# Tag Info

5

They do indeed repel each other. But they are repelled from the point they are coming from even stronger. Imagine having two charged metal balls where one has half the charge of the other. When you connect them with a wire, will charges flow? Yes. Sure, each individual electron feels a strong repulsion from both of the balls, since there already is an ...

4

Electrons do repel each other but they also like to spread out. Quantum mechanics tells us that it costs a lot of energy to localize an electron in a small volume. These two tendencies compete. The quantum mechanical Hubbard model is based on these two effects. It has two parameters: on-site repulsion and transfer energy (transfer Hamiltonian matrix element)....

4

Conventionally, the direction of current is the direction of the movement of positive charges. If the moving charges are electrons, the current direction is opposite to the direction the electrons are moving.

3

First, a DC motor can be made without permanent magnets. The permanent magnets on the rotor can be replaced with electromagnets. Then you have both electromagnets on the stator and the rotor. Second, a railgun's operation is not analogous to a DC motor. It is, however, analogous to an induction motor which also doesn't have magnets, just without the ...

3

If I take cross section close to beggining of the conductor, charges which start moving on one end don't experience as many collisions when they get to that cross section close to the beggining as they will when they come to the other end of a conductor. It seems that resistance should increase from one towards the other end of an conductor. You seem to ...

3

A finite segment of current by itself is inconsistent with Maxwell’s equations. Specifically, it violates the continuity equation. So you will not be able to find such a solution. You will either need to have the current go in a loop or have a changing charge density at either end of the wire. The changing charge density will produce a changing E field which ...

3

The drag is due to repulsion caused by eddy currents induced by the moving magnetic field in the Aluminum metal. The repulsive force opposes the motion of the metal ball according to Faraday's second law of electromagnetism. The same thing will happen if you replace the aluminum with copper metal.

2

Charge build-up is not needed for the explanation of the phenomenon. The easiest way to understand why current "chooses" the path of least resistance is to stop thinking of current "choosing" anything, and think rather about electric field as a cause of all currents. Two wires, connected in parallel to each other, have the same voltages ...

2

This question has already been well answered, and in particular several of the correct answers referred to what you call "charge buildup" in the comments where you say: The "charge buildup" theorem seems to be a point of disagreement among the many answers, while not much information exists on this concept anywhere else. The "...

2

It is not solvable because you allow the current to enter the points $A$ and $C$, but don't have any equation connecting the current $I_{A/C}$ and voltage $U_{AC}$. In other words, you haven't chosen a loop that comes through $E$ (the battery). In Kirchhoff equations, it's important to choose independent loops. But in your case $ABCDA = -ABDA+DBCD$. To solve ...

2

I suppose you are concerned because the leads are shorted by the water, and hence you expect the circuit breaker to engage. Sure, the water contains ions and thus conducts electricity, but even a saturated salin solution is nowhere as conducting as e.g. copper. That means, by Ohm's law, that the current through the water is very limited. In fact, the ...

2

The present statement of the OP's question(s) is (are): Why doesn't current pass through a resistance if there is another path without resistance? How does it know there is resistance on that path? The short answers are: Why: Current DOES flow through a resistance EVEN IF there is a path of lower resistance present, albeit this current may be miniscule ...

2

Kinetic energy of electrons due to electric current $I$ in an inductor is much smaller than magnetic energy $\frac{1}{2}LI^2$ (provided the inductor has large enough $L$, which is usually the case). So yes, strictly speaking total energy stored in the capacitor is transformed into magnetic energy and kinetic energy of current-carrying charges, but the latter ...

2

If we use arbitrary charge/current densities in Jefimenko's equations, then the resulting fields generally do not satisfy Maxwell's equations. The derivation of Jefimenko's equations assumes that the current is conserved. Here's a quick review of the derivation, which I'll refer back to when answering Q2. Half of Maxwell's equations (the metric-independent ...

2

Under the conditions you have given, this is very straightforward to calculate. You measure the energy to switch the light on and call it $E_\text{switch}$. Then you measure the power the light consumes during normal operation, which is energy per unit time, call it $P$. Then just compare $E_\text{switch}$ to $t_\text{off} \cdot P$, with $t_\text{off}$ the ...

2

Why does the change in the speed of a moving electron not change the current? First, let's consider an electron beam in a vacuum, such as in a cathode ray tube. In a cathode ray tube, there is a strong electric field between the cathode and anode. As a result, electrons accelerate as they travel from the cathode to anode. Because of this acceleration, the ...

1

Using a simple model, for an inductor with resistance the phase would be determined as for an ideal inductor in series with a resistor and for a capacitor with resistance you would be analysing an ideal capacitor in parallel with a resistor.

1

Yes, there is a difference. If you made the wire as you mentioned, into a spiral, like this: then there is quite a big difference between this and a straight wire. The difference between a straight wire and a coil or spiral wire is that the spiral wire resists changes in current flow. This is called an inductor or solenoid. It resists changes in the ...

1

It isnt entirely clear to me where you are getting stuck but I will take a guess. You see the circuit as having a battery, a resistor, and 2 wires and you probably have a schematic drawn to this effect. You are confused about how after the electron loses all of its energy in the resistor it is still able to travel down the wire. That would be a problem ...

1

In the small loop, after $S_2$ is closed: iR = - L(di/dt). (Negative because, i, is decreasing with time.) Experience shows that this type of equation has a solution of the form: i =A$e^{αt}$. Find (di/dt) and solve the equation for α. The constant, A, will be the initial current.

1

Yes the strength of the magnetic field produced by a (constant) electrical current only depends on the current. It does not depend on the electrostatic voltage which put those charges into motion in the first place.

1

There was a field across the open switch. When the switch was closed, that field did not instantaneously disappear, but migrated. Zero field in an ideal conductor is true statically, but not dynamically. Transient fields can exist within an ideal conductor.

1

The fact that the electric field in an ideal conductor is zero is not in conflict with the fact that the current in the wires is zero, but rather the opposite: even a negligible electric field in the conductor produces an arbitrary big current. You can interpret your experiment in this way. Initially, there is an excess of electrons on the negative armor of ...

1

In an ideal transformer (with no resistance), that has an open circuit on the secondary, the current in, is limited by the reactance of the primary, and is out of phase with the voltage. There is no power in or out. If current flows in the secondary, that changes the flux through the magnetic core. That shifts the phase of the voltage in the primary ...

1

So if the power absorbed from the voltage source is 0 so how then power can be transformed to the secondary circuit? The power can be transferred to the secondary circuit precisely because it was not absorbed by the primary! Power that is absorbed is power that comes in but does not leave. In order for the power to be transferred to the secondary the power ...

1

The question is a bit misleading because it assumes that there is a « all or nothing » answer. The correct question is: « If you have two resistances in parallel, with values R1 and R2, what relative fraction of the the current will flow through R1 and R2, respectively? » The answer is : I2 / I1 = R1 / R2 So, basically, if R2 is a lot smaller than R1, if ...

1

Basically electrons pass slowly through a resistor so it causes an accumulation of electrons in the resistor which then repel further electrons redirecting them into the other resistance free path.

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