Hot answers tagged

17

In a superconductor, the current can keep flowing "forever" since there is no resistance. But since conductors have inductance (in fact, superconductors are used most often to create magnets like for an MRI scanner), applying a voltage would not (immediately) cause an infinite current to flow. It is instructive to see how an MRI magnet is "ramped" (turned ...


17

Capacitors and inductors are images of one another under the self-inverse mapping that transforms a linear electrical network to its dual network. The network duality transformation maps the network's graph to its topological dual graph,then all the impedances (either as lone-frequency complex scalars or as Laplace transfer functions) in the dual graph ...


13

Another term is thermal resistance, This is incorrect. Thermal resistance is something that prevents heat flow. It is an entirely separate concept from electrical resistance. How is contact resistance explained? To obtain very low resistance in a material like most metals, the electrons must be delocalized from the individual atoms, and free flow ...


10

Short answer - yes, everything in the circuit can contribute. But usually, an ohmmeter is zeroed with the probes in place - in other words, whatever resistance the probes represent is taken out by the meter. There are two other factors that play a role, especially when you try to measure small resistance. The first of these is contact resistance: it is ...


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

There IS a potential, and all three bulbs will be on. First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V. If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V: So, bulb 1 ...


7

A battery connected to a capacitor is an RC circuit in the limit $R \to 0$ (i.e., there is no resistor and the resistance of the wire is negligible). One might think that the energy loss is zero in this limit, but this is not the case. For an RC circuit with a battery and an initially (i.e., at $t=0$) uncharged capacitor, we have \begin{equation} Q(t) = CV ...


6

Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric ...


6

It's a sad fact that pickle/nail interface technology is an intellectual backwater, and has had little support from the military/industrial complex, which means that quality control for pickle illumination systems is generally minimal to none. Thus, it will be quite common that one end's resistance will be significantly higher than the other end's. Call the ...


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 ...


6

If they have 0 resistance then I (V/R) should be infinite? According to Ohm's law, the voltage and current associated with a conductor are proportional: $$V = R \cdot I$$ where the resistance $R$ is the constant of proportionality. This equation holds for an (ideal) ohmic material. We can rearrange this equation to be $$I = \frac{V}{R}$$ except ...


6

This comes from classical electrodynamics, there is no need to go to Standard model theory or quantum electrodynamics for this. The simple answer is that electric potentials, like electric fields, are just a way of characterizing the way charged particles interact with each other. So, charged objects create voltage analogous to the way that they create ...


5

This is a very interesting question, especially considering the very recent history of scholarship on electrical contact resistance (a term first coined in 1964 by William Shockley, one of the inventors of the transistor), as well as thermal contact resistance. For the following explanation, I will use this research paper on electrical contact resistance ...


5

The voltage is zero. That's the point. The main way current gets started, like in an NMR magnet, is by inductive coupling.


5

Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You ...


5

Electrons do "fill up your body" when you jump up and hit a high voltage wire - there is a property called the capacitance of the body that determines how much the voltage increases when you add a certain amount of charge - mathematically, $C = \frac{Q}{V}$. But it's not charge that kills you, it is current: charge flowing per unit time. Since it takes ...


5

I understand also that there would be a tiny minuscule resistive loss through the wire, but really it's not enough to say anything about. On the contrary, it's crucial. Assuming an ideal voltage source (can supply unlimited current) of voltage $V_S$, an ideal resistor of resistance $R$, and an ideal uncharged capacitor of capacitance $C$, are ...


4

A study undertaken by Nutting and Nuttall at the University of Leeds found that "gold is not inherently more ductile than other face-centered cubic metals", such as copper. The authors found by experimentation that "gold is considerably less ductile in tension than silver." But when beaten foil becomes very thin, other metals tend to fragment, whereas gold ...


4

When a capacitor of capacitance C is charged to a voltage V, and discharged through a resistor R, then the current will decay exponentially: $$I = I_0 e^{-t/RC}$$ The voltage on the capacitor will follow the same exponential decay, $$V = V_0 e^{-t/RC}$$ To answer your question one would have to make some assumptions. You will have to do the calculation ...


4

If you take a perfect conductor, there cannot be a field across it since if there were, the particles would arrange themselves in a way to cancel out the field right? Correct, for a perfect conductor, there can be no electric field within the conductor period. Yet, why does the same not hold true for a wire. It is true in the electrostatic ...


4

Use kirchoff's loop and junction law. :)


4

The ammeter measure the current flowing through itself. If you want to measure the current flowing through another component, then you must make the current through the ammeter equal to the current through the component. If you wire it in series, that's true. If you wired it in parallel, the current would be unevenly divided between the component and the ...


4

They are obviously talking about "conventional" current, not the flow of electrons. When you want to know the direction of magnetic force on a current you need to use "conventional" current direction. There is an interesting corollary to this relating to the Hall effect - if current in a semiconductor is carried by "holes" the polarity of the Hall effect ...


3

... but what happens if you connect it in series? Consider a circuit which has a 3-V (ideal) battery connected in series with a 100-$\Omega$ and a 200-$\Omega$ resistor (series resistors). The voltage across the 100-$\Omega$ resistor will be 1 V, and across the 200-$\Omega$, 2 V. Next, connect a 1 M$\Omega$ resistor (a voltmeter) in parallel with the ...


3

You can think in terms of the current density vector. Its definition: $$ \mathbf J = \sum_i n_i q_i \mathbf v_i $$ Where $n_i$ is the charge carrier density in the media, $q_i$ is the charge of the charge carrier, and $\mathbf v_i$ is the average velocity of the charge carrier. Assuming we have several charge carriers (electrons, ions, etc), you have to sum ...


3

Your teacher's explanation makes sense - IFF there is a "missing component" in your circuit diagram. For example, if there is some series resistance (possibly inside the power supply). In that case, the current through $R_1$ will indeed depend on the current through $L$ - since the voltage source plus internal resistance behaves "a little bit" like a ...


3

Use kirchhoff's first law, so for two resistors in parallel: $$ I_\text{total}=I_1+I_2 $$ Then just use I=V/R $$ \\\frac{V}{R_\text{total}}=\frac{V}{R_1}+\frac{V}{R_2} $$ The voltage across any component, whether it be across resistor 1, 2, or the whole parallel portion of the circuit, is the same. It just cancels out so you can divide both sides by V.


3

Resistance is Voltage per Current. $R = V/I$ or $V = RI$ if you like. So if you put two resistors in series, the voltage is that due to the first, plus that due to the second. You understand that perfectly. Now, what is the inverse of Resistance, Current per Voltage? Give it a name, call it Conductance, perhaps. $C = 1/R = I/V$ Then $I = CV$ if you like. ...


3

It perhaps is not as rigorous as you want, but it is simple and intuitive: $$ R = \frac{\rho L}{A} $$ Where $\rho$ is the resistivity, $L$ is the length, $A$ is the cross-section area. When you plug resistances in series, you "are" increasing $L$, and thus $R$ increases. If you put in parallel, you "are" increasing $A$, and thus $R$ decreases.



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