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36

In addition to the other answers, here is something for the intuition: $$V=RI$$ More "pressure" $V$ is required to keep the flow $I$ of charges constant when the flow is resisted by $R$. A thin wire has higher resistance than a thick wire, $R=\rho L/A$) analogous to a "bottleneck" in a traffic jam.


20

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, ...


16

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 ...


15

Start with the initial diagram, but let's color code everything: Now move some wires around, without actually changing the connectivity: Finally, rotate the left and right blocks while again not changing the connectivity:


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 ...


11

If you express power loss in a power line as $V^2/R$, the $V$ in that expression is the voltage difference between the two ends of the power line, not the voltage difference between the power line and ground. To supply a fixed amount of power $P_L$ to a load, if the voltage at the load $V_L$ is larger, the current $I=P_L/V_L$ can be smaller. If the power ...


9

Think of plumbing for a close analogy. Voltage is how hard you are pushing, and current is how much flows. The relationship writes itself: why would you get more or less flow from the same pump? The measure of how much effort is used to get flow (it makes more sense as the reciprical: how much flows for a unit of effort) is the interesting property, and ...


8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


7

Have you looked at Drude's model? I was taught something like that back at school and have kept it in mind as a intuitive way of understanding it. We want to understand why the current (rate of flow of charge) should be linear with the potential difference. The Drude idea is, as you noted, related to friction. Firstly, the EM field is linear in the ...


7

Your original text admitted three interpretations, and I'm leaving the answers here: 1: What happens with a toy model when there's a circuit with an ideal battery and no resistance? All the charge moves around the circuit at one moment in time (infinite current). The energy must leave the system as Electromagnetic radiation - accelerating charges radiate, ...


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. ...


7

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 ...


7

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

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 ...


6

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 = ...


6

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. ...


6

Suppose your pipes form a loop i.e. water can flow through the pipes and get back to where it started. As the water flows round the loop there will be some places where pressure rises (e.g. a pump = battery) and some places where pressure falls (e.g. a restriction = resistor). However if the water goes all the way round the loop back to its starting point ...


6

The instantaneous power expended in a resistor is proportional to $V^2$ (i.e. independent of its direction), so the average power expended is given by the mean square voltage! The square of the average absolute voltage will not yield the power expended in the resistor. Edit: And actually this close-to-duplicate question does have more extensive answers ...


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


5

If you go in depth of resistivity, it will be easy to get through the point. Voltage is the reason for the movements (flow) of electrons that produce current (charge divided by time). If you have many electrons and atoms in the way (like barriers, like when you are running in a crowd!) they reduce the rate of charge flow. Now it is clear that if there are ...


5

If you set up a circuit with any component (not just resistors) connected to a voltage source, you will find out that the current which will flow through the component depends on the voltage. In most cases, the higher the voltage the higher the current you will get. Conversely, you can ask: how large a voltage does it take to get a certain current through ...


5

Think about current flow. If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there ...


5

When the switch is opened, the circuit is the equivalent of this, so I think you can clearly see that the resistors are in series and so are the capacitors. It seems that you already understand how to calculate series resistance, so I'll show how to understand series capacitance. First, you probably already know that capacitance is defined as the ratio ...


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

Does that mean when I apply a voltage, the current will be infinite large? No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid. When those assumptions don't hold, we have to see what the 'infinity' (division by ...


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

… 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 ...


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

The $KV$ in $30KV$ refers to the motor velocity constant $K_v$. It does not refer to kilovolt $kV$, cf. comments by John Rennie.



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