Hot answers tagged

59

Leaving it on would use more energy, absolutely. Sometimes, people try to convince themselves that turning a light on and off uses more energy because there is some high inrush current, or some such thing. Firstly, incandescent lights hardly have any inrush current, because they don't have any capacitors to charge, and they need not strike an arc in the ...


44

The base current in a transistor controls the collector current. The energy comes from the power supply. It is not generated within the transistor.


40

Okay, let's set up a simple simulation: According to the Wiki page on incandescent bulbs, for a 100W, 120V bulb, the cold resistance is ~9.5Ω, and the hot resistance ~144Ω. It takes around 100ms for the bulb to reach the hot resistance on turning on. So armed with this info, we can simulate and prove the initial surge would be absolutely ...


35

The problem assumes you understand something that is not clearly spelled out: the wires and the (unknown) load are in series. Therefore they share the current, not the voltage of the battery. That's the situation: simulate this circuit – Schematic created using CircuitLab As other have pointed out, the voltage drop across the wires is small given ...


33

60A through a 0.01ohm resistance gives a 600mV drop. That is the voltage you need to use in the equation.


29

For most of the copper wire and traces you see, about 60% the speed of light in a vacuum. The energy is the signal so they are the same. The speed of electrons is much slower...slower than walking pace. The electrons aren't moving to and from the powerplant 60 times per second. Think the difference between the speed of sound and the wind. The wave (energy) ...


28

Simplest Inductive Power Transfer Circuit. While it is theoretically possible to make a very slightly simpler circuit than that one, this circuit approaches being utterly brilliant in its simplicity. If it works as well as it appears it should then it's an excellent starting point. The main point to watch is that the transmit and received circuits should ...


25

The Simple Answer DC Power is defined in terms of W = 1 V * 1 A - that is, the power that is delivered by sustaining 1V potential with 1A of current. Thus, a battery pack that can deliver 5400mAh, that is 5.4Ah, while sustaining voltage of 10.4V (this happens to be running in my laptop right now), can in theory deliver up to 5.4 * 10.4 = 56.16 Wh = ...


24

The problem with these theoretical examples lies in the fact the current is assumed infinite for 0 seconds. Crudely substituting this in the conservation law: $$ \frac {\partial \rho }{\partial t} +\nabla \cdot \mathbf {J} = 0 $$ $$ \frac { \rho }{ 0 }+ \infty \neq 0 $$ Since charge is conserved, the assumption of infinite current in zero time is wrong. ...


24

This is like asking "how bright is a red light?". It's as bright as it is. You can make a bright red light or a dim red light. As another answer points out, the energy per photon of an electromagnetic signal depends on the frequency of the signal. But you can make a brighter or dimmer (higher or lower power) source at any frequency by emitting more or fewer ...


23

The field strength at a distance from the inductor is critically important. If the inductor is well shielded, with zero field in the space nearby, then it won't act like an antenna. Obviously. So, how can we maximize an inductor's distant field and create a good radio antenna? Well, first we should wonder about the distance involved. The field must be ...


22

Indeed it can be a very good antenna. Look no further than the transistor radios and AM band receivers. In those ubiquitous consumer goods the antenna consisted of a piece of very low loss ferrite with a very high permittivity. This was wrapped in many amp*turns of very fine copper wire. The high permittivity gave the antennas an effective cross-...


22

(added) The wave speed ratio of electricity v/c is limited by the relative permetivvity, \$ε_r\$ of insulation around the conductor for the speed, v relative to speed of light, c in a vacuum. It is also limited by the relative magnetic permeability, \$\mu _r\$ in the wire or closely coupled around it, as this would occur in magnetic components. \$v/c = ...


21

If i remove both terminals rapidly enough to avoid any arc formation The more rapidly you remove the terminals, the more an arc will want to form. Ideal inductance is defined by: $$ v(t)= L\frac{\mathrm di}{\mathrm dt} $$ If you remove the terminals "instantly", then the current must stop "instantly". That means the di/dt term will approach infinity, and ...


19

According to a Mythbusters episode summary on Wikipedia: " The MythBusters calculated that the power surge from turning on a light would only consume as much power as leaving it on for a fraction of a second (except for fluorescent tube lights; the startup consumed about 23 seconds' worth of power)". So in fact it is possible that on/off would consume more ...


