is, circuit theory is a much simplified version of electromagnetism. In circuit theory, we speak of the "voltage" of a point, in reference to the ground, and act like it is a well-defined quantity. In fact, what is called the electrical potential is ...


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The answers are really good, but I have (as a former long-time student who has not forgotten the practicality of things) a more practical approach without considering currents or voltages, just the topology of the circuit itself: When trying to reduce a circuit, the algorithm is as follows: Any number (> 1) of resistors on the same wire. That is, no ...


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Hyper physics website says this Electric current flow is proportional to voltage difference according to Ohm's law, and both the bird's feet are at the same voltage. Since currentflow is necessary for electric shock, the bird is quite safe unless it simultaneously touches another wire with a different voltage. ...


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In the book i have says that the non-electrostatic force of the source moves charge form the negativive terminal to the posotive terminal. This is true for an open circuit or a close one?


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I guess the question is whether the LED will also act as a photodiode, i.e. whether incident light will excite electrons and therefore generate a current. If so, then reflect the diodes light back onto it will indeed reduce the current flowing in the diode by generating an opposing EMF. This turns out to be surprisingly hard to Google, but Wikipedia ...


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The live found a route to earth, discharging it's energy on whatever load was connected between live and earth, and then luckily the RCD cut off the supply before you cooked yourself.


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Pneumatic circuit components are close to what you are looking for. You can buy pneumatic components that allow you to do a wide variety of operations online from places like McMaster-Carr, or directly from the manufacturer, e.g., from Clippard. I'm not so sure there are too many off-the-shelf components that are direct analogues to a particular electrical ...


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You seem to be asking about mode transformers; these are extensively used in antennas as they connect to waveguide feeds. The waveguides usually employ TE10 (rectangular) or TE11 (circular) modes, but if you want to feed a horn, say, then you have to shape the field properly to avoid reflection, reduce sidelobes, and reduce cross-polarization coupling, etc. ...


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The answer is "maybe" and you will not know which one to adjust, if any, without a circuit diagram. Just record their positions so you can undo any changes you make, and experiment. As you said - it's cheap.


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The norm CEI 287 will give the method of assessment of the resistance of the conductor including the parameters you mentionned. Basically, the method that is used (to my knowledge) is the ohmic losses formula with a corrected resistance due to skin and proximity effect. About the skin effect, you can assume that the current is only flowing in an external ...


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The diagram is a network of wheatstone bridge, but it should satisfy the condition after which it is easy to calculate the net resistance between the points The R2 resistance will be in the middle and will be neglected if the condition of wheatstone bridge is satisfied, i.e. R1.R4=R5.R3


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It depends. The time evolution stays the same: $V(t)=V_0 e^{-5}e^{-t/\tau}$, the "non stabilized" part of the total voltage is just smaller. Often this part is considered constant for practical purposes.


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After $5 \tau$, the voltage across the capacitor is about $.7$% of what it was originally. There are definitely situations where this $.7$% could be significant, so when the problem says you can consider this voltage to be zero, it probably means that the accuracy of any tool you would use to measure the system is low enough that it wouldn't be able to ...


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1) If the n and p doped regions are externally connected using a perfectly conducting wire, why will not any current flow? In thermal equilibrium no current can flow if one connects the two sides of the junction using a perfectly conducting wire. The built-in potential existing at the junction will remain the same, drift and diffusion currents will ...


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I have never heard about an energy field. However, I see two things. First, what causes the energy transfer : the two terminals of the battery have different potential. This means that the electrons will try to go from the - to the +. If they are allowed to do so (if there is a wire), the difference of potential tends to decrases, and the chemical reaction ...


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The "whole" electricity will not travel through the whole network. First of all, on a big network, the common physical quantity if the frequency. The voltage is also shared, but may fluctuate locally (geographic viewpoint) depending on reactive power consumption/production. However, even if the voltage is overall constant, you can draw the flux of energy ...


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Okay, in the first place, this is not really a trivial topic, it's an entire subtopic of electrical engineering. But in this case it should suffice to say, that in ordinary conductors, you have resistive losses, they're not ideal. You can find that the entire resistance of a grid element is proportional to the inverse of its cross-section and to its length. ...


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Physicist137 took a crack at showing why this follows from the structure of the electric field in the absence of time-dependent magnetic fields. That doesn't hold once you have induction in the system, however. So let's look at a simpler approach. Anything that you want to treat as a potential (whether it is a real potential or not) has to have a simple ...


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I believe that the energy in a circuit flows OUTSIDE the circuit as electromagnetic fields surrounding the wires. Better people than I have produced some pretty diagrams to help visualize this. See: "In a Simple Circuit, Where does the Energy Flow?" "Energy Transfer in Electrical Circuits" (PDF) "Understanding Electricity and Circuits" (PDF)


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The rotational of the electric field: $$ \nabla\times\mathbf E = -\frac{\partial\mathbf B}{\partial t} $$ Using Stokes' Theorem on this equation, we get the integral form of this equation: $$ \varepsilon = \oint_{\gamma}\mathbf E\cdot\mathbf{dl} = -\frac{d}{dt}\iint_S\mathbf B\cdot\mathbf{dS} = -\frac{d\Phi}{dt} $$ Which means, the electric field in a ...


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The battery generates an electric field across the circuit, you can visualize a field as force lines. This is because the two ends of the battery are at differnt energies. If you have electrons that can move freely in response to an electric field (the electric fiels makes force on the electrons), such as in a conductor, then the electrons will move around.


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I think you understand that the series has more resistance than the parallel, so more current should flow in the parallel case. Since there is more current in the parallel case, the battery has to supply more current so it is more stressed, and it gives out a lower voltage. So its $V$ (notice that $V$ is really just measuring the voltage drop across the ...


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As you said, voltage is constant across each 'branch' then, remember that the current $I$ is: $I=V/R$ so the current did not change in that branch. The total current will decrease though, because the current through the other branch did diminish (same equation, but $R $ increases).


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There is a 90 degree phase difference between the input current and the induced voltage in the ring because of Lenz's law. Then, there is a 90 degree phase difference between this induced voltage and the induced current because of the self-inductance of the ring. This amounts to a 180 degree phase difference between the input current and the induced current ...



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