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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. $$V = vBL\sin{\theta}$$


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 ...


4

In the limit of long times, the currents are steady, so the magnetic fields they create are steady so there is no induced EMF. This situation is usually tagged "steady state". That said, there will be a period of time when you have just switched a circuit on or off during which things have not settled down and then there will in general be effects not seen ...


4

… an ideal power source capable of providing infinite current with no drop in the voltage it supplies. … Let's ignore the effects of current density on superconductors for now. … In these phrases is the explanation for the contradictory possibilities you have computed: you have supposed an impossible circuit. As a mathematical model, the behavior of ...


3

As Kevin Reid aptly explains, the circuit you have drawn is not realizable. But, let's take the closest physical thing you could build, assuming: your voltage source can supply enough energy that we don't hit its limits like all physical things, this apparatus has non-zero size Then, the circuit you actually built is this: simulate this circuit ...


3

why doesn't the receiver of this device happen to catch another wave of the same frequency instead of the one that was intended? It does catch other waves at the same frequency. This is called noise. Communications are engineered so that the signal is significantly stronger than the potential noise such that it can still be reliably demodulated. More ...


2

Radio wave receivers are designed to resonate at a particular frequency. If you look at the response of a resonant device as a function of frequency you get something like (this image is from the Wikipedia article): This is a rather busy plot, but the point to take away is that the response of the resonant system is greatest when the frequency matches the ...


2

$P = IV$ applies to all circuit branches. $P = I^2R$ or $P = V^2/R$ are restatements of the general rule that apply when we are considering power delivered to an ideal resistor that behaves according to Ohm's law $V = IR.$ I have seen in some circuit $V^2/R$ is not equal to $I^2R$ (like when there is capacitor or inductor). Why is that? Those ...


2

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 = ...


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 ...


1

take only first 2 resistors and rest as $x$ now the vertical resistor and your $x$ will be in parallel, effective resistance would be $Rx/R+x$ with series in horizontal resistor. Now equivalent resistance would be $$Req. = R+ (Rx/R+x). $$ take Req. as $x$ again Form quadratic equation and solve for $x$. This will be the answer.


1

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 ...


1

Intuitively, inductors create back EMFs that "resist" changes in current. Anytime there is an inductor in a circuit, it will resist such changes. Mathematically, inductors force the current in a circuit to be continuous. Let's consider an LR circuit where the inductor and resistor are both in series. Regardless of the initial voltages or current, we can ...


1

As radio amateurs we've all learned the various relationships of power, voltage, current and resistance as expressed in Ohm's Law Ohm's law is: $$ E = IR \tag{1} $$ This doesn't directly say anything about power. There is the related Joule's first law, which relates to electrical power converted to heat in resistive materials: $$ P = I^2 R \tag{2} $$ ...


1

Ohm's law won't get you very far when dealing with RC circuits (R=resistor, C=capacitors). The general way to deal with such circuits, as described in the WP page on RC circuits, is to use Kirchhoff's circuit laws to write down a differential equation. Solving the differential equation will give you $I(t)$. The basic idea to applying the appropriate ...


1

Once the capacitor has fully charged the current in the circuit will be zero, so the voltage drop across the resistor is zero and hence the voltage across the capacitor is equal to the cell voltage. Having said this, the current falls exponentially with time so in principle the current takes an infinite time to fall to zero, and the voltage across the ...


1

What is the flaw in my thinking? The voltage across the capacitor in the series RC circuit given, assuming zero initial capacitor voltage, is given by $$v(t) = E\left(1 - e^{-\frac{t}{RC}} \right), t \ge 0$$ Note that $v(t) \rightarrow E$ as $t \rightarrow \infty$. The energy stored in the capacitor, as a function of time, is $$U(t) = ...


1

There is an ambiguity. Although I did not understand your analysis of the problem completely, charge carriers certainly can run against the (averaged) electric force due to difference in available bands and other particle statistics effects. The gauge freedom is irrelevant. There are two cases for the “ubiquity”. First, these non-Maxwellian deviations ...



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