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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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).


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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. ...


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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 ...


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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...


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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, ...


1

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 ...


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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]"


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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 ...


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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 ...


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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.


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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 ...


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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 ...


5

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)


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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) ...


1

One way of doing is parametrizating a ring. For instance, this is a ring: $\gamma(t) = R(\cos t, \sin t)$. Actually it is a circunference of radius $R$. It is charged with linear density $\lambda$. The potential in the point $P(0, 0, z)$ is $$dV = \frac{kdq}{r} = \frac{k\lambda d\gamma}{r}$$ where $r$ is the distance between an element of charge $dq$ and ...


1

Consider the current as the flow of positive charges. Positive chrges flow from higher potential to lower potential naturally if they are free to do so. But you can force them to move in the opposite direction too with the help of an extenal agency. Actually this is the work of the battery, it will move the charges from lower potential to higher potential ...


0

Idealizations are just that, approximately true in many situations. If you connect a wire across a battery, the non-ideal nature of the wire and battery will show-the wire will have some resistance and the battery will not be a perfect voltage source. In most circuits, the resistance of the other circuit elements are much higher than the resistance of the ...


1

What you are missing is called conservation of charge. The fundamental fact is that electrons do not simply 'disappear' or vanish into thin air inside a resistor. This means that, if there is a certain number of electrons flowing into the resistor, these electrons must also flow out on the other side. In particular, in a steady state (i.e. the flow of ...


0

But your height on the hill is Rearth + H, where Rearth is the radius of the Earth and, so would U = mg(Rearth + H) (never mind the integral needed for now)? But wait, the Earth is in the potential well of the Sun so shouldn't U = mg(sun)(Rorbit - H)? Luckily, you don't need to know an absolute potential which would require knowing the distances, masses, ...


3

The definition of current density is $J = \frac{I}{A}$, or more precisely, $J = \frac{\mathrm{d}^2 I}{\mathrm{d}^2 A}$. It is always true, by definition. $J=\sigma E$ is a different equation: it's equivalent to Ohm's law, which you know better as $V = IR$. Ohm's law is not universal; it only works for certain materials, called ohmic materials. For an ohmic ...


1

Keeping it simple: You can think $J = I/A$ as one definition of the $J$. Since the current $I$ is related with the eletric field, then the $J$ must depend from the electric field as well. As a first approximation we guess a linear relationship between $J$ and $E$. Naturally it is not a linear relationship, but works well as approximation in some cases ...


2

The electric field strength can be represented by the equation $$E=F/q$$ where $E$ is the electric field strength in N/C, $F$ is the force in Newtons, and $q$ is the charge in Coulombs. If we take the equation that you provided: $$\Delta p=qEt$$, and substitute $F/q$ in for $E$, then $$\Delta p=q\frac Fq t$$ which simplifies to $$\Delta p=Ft$$ This looks ...


1

The equation simply says that the variation of momentum is equal to the force impulse. For a constant electric field $E$, we have that the force is $F=qE$. Then, we have $$\Delta p=m\Delta v=mat=Ft=qEt$$


1

Your basic intuition that anything that uses power will heat the room and reduce the consumption of the space heater is a good one. To first order it is correct. One can quibble that the heat is released in a different place, so may not heat the thermostat as effectively, causing a (slightly) higher room temperature and more total consumption, but that is ...


0

Incorrect. The heat produced by lighting device is very low (also depend on device type: LEDs, bulbs, etc) because it is made for produce light and heat is a sign of inefficiency. For example, the most inefficient lighting device is light bulb with 75% efficiency. It means that only 75% of the power converted into visible light, and another 25% converted ...



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