Hot answers tagged

23

You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...


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 ...


11

Question: Which of two pipes of equal length offers less resistance to the flow of water, one of which has twice the cross sectional area of the other? Answer: The one of twice the cross sectional area. But the one of twice the cross sectional area can be thought of as two of the smaller cross sectional area pipes in parallel. This analogy gives an idea of ...


8

There is another speed limit, though The problem with ultra-high speeds could be the doppler beamed and boosted cosmic microwave background (CMB), which may well fry your circuitry. I don't know enough about radiation shielding to give you an exact number where problems would begin. This limit would still apply even should the spacecraft be moving through ...


8

Suppose you have a voltage $V$ between two points A and B in a circuit. If initially you have a resistor of resistance $R_1$ between A and B, the current flowing through the resistor is $I_1=V/R_1$. Now if you connect another resistor $R_2$ in parallel to the resistor $R_1$, then the former will have the same voltage $V$ across it (since it is connected to ...


8

Imagine a block sliding down a slope and that there is an amount of friction between the slope and the block such that the block slides down the slope at constant speed. As the block slides down the slope it loses gravitational potential energy and an equal amount of heat is generated due to the friction between the slope and block. The block does not ...


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

Yes, dimmers save energy. Modern dimmers save more energy than older (rheostat) dimmers, but both save energy. Modern dimmers use a "triac", which is an inexpensive way of reducing power by pulsing the current. These pulses result in lower average current going to the bulb, and therefore lower average power. They are very efficient, so providing current ...


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 suddenly ...


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 can ...


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

Current is the movement of charges through the wire. Since charge is conserved, it has to be the same at any given point in a closed circuit without branches, otherwise the circuit would be leaking. An often cited analogy would be a system of water pipes. It might be more difficult for the water to squeeze through narrow pipes (high resistance) but the ...


5

As drawn, the circuit, assuming ideal circuit elements, is problematic for the reason you've deduced (KVL gives a contradiction). One interpretation is that there is infinite large current for an infinitesimal time which instantaneously charges the capacitors to their final steady state voltages. To gain some insight, add a resistance $r$ in series with ...


5

A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor. When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance ...


5

If I have a circuit with resistance R and voltage V, I get a certain current - that's Ohm's law, $I = \frac{V}{R}$. Now imagine you have two such circuits - completely separate from each other. Each will have the same current. Let's say the voltage is 1 V, and the resistance is 1 A: Now if I connect the terminals of the two voltage sources together (...


5

I'm surprised that no one has yet mentioned the hydraulic analogy for electricity to help the OP understand better. A brief summary of this analogy is: Electricity is like water flowing through pipes. Current = amount of water flowing through pipe Voltage = pressure of water Power = water pressure x water flow (voltage x current) Resistors = constrictions ...


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

The electric field in a conductor is zero if the charges are not moving. The electrons do re-arrange themselves to (try to) cancel out the electric field. That is what is happening in an electric wire; there is no contradiction. The difference between an electric wire that is part of an electric circuit and the same wire isolated in space (when there would ...


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 case. ...


4

There is also a second question you should ask: where do all these electrons go? Since you need to close the circuit to have an electric current, the answer is that the electrons just circle through the circuit indefinitely. If the circuit is not closed the electrons can flow only for a short amount of time. When there are no more electrons available, the ...


4

Manganin, like constantin, are alloys invented in the late 1800s to solve a specific problem: resistance varies with temperature, and every resistor passing a current is subject to Joule heating. So if you are building precise electrical metering equipment their is a design advantage to using materials that show a stable resistance with temperature ...


4

It's all about time constants. Kirchhoff's Laws assume that the steady state has been achieved. How do you know if you can make that assumptions in the presence of of time varying driving signals? By comparing the time scale of the external variation with the time-scale of transients in the circuit. There are two cases to check first Capacitance and ...


4

Crudely: When negative charges accumulate on one capacitor plate, they repel electrons on the other plate and charge in motion is current.


4

The curves show a charging that is proportional to $1-\mathrm{exp}(-t/\tau)$. Essentially, you should flip the exponential decay graph upside down.


4

First answer this: does it take more power to run one light bulb or two? Transmission lines are designed to be low loss, but they run a long way. Lightbulbs are designed to be "lossy" because that's how they work. Let's say for the sake of argument that the whole transmission circuit loses 100 watts and you're talking about a 100 watt bulb: those numbers ...


4

Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ($\...



Only top voted, non community-wiki answers of a minimum length are eligible