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You are increasing the number of electrons in the wire, but only by a very small amount. There's a somewhat clichéd but still excellent analogy for electrical circuits called the hydraulic analogy. In the hydraulic analogy the power supply is a pump, and the pressure is the voltage. The water represents the electrons, so the pressure generated by the pump ...

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No. The voltage is the chemical potential of photons instead to relating to electrons, in other words, an electromagnetic field is established across the wire which moves the electrons. Back to your question, the number of electrons does not change in the wire at the open-circuit scenario, aka the electrons in the wire can't flow into or from anywhere.

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Voltage is potential difference. A higher voltage means that there is more energy that can be used from the same amount of current. In effect, increasing the voltage is roughly analogous to adding more potential energy per electron, as opposed to current, which is moving more electrons through the wire.

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You make a good point. In a battery a chemical reaction (a redox reaction) creates the potential difference, and the potential difference is calculated assuming the electrons start and finish in well defined energy states. If electrons returning to the cathode have some residual kinetic energy then the could affect the reaction and change the EMF. However we ...

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You're quite correct that there will be some current flowing, so there must be a voltage drop due to the internal resistance of the battery. The EMF measured by any voltmeter will always be less than the true EMF. If the internal resistance of the battery is $R_b$ and the resistance of your voltmeter is $R_m$ then the voltage you measure will be: $$V = ... 1 Of course lightning "has a frequency component". If it didn't have non-zero frequencies, it could never start or would last indefinitely. Lightning is a huge current pulse over a short time, usually several pulses over a few milliseconds. But more importantly, the current is started and stopped very abruptly, which by necessity means it has a broad ... 1 According to wikipedia, average duration of a lightning is 30\,\mu s. If we take a gaussian current splash with \sigma=30\,\mu s, its spectrum will be a gaussian with \sigma_k\approx33\,\text{kHz}. This doesn't look like DC. 1 It is very hard to use a voltage source to induce charge on an insulator. The reason is that by definition, an insulator does not conduct electricity - so if you apply an electrode at one place, you will not move electrons elsewhere, and so you cannot induce a net charge (the best you can hope for is to create polarization, and maybe pull off a handful of ... 1 Voltage has absolutely nothing to do with charge. I can "move" an infinite amount of charge trough a superconductor with zero voltage. Are you asking about the relationship of charge to voltage on a capacitor? That's a linear relationship: Q=C*U. The charges, in that case, are not "created" but merely separated. If you want more charge for the same amount of ... 4 It's not a fundamental feature of electrical potential, but: If you have a polycrystalline metal and you cut and polish a smooth surface, the differently-oriented regions will present a different lattice plane to the outside. Crystals cut along different planes may have slightly different work functions, and so the electric potential very close to such a ... 0 This is because what changes in each resistor is the current passing through and not the voltage difference. One the other hand when resistors are in series they have the same current passing through, but different voltage through each one's nodes. In essence when resistors are in parallel do not share same current path (i.e wire) but share same voltage. On ... 2 Kirchoff's laws tell us that the potential drop across any closed loop in a circuit must be equal to the voltage sources in the loop, from which we conclude that the voltage drop across resistors in parallel must be equal. Ohm's law states:$$V=IR$$From which we conclude that, since V is fixed, if the different resistors have different R's, then the ... 14 For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So ... 7 Voltage is a continuous function. If you are a certain distance from a (point) charge q, the potential is$$V=\frac{q}{4\pi\epsilon_0 r}$$By adjusting the value of r to anything you want (not quantized), you can get any potential you want. And so yes, when you do any analog-to-digital conversion, you will "destroy" a certain amount of information. ... 5 Voltage doesn't come directly from the charge of the electron. It's the energy per charge. The charge carriers may be discrete, but the energy is not. We can easily generate a potential by moving a wire through a magnetic field. The potential is proportional to the speed of the wire, which is a continuous value.$$V = vBL\sin{\theta}$$0 Supposing that the charger gives the voltage greater than 12 V (say, 15 V), we can estimate 15 V × 100 A = 1500 W, a power of a small electric kettle. It is insufficient to effect an actual explosion quickly, but the battery will possibly immediately start to spew the acid mixed with hydrogen bubbles (note that hydrogen is flammable). Another question in: ... 1 The 2nd equation defines the ideal inductor circuit element. It is understood that the voltage v and current i in that equation are the voltage across and current through the inductor. The inductor emf is the opposite sign of the inductor voltage.$$\mathcal E_L = -v_L $$Clearly, when the current 'stabilizes' (the time rate of change of the inductor ... 0 The voltage across the battery when there is zero current (no load connected) is called the open circuit voltage. The emf of the battery is equal in magnitude to the open circuit voltage. I'm not certain that there is a standard term for the battery terminal voltage when a load is connected since, in general, this voltage varies with the load. One might ... 1 DC signals do not induce electro-motive forces, for you would need a change in the magnetic flux through your circuit which can only be achieved with a time-varying current (resulting in a time-varying magnetic field). This, of course, assuming that your circuit is stationary, so basically you are not moving the wire around for this would change the area ... 0 I'm not sure what you are asking, but I'll address one point. Ohm's Law, like many other physical laws, is an idealization. It applies only to ideal systems, and ideal systems do not exist. But Ohm's Law is useful because it accurately describes a very large class of real systems. Even for systems within this class it is only an approximation. One will ... 2 The resistivity of any material is related to the mobility of the charge carriers within it by:$$ \rho = \frac{1}{ne\mu} $$where \mu is the mobility, e is the electronic charge and n is the number density of charge carriers. I've deliberately used the term charge carriers rather than electrons because in semiconductors like diodes the carriers can ... 0 For a constant resistance Ohm's Law is$$V = IR. Now, it happens that it's pretty easy to make a constant-resistance device (we call them "resistors") and that it's easier to make constant-voltage devices than constant-current devices. So most of the circuit problems we encounter have a constant-voltage device like "a wall socket" or "a battery" or "a ...

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Ohm's law can be summarized as this: The drop in electric potential (voltage) across a section of a circuit is equal to the total current flowing through that section of the circuit multiplied by the total resistance of the section of the circuit. By that law, if you have a current flowing and there is some net resistance, there will be a drop in voltage. ...

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