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Ohm's Law is an idealization. No material follows it exactly, even for small applied voltages. But if your voltage is small enough, the deviation from Ohm's Law will be so small as to be unmeasurable ... but it's still there. What we usually mean in Ohm's Law is that the behavior is linear for the purpose of whatever the application is. Once we have ...

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Actually, all materials(non conducting) in this universe can be considered as non-ohmic conductors because on applying certain voltage, they break down because of the breakage of bonds in the lattice. But perfect insulators, unlike resistors, don't allow current to flow through them at all until this voltage is attained. Their V-I graph is as follows: ...

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There are two related questions I answered: Does perfect insulation exist and Graph of Electrical Resistivity of Air vs Air Pressure The crux of it is, that everything will conduct eventually, even nothing (look at the first link). To call capacitors (which are made with dielectrics) non-Ohmic conductors, can apply if you feed them with AC voltage. They ...

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The product of the instantaneous voltage across and current through a circuit element gives the instantaneous power delivered to the circuit element (assuming passive sign convention). $$p(t) = v(t) \cdot i(t)$$ This holds regardless. For particular circuit elements, one can eliminate one of the variables, e.g. Resistor: $v = Ri$ $$p(t) = Ri^2(t) = ... -1 All 3 formulas are true for a V and I that are constant in time. Imagine there is a constant voltage V across a resistor R. By Ohms Law this causes a current I={V\over R} to flow through the resistor. The power being dissipated by the resistor is$$ P=IV $$Ohms Law says I={V\over R} so substitute this into the first equation for I to get ... 0 The force per unit volume exerted by a magnetic field on a charged material is (see e.g. Feynman):$$\frac{\Delta \mathbf F}{\Delta V} = \mathbf j \times \mathbf B$$In particular, for a thin wire, this reduces to:$$\mathbf f \equiv \frac {\Delta \mathbf F} {\Delta \ell}= \mathbf I \times \mathbf B,$$where \mathbf I = I\hat {\mathbf t} is tangent to the ... 1 A general solution is to Assume a potential difference between the electrodes (the two balls in the first problem, or the ball and the plane in the second problem). Calculate the electric field present in the medium surrounding the electrodes. Use the microscopic form of Ohm's Law:$$\vec{J}=\sigma\vec{E}$$and integrate over a closed surface ... 1 According to my understanding, indeed you could define a physical quantity like$$\vec{I} = I\;\vec{n_d}$$where \vec{n_d} is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between I and \vec{j}. The ... 0 Right now, there's nothing connecting the +1V node to anything else on the circuit, so we can delete it. There's nothing connected to the V node, which means we can eliminate that leg. There's nothing connecting the pins (the unfilled circles) to anything, so the leg between the +2V node and the filled circle is a short (it's all the same voltage). So ... 1 In analyzing circuits one must always consider the possibility that things you've ignored in some situations cannot be ignored in others. For instance, there is always the possibility of an internal series resistance in a voltage supply or a parallel resistance in a current supply. Usually we ignore those if the external resistance is much higher than the ... 0 yes, there is \epsilon Volts in the 2 branches, but one is \epsilon' Ohms and the other a lot more. So I = U/R really hugely prefers the very smaller denominator. By they inverse ratio of resistance. \epsilon, \epsilon' : because there is no such things than zero and infinity in real (classical) physical world. 0 So if you know you wanted 10mA for the led, then there would be 3.3 Volts dropped by the LED which leaves 5.7V for the resistor to drop. 5.7V/ 10mA is 570 ohms. 0 The bulb is intended to operate using a particular source voltage and frequency. (For example, 120 V of 60 Hz single-phase AC.) Under these conditions, the bulb is expected to use an average of 10 W of electrical power. (The power usage might be different while the bulb is warming up after being turned on, or while the filament is exploding while burning ... 0 Wikipedia mentions something similar to your situation: The end of life failure mode for fluorescent lamps varies depending on how they are used and their control gear type. Often the light will turn pink (see Loss of mercury) with black burns on the ends of the lamp due to sputtering of emission mix (see below). In thermionic emission, electrons pass ... -1 Its a chemical reaction caused by the heat and the florescent filament that causes the black marks. It only happens near the ends because that is where the energy is more concentrated. Why they flicker? well that can be for a few reasons. Corrosion to the metal connectors, or simply they no longer can hold the energy out put. Or simply they need to be re ... 1 Theoretically, these requirements arise from the way you connect the measurement devices to the rest of the circuit. A voltmeter is connected in parallel, as you said. Say that you are trying to measure the voltage drop across a resistor R through which passes a current i. If the internal resistance of the voltmeter is comparable to R, then the ... 0 a resistor reduces the potential energy of the current across it, then the current that leaves the resistor will have less potential energy and thus, less pressure or voltage. The current does not have potential energy, the electric field does (the potential being exactly its potential as in the sense of differential forms). If one resistor only ... 1 I know I'm a little late, but I'll take a shot at answering this for you. I'm actually very much a beginner at understanding electronics myself, so everyone: please keep me honest! There has been some criticism of your question, as it does not show a complete circuit. I need to agree with this, as any reliable calculations within a circuit require ... 1 The current flows through both resistors. What works for me is to use the analogy of fluid through pipes under pressure. Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The ... 0 Consider resistors R_1,R_2,R_3 connected in parallel (voltage V is across them, I is the total current from V. and I_1,I_2 and I_3 are currents in each resistor) The voltage across each resistor is same, which is V. So current would be same through it, even hundreds of resistances are connected in parallel. So if we add more resistances ... 2 The external force on the whole current-carrying wire in region V can be expressed as$$ \mathbf F = \int_V \rho\mathbf E_{ext} + \mathbf j \times \mathbf B_{ext}\,d^3\mathbf x. $$where \mathbf E_{ext}, \mathbf B_{ext} is the external electric and magnetic field and \rho,\mathbf j is the charge and current density. Stationary and moving wire will ... 0 Well for me it works that since for both charging and discharging, Q is inversely proportional (exponentially) to t, hence it is always decreasing in its decrease or increase with time. Hope it works for you too :) 0 Well in my book it says that " An EMF is induced in a circuit to oppose the flux change, while closed circuit is special case in which since the circuit is complete, a current flows" 0 Random thermal motion (Brownian motion) allows the particles to become ergodically distributed in the their phase space. They scattering off each other and any other particles in the environment, this randomises the motion. Otherwise they would carry on in the same direction until acted on my a force. It is the randomisation nature of scattering events that ... 0 Nothing "causes" diffusion, it is a statistical process. The atoms in any system in thermal equilibrium are constantly moving with velocities of order \langle v^2\rangle \simeq T/m. This motion is randomized by collisions on some microscopic time scale \tau. The simplest case is a dilute gas, where \tau\sim 1/(vn\sigma), where n is the density of the ... 0 To be clear, current is charge passing through a certain area per unit time. This does not imply a second parameter in the denominator of the formula for current dq/dt; just a guideline for how to measure dq. The larger the cross sectional area, the larger the perceived current will be. This is why resistance is seen to decrease for larger cross sectional ... 3 If your power supply is sourcing a positive current toward the ground, that corresponds to a flow of positive charge from the supply to ground. This is equivalent to a flow of negative charge from the ground to the power supply. In a real wire, only negative charges can flow, so the second thing will happen: electrons (which have a negative charge) will ... 0 Given a particle of charge q placed into an electric field, the equation of motion is m \mathbf{\ddot{r}} = q\,\mathbf{E}(\mathbf{r},\mathbf{\dot{r}}). The direction of the electric field comes into play in the difference of potential V_A - V_B, according to whether your ground is at a lower or higher voltage (usually grounds are at lower voltages). ... 0 Conductors, insulators, and semi-conductors all have physical definitions relating to how their valence electrons behave. But from the simpler electrical point of view, the difference between conductors and insulators is simply their conductance/resistance. Insulators conduct very poorly, conductors conduct well. You can make a conductor (say, a metal wire) ... 0 The overall resistance of the circuit is 8 ohms.$$ I = \frac{V}{R} \\ I = \frac{16}{8} \\ I = 2 \textrm{ A}$$2 A will flow through the first four ohm resistor, 1.33 A through the 12 ohm and 0.67 A through the 6 ohm resistor. You can now calculate the power dissipated in each resistor by using P=I^2R. -3 My simple view in layman's terms. Electricity is like water flowing in a river. The river are the wires and the electrons are the logs in the river, copper is full of logs by the way. So the photons are like the water moving very quickly propelled by high to low pressure (volts). As the water in the river passes the logs (electrons) it moves them slowly ... 0 QUESTION Current is constant throughout the circuit with a resistor hence we cannot say that the electron loses kinetic energy after passing through the load. SOLUTION Current throughout the circuit with a resistor is constant , no doubt about that. But to be fundamental, current in a circuit is set up by the electric field, not by electrons. For ... 0 To maintain a current you need to "push" the charge through any obstacles on the path. If there is resistance against the current, then the "push" must be large enough to overcome this. The potential difference is this "push". Of course, as soon as the resistor has been passed, then a large "push" is no longer needed to make the current keep moving. Now ... 1 Any ammeter which uses a shunt resistor will ignore magnetic fields (if it is well-designed), and measure only the voltage across the shunt. Since the shunt is a known resistance,$$i = \frac{V}{R} For instance, this is the standard way a DMM measures current.

