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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 ...


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 ...


7

There IS a potential, and all three bulbs will be on. First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V. If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V: So, bulb 1 ...


3

Basically, you want to find the proportionality between the total current and the voltage difference between cathode and anode. Let's assume that the current flow is radial under steady-state conditions, which basically allows me to ignore the $z$-direction throughout. In a steady-state solution, we will have $\nabla \cdot \vec{J} = 0$; moreover, if we ...


3

If you really mean "points", see the answer to this question. Basically, the logic is as follows: If you try to inject a finite current at a "point" in a bulk, it will necessarily lead to a divergent current density $\vec{J}$ in a neighborhood of that point, proportional to $r^{-2}$ (where $r$ is the distance from the injection point.) A divergent current ...


3

Since Michael has already pointed out that the problem as stated has no answer, I will answer a different question instead: if we have a resistive spherical shell with inner radius $a$, outer radius $b$, and bulk resistivity $\rho$, and the surfaces of this shell are coated with a conductive layer, what is the resistance between the inner and outer surface? ...


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

Related question on EE: Does perfect insulation exist? (especially the part about vacuum) Insulators and conductors The property of a material to carry charges from one point to another is what electric current is. The difference between insulators and conductors lies in the electron band structure they posses. In conductors the Fermi-level ...


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

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

I'm assuming $V_R$ is taken to be the voltage across the resistor in a series RC circuit. The transfer function comes directly for the voltage division rule: $$\frac{V_R}{V_\text{in}} = \frac{Z_{R}}{Z_\text{series}} = \frac{R}{R+\frac{1}{i\omega C}} \, .$$ In this equation $V_R$ and $V_\text{in}$ are phasors, meaning that the actual time dependent ...


1

Bulb H is connected to second battery directly. As long as this bulb is not short circuited, there will be some applied potential difference across its terminals and hence it will be ON


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

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

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 ...



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