# Tag Info

## New answers tagged current

0

Adding extra electrons isn't going to make any difference, at least as a first approximation. Suppose we have a wire with some applied voltage $V$ and some current flowing $I$. What this means is that the power supply is injecting $I/e$ electrons per second into one end of the wire and extracting $I/e$ electrons from the other end of the wire. The ...

0

Let's assume that the power company is supplying a neighborhood with 1000 A of current at 120 V. Since P = IV, the neighborhood is receiving 120 kW of power, which is the "load" seen by the power company. To maximize efficiency, the power company wants to minimize the losses involved with transmitting power to the neighborhood, which occur due to ...

1

A current-current diagram is a diagram in the diagrammatic expansion of a current-current amplitude. For the latter, see. e.g., equation (1.1) in Keith Hamilton, The Standard Model Part II: Charged Current weak interactions I, http://www.hep.ucl.ac.uk/~campanel/Post_Grads/2013-2014/SM-CC-Weak-Interactions-I.pdf

0

Well, I guess you could "derive" it in the Drude model (see my post here), where the proportionality of current density $\vec j$ and electric field $\vec E$ is the conductivity $\sigma$ or inverse resistivity $\rho$: $~~~~~~\vec j = \cfrac{q^2}{m} \tau n \vec E = \sigma \vec E = \cfrac{1}{\rho} \vec E$ Using the current density $|\vec j| = I / A$ as ...

2

The short answer This is not 100% true since it assumes DC transmission, but it gives the simplest form of the idea: even if the transmission lines are themselves at high voltages, that doesn't directly mean anything, since voltages are not defined relative to anything special (they're defined relative to some other line which is in parallel with your ...

2

There are two different $V$'s here. Suppose the power station outputs at 10,000 V. By the time the wire makes it to your house, this may have dropped to, say, 9,000 V. The $V$ in the first equation refers to the voltage difference you can use, which is 9,000 V (between the wire you receive and ground). The $V$ in the second equation refers to how much ...

0

Voltage is a measure of the electric potential difference across two points in a circuit. It may be considered the work done to transport an electric charge. Power lines are made of thick easily conductive material in order to minimize resistance and power loss to heat. But resistance within power lines is fixed, and power is delivered through the line ...

0

Having a magnetic field and moving a conductor non parallel to this field it will be induced a flow of electrons perpendicular to the plane from both the magnetic field vector and the direction of the displacement of the conductor. The induction of electric current takes place because electrons have a magnetic dipole moment and an intrinsic spin. For all ...

0

The electric fields and magnetic fields considered up to now have been produced by stationary charges and moving charges (currents), respectively. Imposing an electric field on a conductor gives rise to a current which in turn generates a magnetic field. One could then inquire whether or not an electric field could be produced by a magnetic field. In 1831, ...

1

The formula is $$R = \rho \frac{l}{A},$$ where $R$ is the resistance, $l$ the length of the medium current is flowing in and $A$ its cross-sectional area. $\rho$ is the resistivity, a property of the material. An intuitive way of understanding the dependence on $l$ and $A$ is the following. The longer the wire (increase $l$), the more collisions electrons ...

0

For slowly moving charges, magnetism is just a relativistic correction, so the relative size of its effect is $O(v^2/c^2)$. Since $v$ is very small for charges in a wire (less than 1 cm/s), the effect will be insignificant. Since parallel currents attract, the current will be attracted to the center of the wire a tiny, tiny bit.

2

In my opinion, the mathematical equation we call Ohm's Law is best taken not as a “law”, a fact about the universe, but as the definition of resistance. $$R \overset{\mathrm{def}}{=} \frac{V}{I}$$ Given this definition of the quantity $R$, we can then make (as other answers have mentioned) the empirical observation that many materials have approximately ...

1

Ohm's Law actually follows the definition of power, current and voltage. Let's begin by defining power $P$, current $I$ and voltage $U$ as $P = \displaystyle \lim_{\Delta t \to 0} \frac{E}{\Delta t}$, $I = \displaystyle \lim_{\Delta t \to 0} \frac{Q}{\Delta t}$ and $U = \frac{E}{Q}$. We then find for a constant current $I$ with a constant voltage $U$ that ...

1

You are partially correct. You are correct about the part that the total voltage in the conductor is the addition of the voltage (S) and induced electromotive force, assuming that: 1. The applied magnetic field doesn't affect the flowing current due to the voltage (S). 2. The induced current due to the time varying magnetic field doesn't affect the magnetic ...

0

I'm assuming the bottom of the conductor is grounded as you've indicated a flowing current? EMF occurs in a conductor when submitted to a time varying magnetic field as governed by Faraday's law, is the field in your problem static or time-varying, or is your conductor in motion? In this case the solid conductor would experience eddy currents, circles of ...

-1

Ohm's law isn't fundamental and holds true only under certain conditions, like constant temperature for example. However, there is a simple way to think about it. Imagine the flow of massive objects through a wide water pipe. This is like a current. The water pressure causes the objects to flow quickly, that's your voltage. If the pipe is narrow then the ...

