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18

The identity $$ V = K \frac{dV}{dt} $$ is only guaranteed with a constant $K$ if your assumptions actually hold. The first identity $V=RI$ only holds for a resistor, while the other holds for a capacitor. So in this sense, the letters $V,I$ in these equations mean something else. In one of them, it's the current through (or voltage on) a particular resistor, ...


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 ...


14

The physical 'meaning' of the imaginary part of the impedance is that it represents the energy storage part of the circuit element. To see this, let the sinusoidal current $i = I\cos(\omega t)$ be the current through a series RL circuit. The voltage across the combination is $$v = Ri + L\frac{di}{dt} = RI\cos(\omega t) - \omega LI\sin(\omega t)$$ The ...


12

$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmM{{\mathrm{M}}}$ $\def\ph{{\varphi}}$ ...


11

Alfred got in before me, but I have a diagram! I've marked all continuous bits of wire in the same colour, and marked the corresponding colours on the ends of the resistors. A quick redraw later and I get: which is a lot simpler!


10

instead of thinking your body is empty and that a charged wire has to push electrons one by one through you and into the ground (blood is actually full of charge carriers), a better analogy would be a very long queue of pushy people. if the entrance to the apple store doesn't open, it doesn't matter how hard the guy at the back pushes--nothing moves. ...


10

If you redraw your diagram as: It should be clear which capacitors are in parallel and which are in series.


9

The resistance of water, even with ions and minerals and such, is still fairly high. So, a tiny current flowed through the water, but not very much. Additionally, the heating effect that often destroys them when short circuited would also be nullified by the cooling water.


8

If you have just given the voltage signal with $$ \def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V $$ then the current at $t=2\rm s$ is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that $$ i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ...


8

I find this sort of thing becomes much more intuitive if you can think of an analogy in terms of water. In this case, we can think of it like this: Here we have water flowing through a hole in a bath tub, into another tub underneath. The stick figure has been given the task of keeping the water level constant, by lifting water back up into the top tub ...


7

Gregsan's and Kieran's answers are insightful analogies and the pushy electrons are certainly part of the answer. There is another aspect to the "decision" process and that is the propagation of electromagnetic waves. There is a chapter in the second volume of the Feynman Lectures on Physics - I don't have it with me but the relevant section will be just ...


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. ...


6

This is true but strictly limited to RC circuits without external sources: that is, a resistor hooked up to a capacitor with nothing else in between. In that case, $V$ is indeed proportional to $\dot V$, with a crucial minus sign in between: $$ \dot V=-\frac1\tau V, $$ where $\tau>0$ is some constant. This equation implies that $V(t)=V(0)e^{-t/\tau}$, ...


6

There does not need to be an magnetic field in the inductor for there to be "back emf" (I would prefer "induced emf"). The induced emf is the consequence of a changing magnetic field and not of a magnetic field itself and hence there can be a changing magnetic field even at zero magnetic field (something like a positive acceleration downwards for a ball ...


6

Basically motors without rotation act as a simple Resistor between the power source, so it is always a closed circuit and current flows. In motors, when forced not to rotate, this is called stalling. Only some special motors can sustain long periods of stalling, called Torque Motors, and are used to apply a force while speed is zero or very near zero. ...


5

The electric field assigns a single vector quantity to each point in space (specifically, the direction in which a positive test charge would accelerate if it popped into existence at that point, assuming it didn't perturb the setup creating the field in the first place). I believe that the difficulty of this question arises from an ambiguity in the problem ...


5

The key thing is that there is NO electric field within the perfect wire. So, there is no force acting on the electron, and thus no work done on it (while it's in the perfect wire). This goes back to the definition of a perfect conductor (which the perfect wire is). Within a perfect conductor, there is no electric field. Instead, the charges (which have ...


5

Voltage is similar to height. It plays the same role for electric charge as height*gravity does for a ball on a hill. So high voltage means high potential energy the same way a ball being high up on a hill means high potential energy. Voltage is not potential energy, the same way height is not energy. However, if you have a certain amount of charge $q$, you ...


5

Kirchhoff's loop rule is also called Kirchhoff's voltage law (KVL). Which is different from Kirchhoff's current rule which is also called Kirchhoff's current law (KCL). KVL is derived from Maxwell–Faraday equation for static magnetic field (i.e. the derivative of B with respect to time is zero). KCL is derived from charge continuity equation which is ...


5

A commonly used analogy is to represent the electric circuit with pipes filled with water. The electrical current is modelled by flow of water in the pipes, and the voltage is modelled by the pressure. This is known as the hydraulic analogy. Anyhow, if you have a pipe filled with water and you suddenly increase the pressure at one end, e.g. by opening a ...


5

Your problem is assuming that the charge transferred through the resistors is different. I don't know where you got that from, so I don't really know how to refute it other than by saying that since the currents must be the same, so must be the charge transfer in a given time. Edit in response to your comment: What you said is plainly not true. ...


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


5

Generally, no. You've written two equations. The first relates voltage and current for an isolated resistor. The second relates voltage and current for an isolated capacitor. Given only those two expressions, there's no reason at all that they can or should be combined. That is, you have no circuit, only isolated components. There are principles for ...


5

How can I prove Thevenin's and Norton's theorem? Here's an outline - you can fill in the dots. Measure the voltage (with an ideal voltmeter) between any two nodes of an arbitrary linear circuit. Call this voltage the open circuit voltage $V_{OC}$ since there is zero current through an ideal voltmeter. Then, place an ideal ammeter across the same two ...


5

Inductors resist changes in current. So in a circuit like the one you describe, for short times after the switch is closed, the inductor acts like a broken wire. This is consistent with the statement you made that there is 2 V across the inductor. For these short times the circuit is essentially reduced (i.e. literally just cut the inductor out of the ...


5

A transistor is not simply a combination of diodes. A bipolar junction transistor is a complex device consisting of three layers of differently doped semiconductor. If a current is pumped through the base-emitter juction, there will be several times greater collector-emitter current because of the increased amount of minority carriers in the base region. ...


5

There is a physical meaning behind the imaginary component of the impedance. You can re-cast the complex impedance $Z = R + jX$ (using engineering's notation $j$ for the imaginary unit) in polar form to get $Z = |Z|\exp(j\phi)$. $|Z|$ is the magnitude of the impedance, and scales the amplitude of the current to get the amptlitude of the voltage. $\phi = ...


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

Thare is only one node connecting elements B, D, E, and G. The dots on the diagram are just to indicate that the lines do in fact connect, so that all of the wires are part of the same node. In this kind of schematic diagram wires are considered as ideal, and all points on the wire are considered to be at an equal potential.



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