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6

Your original text admitted three interpretations, and I'm leaving the answers here: 1: What happens with a toy model when there's a circuit with an ideal battery and no resistance? All the charge moves around the circuit at one moment in time (infinite current). The energy must leave the system as Electromagnetic radiation - accelerating charges radiate, ...


6

Transformers generate oscillating magnetic fields at the mains frequency and the fields produce an oscillating force on: anything nearby that's ferromagnetic (like the core) anything nearby that is carrying a current (like the windings) The sound you hear is because various bits of the transformers are moving in response to the oscillating fields and ...


4

You are familiar with the concept of static electricity. When you have a DC line with many thousands of volts on it, there will be a polarizing effect, and that will attract things to it - especially dielectrics. Now if your piece of paper is thus attracted to the line, and you try to grab it, it is quite possible that you get peripheral nerve stimulation ...


4

You are being imprecise about electricity. It's probably better to just think of electricity as current. You have a current whenever you have a charge moving. To your question, yes, positrons are just as good as electrons for carrying a charge. There is no difference between "positron electricity" and "electron electricity". Another way to see this is ...


2

The first mistake is directly taking $dq=\lambda dy$. If you look closely, as you go up from the origin to higher y, you find that charge density per unit y increases. Take instead $\lambda$ as charge per unit circumference of the wire. i.e $\lambda={dq \over dl}$ = $\lambda={dq \over rd\theta}$ Now, $$dE= {k\lambda dl \over r^2} \cos{\theta}$$ ...


2

In addition to the good answers above here is something to think about. If the resistance is zero then for a current to flow there does not need to be a battery - the emf can be zero. No work is done as the current flows. This may sound like a strange case, but it is how many strong magnets work in nmr (nuclear magnetic resonance) spectrometers. The have a ...


2

The line itself does not change much over years. What changes and therefore needs maintenance on power transmission lines is insulators, connectors and spacers. Insulators get dirty or simply break, connectors work loose due to thermal expansion and contraction, mechanical stresses and oxidation, and spacers can be damaged by wear due to these same ...


2

I think EnergyNumbers's Answer is an excellent one, but could leave some people a bit mystified by what exactly a "direct ray" is and what exactly its relevance is. The essential idea here is that a Fresnel lens is an imaging machine: it puts a curvature on a low aberration wavefront so that that wavefront converges. Its working depends on there being ...


2

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 ...


1

In 1999, the president of the IEEE Power Engineering Society, Robert Dent, noted that: "The degree or intensity of the corona discharge and the resulting audible noise are affected by the condition of the air--that is, by humidity, air density, wind and water in the form of rain, drizzle and fog. Water increases the conductivity of the air and so ...


1

But what about the 180° gap in the middle? This implies a negative real part of the impedance, i.e., a negative resistance. Express the impedance $Z$ in polar form $$Z = R + jX = |Z|e^{j\phi}$$ where $$|Z| = \sqrt{R^2 + X^2}$$ and $$\tan \phi = \frac{X}{R}$$ For $R > 0$ $$-90^{\circ} \lt \phi \lt 90^{\circ}$$ but for $R < 0$ $$90^{\circ} ...


1

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 ...


1

Some corrosion always takes place (pure gold is not used for transmission lines, AFAIK:-) ), so the conductivity decreases with time, although for some materials this effect can be very small ...


1

I think the range of answers you have got from friend, teacher and web reflect that there is not a straightforward response. Without high voltage it would not be possible to drive the dangerous current through the body, but high voltage itself is not lethal - it depends how much current can be delivered at high voltage. Another question is how high does the ...


1

Cores of the transformers a made of ferromagnetic material, which can change the shape due to magnetostriction phenomena. When transformer works with typical grid frequency of 50Hz..60Hz, it can be heard. As it was answered above, generated noise depends on mechanical construction. New power supplies use frequency converters, which drive the transformers ...


1

But recall that power dissipated $P= VI$ is also , from Ohm's law, expressible as $P = I^2 R$ So the dependency of power dissipated is linear in voltage, but quadratic in current, given the same resistance. Also remember that the voltage supplied from the power station, and the voltage drop across the transmission line - which is what is important in ...


1

The pendulum motion can be converted to energy any number of ways. For example, if the pendulum bob is a magnet simply placing a coil of wire near it as it swings will induce a current which can be siphoned off and used.


1

You've begun with this: $P_l = I^2R$ $P_l = IV$ This is correct, but the $V$ here is not the line voltage, but instead the voltage drop across the resistor under consideration. Increasing the line voltage does not increase the voltage drop. Your diagram with a single resistance and a power station implies that the current in the line depends on that ...


1

The speed of an individual electron cannot be c: an electron is a massive particle and can therefore never achieve the speed of light Furthermore, an individual electron moves very slowly in a current (mm/h scale). The question i think you're trying to ask is how fast do electric signals travel through a wire. (e.g. if I had two lightyears of copper wire ...


1

The exact equations for I-V characteristics of transistors are derived using quantum-mechanics. Several approximations can be used, one of which is based on the shottky barrier analysis This reference here derives the I-V linear and quadratic approximation (in saturation) for FET transistors. Another reference here UPDATE: As @QMechanic pointed, ...



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