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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 deal with circuit elements (resistors, inductors, capacitors, batteries, meters) which have two connection ports. Two components are connected in series when they are connected to each other only once and not connected to other components at that connection point, also called a \textbf{node} (IMPORTANT term). If a third component is in series with ...


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


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 ...


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 ...


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 ...


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

Does it take the same amount of time to charge a capacitor as it does to discharge to get to the stable state? Yes, in an RC circuit both of these are $\infty$. will it take 5Tau to discharge a charged capacitor to its stable state like it would to charge a capacitor when the capacitor is completely uncharged? Well, that's not "completely ...


0

First of all, it will never get fully (dis)charged, i.e., its voltage drop will never match the supply. The time constant is always the same, because it does not depend on the boundary conditions of your equations. To answer your question, all time scales remain the same. That means if you define the system to be stable after $5\tau$ (which is reasonable), ...


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

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

Consider a metal sheet placed in a uniform electrical field normal to the sheet surface. The electrons will be "dragged" by the electrical field and form an excess of negative charges on one side of the metal sheet and an excess of positive charges on the other side of the metal sheet. The excess charges produce an electrical field that cancels the applied ...


0

This problem can be solved using a simple integral. First, we take the formula for resistance and rewrite it small lengths of wire. I also break up the cross sectional area in terms of thickness $t$ and circumference $2\pi r $ $$dR = \frac{\rho dl}{2\pi rt}$$ Think of this as the resistance of a single ring of plastic sheet. Now we integrate over the ...


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? ...


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

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 ...


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 ...


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 ...


0

Your answer would be correct if there was no voltage drop across the resistor on the right; but since a voltage will be developed there because of the return currents, the voltage across the resistors is less than the 30 V, 20 V of the respective batteries. It might be easier to see this is so if you drew the circuit in a slightly different form:


0

Your awnser is not correct because in the equation V=RI , V is the potential difference of the resistor but in the question The potential difference of the resistors are not 30 and 20 volts. 30 and 20 volts are the potential difference of the batteries.


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 $\frac{V_A}{2}$. ...



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