23

You could start from Drude in zero magnetic field, that equates the derivative of the momentum $\vec p$ by the electrostatic force $\vec F_{el} = q \vec E$ as a product of charge $q$ and electric field $\vec E$ minus a scattering term (with time constant $\tau$; compared to Newtons second law that does not feature the latter, crystal term): $~~~~~~\dot ... 33 Ohm's Law is not a construct which can be derived. It is essentially a generalized observation. It is only useful for a few materials (conductors and medium resistivity), and even then virtually all of those materials show deviations from the ideal, such as temperature coefficients and breakdown voltage limits. Rather, Ohm's Law is an idealization of the ... 0 No current, no back emf, no reaction force if the solenoid circuit is open. But if you close the circuit by shorting the solenoid input leads you will see current, and a reaction force. You need a closed loop to have current flow 0 You seem to be asking two questions: why is the current on either side the same, and what happens to the energy. To clarify where the energy goes, the kinetic energy is transferred entirely into electrical potential energy of the capacitor. This is the energy stored as a result of all the similar charges being close together on either capacitor plate. There ... 0 You are right - the electron might make it all the way to the other plate. But then it will have zero velocity, and it still feels the electric field. So it will "fall back" to the plate it came from - and the net current is zero. In reality the velocity distribution of the electrons is continuous (they don't all travel directly to the other plate with the ... 0 First of all power will alway be same i.e VI (primary side of amplifier) = VI (secondary side of amplifier). However V and I individually can change. So in your case: Case 1: Voltage= 20 current will become I = 0.5 amp so net power = 10 watt Case 2: If current I = 8 amps V will be = 10/8 Volts so net power = 10 watts 2 Without mathematics: The divergence operator tells you the net flow into or out of a volume element. Imagine a car park. They count the number of cars that are coming in and the number going out, and a sign says "there are 5 spaces". Not because they count spaces- they count in and out flow. For a steady state (same number of cars in the car park) the ... 2 This is arising from the charge continuity equation. This condition occurs when$\partial \rho / \partial t$= 0 because the full equation is given by: $$\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0$$ if there are no sources or sinks. So without charge creation, this holds. Example Application Consequently, the restriction that$\nabla ...

3

He is saying that the surface integral over any closed surface must be zero for time-independent current distribution, because otherwise there is a net flux of charge into or out of a volume, and we can't have that going on indefinitely. If $\Sigma$ is a volume with surface $\partial\Sigma$, we have by Stokes' theorem that $$\int_{\Sigma} \vec\nabla ... 0 Power(P) = Voltage(V) * Current(I) That law describes the relationship between power, voltage, and current in a conductor. It means that, if you measure the current flowing in the conductor, and you measure the voltage difference from one end of the conductor to the other at the same instant, then the product of voltage and current will be the rate at ... 1 After twisting my own brain for a LONG time about this, I have come to this conclusion: Resistance is a measure of how FAST a load is able to absorb ALL the potential/kinetic energy of a given number of electrons passing through it. (I say all, because with 1 load the V drop will be equal to the V of the source. Which tells me that all energy will be gone ... 2 Why is resistance NOT calculated simply by looking at how much voltage drop is created by a certain amount of charge passing through, as in R=V/Q. Your statement in bold is one way to define resistance. But your words do not match the expression V/Q. We're interested in the voltage drop per charge passing through, right? Well, how do you measure how ... 0 Resistance is defined as$$ R(i) = \frac{dV}{di} $$and, as all definitions, is a matter of terminology. Also, as you stated, since it has to measure the drop against the flow of electrons it makes sense to take the derivative with respect to the current, which is exactly what the flow of electrons is. Since the difference of potential must be calculated ... 1 There is too much confusion here to answer meaningfully. Current always flows through the battery. What goes in one end comes out the other. Some types of batteries can be charged by forcing current thru them backwards. Those are called rechargable batteries. Others don't work that way and can be damaged by reverse current. Those are often called ... 1 Since the circuit is fully symmetrical (left/right symmetry) the potential at C is exactly half the potential between A and B. This means that there is no current flowing across the point (from the "tip of the V" to the middle of the two resistors at point C), and you can break it without changing the underlying equations describing the current flow. 0 I assume you mean, how do we know we can transform the circuit this way without changing any of the node voltages or branch currents? Let's use your node "B" as the reference node, and assign it a potential of 0 V. Then we can see that V_A is just the battery voltage. From symmetry, we can tell that the voltage at "C" must be \frac{V_A}{2}. ... 6 There are really two questions here (I think): why is the voltage drop so different for the cold LED (notice it ranges from 3.5 to 4.5 V at LN temperature, but from 2.0 to 3.2 at room temperature) why does the LN curve exhibit the strange curvature? CuriousOne already hinted at the answer - this has to do with the temperature of the LED. In particular, ... 1 Be careful because that formula is only valid for a very limited set of field geometries. It is always better to derive EMF from the change of magnetic flux. To answer your question, the induced voltage at zero current does not depend on the resistance of the conductor. As soon as a load is connected to it, the effective voltage measured on the conductor ... 5 The temperature of the LED increases. Do the same experiment with short pulses of current (1ms repeated once a second) and you will see that it's a temperature effect. 1 The electrons are not moving in a curved path. They are moving according to the solutions of the Newton's equation$$ m\textbf{a}=\textbf{F}(\textbf{r},\textbf{r}')=q\,\textbf{E}(\textbf{r},\textbf{r}')  As the above being a Cauchy problem, the form of its general solution explicitly depends on the initial conditions for position and velocity and in ...

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