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## Hot answers tagged electricity

5

Definitely! Consider Ohm's law, $\vec j=\sigma\vec{E}$, where $\vec j$ is the current density, $\sigma$ is conductivity (the inverse of resistivity) and $\vec E$ is the electric field. Anisotropic conductivity corresponds to turning $\sigma$ into a tensor-valued quantity, a $3\times3$ matrix. Ohm's law is then given by $$j_i=\sigma_{ij}E_j,$$ where ...

4

In principle, sure. That's what microphones are, as ACuriousMind points out. But if you want to power anything substantial, an important issue to overcome is the relatively small amount of energy contained in sound waves. According to this website, the front rows of a rock concert have a sound intensity of $10^{-1}~\text{W m}^{-2}$. So even if you had a ...

4

The radio waves or microwaves that are used for communication don't contain just one photon. They contain a bunch. (Maybe someone will do the math for how many photons a standard radio broadcast antenna is producing each second; it'll blow your knee-high off even if you're wearing sandals over them.) Consider for example a frequency-modulated signal. The ...

4

The analogy of electricity to flowing water may come in handy here. In this analogy, a potential difference is like a difference in height. One lake on top of a mountain and another in a valley, for example, might represent the two terminals of the battery, which are at different potentials. If you think about that situation, it's clear that no water flows ...

4

I need to know why that is the answer. (Jump to the summary if you want to skip the details). As you may know, a transformer is essentially two inductors that share most if not all of the magnetic flux linkage. Recall the formula for the voltage across an isolated inductor $$v_L(t) = L \frac{di_L}{dt}$$ Now consider two inductors, $L_1$ and $L_2$ ...

4

This table should answer your question which is about intermolecular interactions. The largest distance that the electromagnetic forces can be effective for neutral atoms and molecules is in fractions of a nanometer. When in classical distances then the electric field coming from neutral atoms is effectively zero. Nano structures though, if you make your ...

4

To a degree this is matter of terminology. Resistance is a scalar quantity, but it is derived from the resistivity, which is a second rank tensor and is anisotropic in many materials. For an isotropic material we have the usual formula: $$R = \rho\frac{\ell}{A}$$ and the usual: $$V = IR$$ where $\rho$ is the (isotropic) resistivity, $\ell$ is the ...

3

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there ...

3

The individual resistances are all positive, so the sum $$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances. No mucking around with the two-resistor form required.

3

It's not a dumb question. The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) ...

3

It is quite easy to use a step-up converter to generate almost any voltage. The question is "voltage at what current". The power a battery can deliver is finite (power = voltage times current), but you can convert voltages in many different ways. The most obvious is an oscillator (inverter) followed by a transformer and a rectifier, but there are more ...

3

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.

3

The electric potential is always a function of both spatial position and time, in both circuit theory and electrodynamics. $V$ is constant along a wire, so in circuit analysis you can take a short cut by specifying the position by just specifying which wire you're talking about, like $V_{1}(t)$, instead of specifying three spatial coordinates. $V$ doesn't ...

2

Another useful analogy, apart from the gravity one described by David Z, is temperature. You can think temperature as your potential, and the heat flow as your current. Two points of space may be at different temperature, but if they are correctly insulated, they won't exchange heat. The heat will flow only if they are connected somehow. For the current is ...

2

The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue). The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so ...

2

We can prove it by induction. Let $$\frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n}$$ Now, when $n=2$, we find $$\frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}}$$ Since $\frac{R_1}{R_2} > 1$, we see that $R^{(2)}_{eq} ... 2 The battery doesn't do any significant work if there's nothing connected to it. You can think of a battery with nothing connected to it as being a battery with an extremely large resistor connected to it. That's even essentially physically accurate, instead of just being an idealization, because although the resistivity of air is extremely high, it's not ... 2 Short answer: even in non-uniform fields, the speed won't change, but the guiding center can drift with some velocity. In a magnetic field (uniform or otherwise), the force on a charged particle is: $$F = q\; \vec{v} \times \vec{B(x)}$$ The direction of this force will be tangential to the velocity vector because of the cross product (the result of the ... 1 It comes from the fact that the voltage in any loop is required to sum to zero. Here is a description of the rule (Kirchhoff's Voltage Law). Your instructor may have told you to follow an electron for the current law, but it is not ideal if the current flows in the opposite direction on one leg of the loop. Determining the voltages on each leg is needed, so ... 1 I have to cite @CarlWitthoft from the comments. Intensity is not a term used in electric circuits. Intensity of current would not make any sense. However, I can vaguely make out what you are trying to ask: What you are trying to say is: "Two circuits are identical to the one in the figure, except for the fact that one has a nichrome wire and the other has a ... 1 In the quantum mechanical description of a conductor all energy levels of the conductor are filled up to some specific energy level, called the Fermi level. This is because of the Pauli exclusion principle, which says that electrons with the same spin cannot occupy the same energy level and thus causes higher energy levels to be populated. Therefore, even at ... 1 The key think you need to know is that if you apply a voltage$V$and a current$I$flows then the power dissipated is: $$P = VI \tag{1}$$ You know the power at 220V so you can calculate the current, and if we assume the kettle behaves as a simple resistor then it will obey Ohm's law so you can calculate the resistance. $$R = \frac{V}{I} \tag{2}$$ ... 1 The answer-as requested by HDE. So my foolish misconstruct of the electric field is that since it accelerates charged particles more or less depending on the distance from the source, it should create less motion in electrons a further distance from the source. I'm not sure why I thought resistance had much to do with it-on further analysis resistance ... 1 Dielectric heating is the priciple of a microwave oven. Water$H_2O$has a strong dipole moment. Since the water molecule is not linear and the oxygen atom has a higher electronegativity than hydrogen atoms, the oxygen atom carries a slight negative charge, whereas the hydrogen atoms are slightly positive. As a result, water is a polar molecule with an ... 1 In a perfectly ideal circuit without any resistive losses (even internal to the battery), energy will flow to fill capacitance with charge, and be stored as an electrical field but the transient flow will also create magnetic fields in any inductance. Without loss, the energy will bounce back and forth between the inductance and capacitance as magnetic and ... 1 yes, you will have to pay more if your load is inductive. most of energy meters work with the voltage and current to calculate energy. When we have inductive load it takes more current than resistive load to produce same power or output. P=V.I.Cosx where x is the phase angle between voltage(V) and current(I). if the phase angle x goes higher, the power ... 1 I seriously doubt that the batteries were putting out 30 kV. You surely misread or misheard something. The chemistry of batteries is such that individual cells produce from a few 100 mV to a few volts. A 30 kV battery would require 1000s of cells, which would make no sense at all. In addition, 30 kV is much more difficult to handle and would be much less ... 1 The two resistors are in parallel. This means that at$A$the current splits between them relating to their reistance. So the current throw the top resistor is$3\,A$and throw the bottom resistor is$1 \,A$. If we use Kirchhoff's current law which states, that in any node (like$A\$) the current flowing into the node is equal to the current flowing out of ...

1

This is a parallel circuit, not a series one. So the currents not need be the same, but the potencial difference is the same instaed. In a series circuit the current is the same but the potencial is different in the many elements of the circuit. You can think about this like the current is a flow of water, since is parallel, the current (the flow) "divides" ...

1

The transformer operation is based on Faraday's Law $$\nabla\times\mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t}$$ This law relates the generated electric field (resulting in electromotive force in a circuit) with the variation of the magnetic flux density. Also indicates that a changing electric field generates a density also varying magnetic ...

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