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Yes. There are ammeters which are usually used in AC circuits which measure current on the basis of Joule heating of the ammeter wire. You can use a simple compass needle. If the deflection of the needle is appreciable, say about 3 degrees, it will probably interfere with the reading of a shunted-galvanometer ammeter.

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EMF is actually the potential difference across the conductor. In your case, since $V = \int \Sigma .dr$, therefore the resulting EMF is $\Sigma$ multiplied by the length of the conductor in direction of applied electric field.

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It appears to me that you are slightly confused with regards to the concept of current in conductors. Now, if I only choose one side of this rectangle, and apply external electrical field ∑ only to it, what EMF would I create on the conductor? I would simply say ∑, however then I had the following idea, and I started to doubt if I create 2∑ instead ...

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How is the damping force produced by eddy current? As in the picture given below, the moving conductor experiences a change in magnetic flux as it passes through the magnet. This change in magnetic flux, in turn, by Lenz's Law produces an induced EMF on the conductor. Hence an eddy current flows in clockwise\anticlockwise direction thereby ...

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It can be analytically easily shown. Let $T$ be the period of the ac signal ($T=1/2\pi\omega$). The square of the RMS current will then be given by: $I_{rms}^2 = \frac{1}{T} \int_0^T (a+b \sin \omega t )^2dt=\frac{1}{T}[ \int_0^T a^2 dt + \int_0^T 2ab \sin \omega tdt +\int_0^T b^2 \sin^2 \omega tdt]$ Given that the second integral vanishes we have: ...

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