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2

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 ...


1

You should be able to get the same result by using the electrical work formula - but note that you need to integrate since the force changes with position. That's really all the potential is - it is the integral of force for unit charge. That's why force has the $1/r^2$ relationship while potential has $1/r$ (with appropriate signs and constants...). ...


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Do not open the cone. Think of it in the profile view : You have an isosceles triangle. Now move along the axis of the cone, say a distance $x$ and take an element $\mathrm{d}x$. Somewhat like this : This small element is similar to the rectangle you described. With length as $2\pi r(x)$ and width $\mathrm{d}x$. You also know the velocity with which the ...


0

Others have provided mathematical explanations, so I'll try an intuitive explanation. You've probably seen the electric field represented like one of these: (image source) At the top of the mountain there is a positive charge, and at the bottom of the valley is a negative charge. A positive test charge in this field experiences a force analogous to a ...


2

As you said, current is like water flow similarly voltage is like water level and voltage difference like difference of water level. We know that water flow from higher level to lower level like current flow from higher voltage to lower voltage. Voltage difference means there is a difference of charge i.e. difference in number of electrons. Now we consider a ...


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Suppose your pipes form a loop i.e. water can flow through the pipes and get back to where it started. As the water flows round the loop there will be some places where pressure rises (e.g. a pump = battery) and some places where pressure falls (e.g. a restriction = resistor). However if the water goes all the way round the loop back to its starting point ...


1

The Kirchoff Voltage law states that the sum of emfs in a circuit is equal to the total potential drop in the circuit. So for a simple example, where you a 6V cell, for example, and 2 resistors in series. The 6V cell can be seen as a place where the water is given potential energy - if we imagine a ramp, it would be the water being pumped up to the top of ...


1

AC can be made from DC, and vice versa. DC is a steady voltage, with AC the voltage fluctuates from negative to positive and back, many times per second. Most electricity generators generate AC. The reason for preferring AC is that, historically, it is easier change voltage with AC: all you need is a transformer. The generators generate a high voltage, ...


0

ahh, i made this very same question when learning about electricity in high-school If resistance is the slowing down of electrons and the flow of electrons is the current, why does the current stay the same when passed through a resistor? The answer (the physics teacher was good) was this: Indeed a resistor (in a sense) alters the passage of electrons ...


1

What you could say is "if energy is lost in a resistor, then why doesn't the velocity of the charged particles increase, as per Work-Energy theorem?". So you're question should really go something like "Why doesnt current which is $Q/t$ increase if the velocity will increase after voltage drops"? The answer to this is that we assume all potential energy ...


0

What the resistor indeed does is to "slow down" the current. The more resistors you put the smaller the current (the current drops). If there were no resistor you would not have any voltage drop and assuming the wire does not have resistance, you will have a short circuit (maximum current flow). The circuit equations for a resistor already take into account ...


0

The original poster explained unambiguously that the question is about spatial dependence, not temporal, i.e. a partial case of the Kirchhoff’s current law was interested about. Tiny electrons are not things providing high-level integrity of electric circuits; they are only charge carriers. It is a good metaphor: when you send a mail, you do not care about ...


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Part a.i: It would be a clearer question if it specified the orientation of the solar panel, but since it states the battery is being charged, it makes sense to assume the solar panel is oriented in such a way that the current is flowing into the positive terminal of the battery (the opposite way from the way it would be flowing if the battery were ...


2

Your reasoning is correct, it's just a lot harder to do with the surfaces you've chosen. Draw a small, elemental ring at some arbitrary height above the charge. A line from any point on the ring to the charge subtends the polar angle $\theta$ with the z-axis and is a radial distance $r$ from it. (i.e. $r$ and $\theta$ are the usual spherical polar ...


2

You probably did some wrong calculation, because your reasoning works. Take a circle of radius $r$ a distance $a$ above the charge. The increase $d\Phi$ of the flux by increasing the radius by $dr$ is given by $$d\Phi=\frac{q}{4\pi\epsilon_0}\frac{2\pi r \cos \theta dr}{a^2+r^2}=\frac{q}{2\epsilon_0}\frac{ar dr}{(a^2+r^2)^{3/2}},$$ where $\theta$ is the ...


1

The Gauss Law indicates that the field lines $\vec{E}$ should be normal to the Gaussian Surface taken $dA$. Thus we take the dot-product to take the normal component of field $\vec{E}$ with the area. $\vec{E}\cdot \hat{n}dA = EdAcos\theta$ The reason to take Gaussian surface as a sphere, with the point charge being its center, is because 1) the field ...


0

So you might want to read this wiki article I like to think of a piezo as a voltage source (pressure dependent), with a series capacitance, (and then some leakage resistance.) When you squeeze it you generate some voltage on the cap, but it leaks off. When you release the pressure then you get opposite sign of voltage for a time. (That assumes you are ...


