# Tag Info

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): $~~~~~~\dot ... 6 There are really two questions here (I think): why is the voltage drop so different for the cold LED (notice it ranges from 3.5 to 4.5 V at LN temperature, but from 2.0 to 3.2 at room temperature) why does the LN curve exhibit the strange curvature? CuriousOne already hinted at the answer - this has to do with the temperature of the LED. In particular, ... 5 The temperature of the LED increases. Do the same experiment with short pulses of current (1ms repeated once a second) and you will see that it's a temperature effect. 3 He is saying that the surface integral over any closed surface must be zero for time-independent current distribution, because otherwise there is a net flux of charge into or out of a volume, and we can't have that going on indefinitely. If$\Sigma$is a volume with surface$\partial\Sigma$, we have by Stokes' theorem that $$\int_{\Sigma} \vec\nabla ... 2 This is arising from the charge continuity equation. This condition occurs when \partial \rho / \partial t = 0 because the full equation is given by:$$ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0 $$if there are no sources or sinks. So without charge creation, this holds. Example Application Consequently, the restriction that \nabla ... 2 Without mathematics: The divergence operator tells you the net flow into or out of a volume element. Imagine a car park. They count the number of cars that are coming in and the number going out, and a sign says "there are 5 spaces". Not because they count spaces- they count in and out flow. For a steady state (same number of cars in the car park) the ... 2 In my opinion, the mathematical equation we call Ohm's Law is best taken not as a “law”, a fact about the universe, but as the definition of resistance.$$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 ... 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 ... 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 ... 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 ... 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 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 ... 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. 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 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 ... 1 Point particles as the electrons (which are the charge carriers) move according to Newton's law \textbf{F}=q\textbf{E}=m\textbf{a}. Whenever an electric field is present it generates a difference of potential between two points A and B given by its differential form calculated between the two points$$ V_A - V_B = \int_A^B \textbf{E}\cdot d\textbf{s}. ... 1 The alternative to moving in a curved path is moving in a straight path. Which your electron will only do only if all the forces are parallel to its velocity vector. So an electron starting from rest in a uniform electric field will travel in a straight line - but that is an exception, not the rule. But you could think of that as a "curved path with infinite ... 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 ... 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 ... 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 ... 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 ...