18

This is mostly a con. The circuit is probably only to light the LED so that it looks like it's doing something. Most power grid loads, like your house, are inductive. Inductive power factor (current lagging voltage) can be offset by adding capacitive power factor (current leading voltage). If you get it just right, then the result looks resistive, which ...


18

Taking, at random, Overland Park, Kansas, as an example: Population 191,278 (2017). Area 195 km2. Annual energy demand (per capita) 13,500 kWh = 37 kWh/day. World Bank. City demand = 191278 x 37 = 7 x 106 kWh/day = 7 x 109 Wh/day = 3600 x 7 x 109 = 25.5 TJ/day. For pumped storage the formula for energy stored is \$ E = mg\Delta h \$. Assuming we could ...


18

I am not sure exactly what you are expecting as an answer, but storage has already started to be used to supplement all energy sources. Utility-connected battery banks have already proved superior to any alternative peaker plant. This report on the Australian 129MWh Tesla battery installation's first year of operation could shed some light on it. In just ...


17

TL,DR; What you are seeing is primarily the result of reactive load droop sharing, with secondary loading voltage drops in the transformers and lines. When generators run in parallel like they do to supply national grids, there are two mechanisms at play that largely go unnoticed by the local population. Perhaps the most critical one is speed droop load ...


16

Simple thought or practical experiment: If he's right then the heating time to bring water to boiling point is independent of the quantity of water. One cup will take as long as two. If you're right two cups of water will take twice as long to boil.


16

Capacitors don't have power ratings because, ideally, they don't dissipate any power. They store energy unlike resistors which consume energy, giving it off as heat. Instead, you need to consider the following: The voltage rating needs to be at least that of the maximum voltage they will see in service. For power regulation and loudspeaker connection ...


14

It seems you want a dummy load that can dissipate 800 W. Normally I use incandescent light bulbs for things like this. They have some nice properties: Are designed to handle the heat. Are cheap and plentiful, so a combination at the desired power can usually be found. Are self-indicating. You know when they are on or off just by looking. However, ...


13

It depends on the type of lightbulb! Halogen, incandescent, florescent, and vapor lights all use tungsten filaments that heat up and emit electrons via thermionic emission. In that sense, they are similar. However, the method to "turn on" the lights varies. Incandescent bulbs are simply turned on once and left on. The inrush current is on the order of 12 ...


13

The 'catch' is that a transistor only controls the flow of current; it does not itself generate power. The power would come from some other part of the circuit, perhaps from the electric company via a power supply or from a battery. Now, one possible point of confusion is that transistors can be represented as equivalent circuits that contain a so-called ...


13

When masses collide in an inelastic manner, momentum is conserved but energy has to be lost. It's the same with the two-capacitor paradox; charge is always conserved but, energy is lost in heat and EM waves. Our schematic model of the simple circuit isn't sufficient to show the subtler mechanisms at play such as interconnection resistance. An elastic ...


13

You need to separate two slightly different concepts. There is the electron drift velocity which is the speed at which the charge carriers in the wire move due to the electric field (and is basically negligible, think mm/s) and the propagation velocity of the EM wave between the conductors which is a significant fraction of lightspeed. It is the em wave ...


12

"Turned Off", can (unfortunately) mean different things, it may refer to the device still having power available but in a low power standby mode, where consumption is minimal. Many consumer devices do this for various reasons, such as waiting for a remote signal to turn on. If power is completely cut off (e.g. the circuit is broken) then the device cannot ...


12

It's conceptually similar to an automotive electrical system. There is a generator (driven by the engine), batteries and some other stuff. Small planes tend to use 28VDC for the power bus, which is just double the voltage used in an automotive electrical system (voltage is quoted with the fan up front running, since there will be bigger problems should it ...


12

I used to design electricity meters so I'm biased. There's only one way to measure power (digitally) and that is to simultaneously sample the voltage and current waveforms at a reasonably high rate such as 1kSps or above. You multiply the v and i samples to get instantaneous power samples. You then average these power samples to get true average power i.e. ...


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