1

The short answer is no, the diameter of the wire doesn't affect the bandwidth. Bandwidth can be a tricky subject. If you are talking about injecting a very high frequency sine wave at one of the wire, and seeing if it is detectable at the other end, then wires of all diameters have a surprisingly high bandwidth. But if you are talking about the ability of ...


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Bandwidth is a function of the frequency of the signal carrier. As the diameter of the copper increases you start to lose "conductivity" because of skin effect as the current tends to be carried in thin surface areas as the frequency increases. This does not directly impact bandwidth but it does hinder its application and forces the use of thin wires for ...


1

Electricity is distributed at high voltage to minimise losses through heating. The amount of power lost through heating is $P_{loss} = P_{total}^2 R / V^2$. One can use a transformer to change a high current, low voltage signal to a low current, high voltage signal. At the receiving station, this is converted back to lower voltage for distribution by ...


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The answer to your question is, unfortunately, more complicated than you would like. The good news is, that you can start with Maxwell's equations and you can set everything with a B or an H to zero, because magnetic fields do not matter in this case. This allows us to treat the entire problem as a potential problem with one scalar potential, where ...


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Grin, Let me start with an old physics joke. So I have no idea, but I might start by treating the brain as a spherical ball filled with salty water, and then ask how the currents will flow, and that would give me a first guess. (Maybe not a very good one.) I just measured the conductivity of water, see my answer here. Do you really push 4mA through the ...


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If you want to transfer energy using the electromagnetic interaction, you really have three choices: Use a directed stream of photons: microwave, focused light, maybe laser. This is a "far field" solution, you need to efficiently couple energy into the electromagnetic field. You either need properly designed antennas, or efficient light sources, and good ...


0

Capacitative coupling. IR LED, PV cell. All varieties of RF. Ultrasonic, piezo receiver (tuned). Ion wind. Ordinary wind, microturbine. Vibration (see piezo).


1

Arc flash is a separate issue from electrocution. It has only been integrated fully into the electrical codes over the last decade, and is not well recognized by folks not up to date on the latest requirements. Arc flash is now covered well in the various electrical code documents. A rapidly expanding plasma generated by high voltages and currents rapidly ...


1

Is the Ohm's law violated here? No, the Ohm's law is not violated since Ohm's law relates the voltage across ('between the ends') of a conductor to the current through the conductor. But the voltage on a transmission line (the voltage referenced to ground) is not the voltage across the transmission line (the voltage difference between the ends of the ...


1

The current required to carry a given power decrease when you increase the voltage because the power is the product of the current with the voltage (and power factor).


0

If you are talking about the open circuit voltage of the battery, then internal resistance and resistance of any conductors connected to the battery are irrelevant. The voltage across a resistor is proportional to the current thru it, see Ohm's law. When there is no current, there is no voltage drop across a resistor, so both ends are at the same voltage. ...


1

The "explosion" is actually a physiological response of the body of the person electrocuted. Our muscles are basically commanded by electrical signals through the nerves to expand and contract. When large currents flow in the body from an electric shock (in the form of ion currents), these flows "hijack" the muscles, and the latter "thinks" they are being ...


0

Almost all electrical machines can be run in both ways (generator or motor). If you're talking about the direction of rotation, it will work too. However, your message is so vague, we can't help you : we don't have any clue about the kind of machine, about the frequency, voltage, use...


2

CuriousOne's comments basically answered your question. I will add that if enough current is allowed to suddenly flow it can vaporize materials very rapidly, including metal. This sudden vaporization can create a rapid expansion and if that expansion is restricted by something then it can explode in the same fashion as a bomb. You often hear of ...


1

Yes, electric current is movement of any kind of charges. The problem with your particular example is that most liquids containing ions are also conductive. Electrons will hop between molecules and equalize the ionic charges, then end up providing most of the conduction themselves. There are cases where actual ion migration results in much of the current, ...


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Well, technically yes (assuming you mean ionized and not just "charged"). Current (in simple terms) is only the time rate of charge flow, which is not exclusively limited to electrons or any specific charge carrier. Electrolytic conductivity is well documented, naturally being higher for strong electrolytes as compared to the ones that dissociate weakly in ...


1

Ions can indeed carry current (ex. electrolysis). "An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in a plasma.[1]"


0

It is better to think of Ohm's Law as I=V/R. What it is telling you is that if you apply a voltage (V) to a resistive material (characterised by R), then that voltage is capable of driving a current I. The material could be anything, a piece of copper, or the plasma in a star. The voltage is constantly supplying energy to the electrons in the material, but ...


2

how can this be true? If we connect 1.5V cell to a 10 ohm resistor, the current is, by Ohm's law, 0.15A and the power delivered to the resistor by the cell is 0.225W. Now, connect a 9V battery in place of the 1.5V cell. The current is now, by Ohm's law, 0.9A and the power delivered to the resistor is now 8.1W. The 9V battery must deliver far more ...


1

I think you've answered that yourself. If you are putting more work into moving unit of charge, then that unit of charge is going to move faster (all else being constant). Current is the flow electric charge across a surface at specific rate (1 ampere = 1 coulomb per second) and hence - more voltage, more work, faster flow (rate), higher current.


0

We know that the tension between a perfect current source is 0 That's a common misconception held by those first introduced to current sources. In fact, an ideal current source produces a fixed current regardless of the voltage across. Put another way, the voltage across an ideal current source is determined by the attached circuit which is, in this ...


0

I have used a similar two-leads test light in a class-room demo/lab to show the electric current produced by waving plate sized pieces of statically charged styrofoam close to metal pie plates connected by wires. These lamps light up with very small currents. For a single lead tester, the tester's body acts as a one "plate" of a capacitor on the end of the ...


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It depends on the internal resistance of the source. Fist consider a "voltage supply". What does "voltage supply" even mean? A voltage supply is supposed to output a fixed voltage no matter what we connect it to. Is this even possible? Suppose we connect the two terminals of the voltage supply together through a piece of wire, i.e. a really low resistance ...


1

You are right, as is Eoin's answer. I'm only answering to show one way to think about it that is useful to me when people bring up this common misconception. Imagine you have a pair of terminals with some voltage between them (like a common 1.5V battery). Nothing's connecting those two terminals together. Except for air, that is. Is there any current? No, ...


3

It depends on if your power supply is constant voltage or constant current. Usually, it's the former, so it means that the $P = V^2/R$ is the more appropriate one to use. Therefore, a smaller resistor will dissipate more power in this situation. In some situations (electromagnets is one that comes to mind), the load is driven with constant current, so ...


0

Though I don't, by any means, have a collegiate degree in Physics or Electromagnetism, I'd like to attempt to give an answer. First, we need to understand what electricity is. Fundamentally, electricity is the flow of particles with any electric charge. There are two common matter particles with a charge: protons and electrons. Since the proton's mass is ...


4

know that the two conversion coefficients are close but I simply cannot see why the RMS value is the one that conforms to reality. It's true that we can ask for the average of the magnitude of a sinusoid over time and calculate it. This is a useful quantity if, for example, we want the time average voltage at the output of a full wave rectifier ...


2

Waffle's answer shows you exactly why the RMS isn't the average: Here, the average value of $V(t)$ clearly isn't $2V_p/\pi$ that you've obtained, it's zero. The half-cycle business doesn't make much sense given that if you chose $\pi/2$ to $\pi$, instead of 0 to $\pi/2$, you'd have a negative average, so which would be the true "average" $+2V_p/\pi$ or ...


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The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage! The square of the average absolute voltage will not yield the power expended in the resistor. Edit: And actually this close-to-duplicate question does have more extensive answers ...


2

The definition of RMS is Root Mean Square and means the square root of the avarage of the square of some quantity, that is $a_{RMS}=\sqrt{\langle a^2 \rangle}$. So for the case of a sinusoidal current, we would get $$V_{RMS}^2=\frac{1}{T}\int_0^T\left(V_p\sin\frac{2\pi t}{T}\right)^2dt=\frac{V_p^2}{2}\implies\\ V_{RMS}=\frac{V_p}{\sqrt2}$$


0

I think that the problem here is that you haven't properly set up the circuit equations. So if we have a circuit with a capacitor and a bulb connected in series in an AC circuit Since the bulb and capacitor are series connected, the current through each is identical. Denote the series current phasor as $I_s$. Assuming the source is an AC voltage ...


0

Take a infintesimally small element 'dl' ,now the potential due to this element is dv=1/4piεo λdl/r^2 and r =sqrt(z^2 +R^2) ,(which is const) And thus integrate 'dv' And since 'r' and all others in the equation are const except 'l' thus integrating 'dl' will give 2piR (problem solved)


0

The ring can be parameterized by ${\bf r}' = R \left(\cos \theta \ \hat{\bf x} + \sin \theta \ \hat{\bf y}\right)$. Noting that $$ \left|d{\bf r}'\right| = R d\theta \left|\left(-\sin \theta \ \hat{\bf x} + \cos \theta \ \hat{\bf y}\right)\right| = R d\theta $$ and $$ {\bf r} - {\bf r'} = \left(x-R\cos\theta\right) \ \hat{\bf x} + \left(y-R\sin\theta\right) ...



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