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

Congratulations for having the wit to know something was wrong! simulate this circuit – Schematic created using CircuitLab Figure 1. Parallel and series arrangements of batteries will have the same VAh rating. I will effectively have a battery of 20 V with a capacity of 2 Ah. That's the error. In parallel they can supply 1 A each for one hour. ...


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

It means that the battery has a capacity of 25 Ah when discharged in 10 hours. 25 Ah is 25 Amps for 1 Hour which is equivalent to 2.5 Amps in 10 Hours. So if you load the battery with 2.5 Amps it will last 10 Hours. If loaded with a higher current usually battery capacity decreases so that is why the 10 Hours is mentioned, it results in a higher battery ...


35

Anyone who has a clue about how physical units works will of course realize that kWh/1000h means "1000 watt-hours per 1000 hours" which can be shortened to just W. But when it comes to lamps, the unit "W" is already used for the light output. Light bulbs which use more energy-efficient technologies than the classical incandescent light bulb often state ...


32

They don’t bill in watts (power) either. They bill in watt-hours, that is, energy consumed. (kilowatt-hours typically.) Let’s break this down a bit. Current alone doesn’t tell you power. You also need to know the voltage, as power is voltage * current. Then, you tally power over time to figure energy. Now you could estimate power from current if you make ...


28

No. This is basic physics. There is no free lunch (or energy). If the laptop only puts out 500 mA at 5 V, for example, then you get 2.5 W. You could convert this to a different combination of voltage and current, but the result can't on average exceed the 2.5 W you put in (It is possible to get higher power out for short durations, but that's clearly not ...


22

No, the electric bill does NOT say "you have used 5000 Watts". Look at it more closely. It says that you used 5000 kiloWatt-hours. A kilowatt-hour is one kiloWatt (1000 Watts) for one hour. That is a measure of energy, and is the same as charging for Joules. One kiloWatt-hour equals 3.6 MJoules. Or put another way, they do charge by the Joule, just ...


21

First let's do a quick number crunch: 6.528W/10W = 65% (of 10W) Referring to the datasheet: There is about a 165C rise in temp. Do not touch!. As for "Is it a safe temperature for the resistor?", refer to the next figure: I'll admit that the Derating Curve Graph kinda hurts my head. But, if you follow the 10W curve over to 25C (about room temperautre), ...


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 ...


19

I concur with what others have said above. Battery capacity is somewhat dependent on discharge current At higher discharge current, the battery capacity decreases. Here are discharge curves that illustrates this. The same battery was discharged at different rates. ( source ) This chart is for a Li-ion battery. A chart for a Lead-acid battery would ...


17

Why would companies bill for wattage instead of amperes? Because amperes don't tell the full story about energy transfer from a source to a load. If you supplied a load that took 100 amperes at 1 volt, the power consumption (joules of energy per second) is 100 watts. If a different load took 100 amperes at 100 volts, the energy transfer per second is 10,000 ...


15

The energy rating covers all types of electrical appliances including fridges, washing machines, etc. In the case of a fridge the instantaneous current could be zero or full on depending on the thermostat. It makes more sense to put the fridge into a 20°C room, power it up and read the kWh used in, say, 24 h and scale it up. This gives a better idea of the ...


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, ...


14

When you double the batteries, you double the current, and consequently quadruple the power in the resistor. You got all that correct. However you seem to have missed that by doubling the current, you also halve the runtime of your batteries. So it won't run for 10 hours, but 5. So the total energy is only doubled, not quadrupled. When you connect ...


13

Since resistors are in series, the current will be the same throughout the chain. Assuming that the voltage V across the string of resistors is constant, you can calculate the current \$I={V \over R_1+R_2 +R_3+ ...}\$. Power dissipated on each of the resistors: \$P_i=I^2R_i\$. If \$P_i\$ is less than the power rating of the resistor, it should dissipate the ...


13

Assume the power input to the bulb is 10 Watts. Assume for now 100% efficiency from battery output to bulb input. Efficiency of energy storage by the battery of energy supplied to it will vary with battery chemistry and how well the charger is designed. Best case using a Lithium battery of some sort, over 90% efficiency may be achievable. Lower or much ...


12

The replacement fan is rated 44 CFM vs. 70CFM for the original- meaning it moves somewhat more than half the air (not surprisingly since it uses somewhat more than half the current). If you're not operating close to the design limit of the power supply you may get away with it, but it will likely shorten the (remaining) life of the power supply. On the ...


12

The capacity of a battery depends partly on how fast you discharge it. A specification like "25Ah/10hr" says that the battery has a capacity of 25 amp-hours when it is discharged at a rate that takes it from fully charged to fully discharged in 10 hours — in other words, at \$\frac{25Ah}{10 hr} = 2.5 A\$.


12

It's right there on page 1 of the data sheet. The graph below shows the power output curve being 1.5 watts: - What you have calculated is the electrical input power and this does not equal the mechanical output power (\$2\pi n T\$). Take the example at 10,000 rpm. That's 167 revs per second (n above). Multiply it by torque (approximately 1.4 mNm) and you ...


11

If a car is travelling at 40 mph it doesn't mean it travels at 40 mph per minute. If a TV is consuming 140 watts it means it consumes 140 joules per second. What you get charged for by your utility company is joules and that is oddly (but acceptably) converted (without mathematical error) to watt seconds or more conveniently watt hours or kilowatt hours. ...


11

If you can only take measurement at discrete times, then summing up and dividing by the time between measurements is the only way possible – the integral $$E_\text{total}=\int\limits_{T_\text{start}}^{T_\text{end}} P(t) dt$$ really collapses to a sum, it \$P(t)\$ is only known for set of points. For example, assume the power value is constant for amount ...


10

The constantly on setting would consume more energy powering the bulb. A possible counter-argument would be that the turn-on/turn-off cycling would shorten the bulb life, and thus the energy cost of manufacturing, transporting, and disposing of it would be amortized over fewer service hours. But without digging up actual numbers, my gut feeling is that ...


10

Neither. You're confusing energy and power. Power is energy per unit of time, and one watt is one joule per second. So if you have a 10 W device, that means, that it's consuming 10 J each second. Now, the issue with electric energy is, that joule, as a unit, is quite small. Therefore, we use larger units, such as wat·hour or kilowatt·hour. Note the ...


10

I think because they care about energy consumption, energy is what costs money and resources to generate, also you can have different voltages when using 2 and 3 phases which would make the charge measurement by itself useless -- edit, and yes, they measure in watt*hours for obvious reasons


9

"10 amps" and "2.16 kilowatt-hours" are not the same units and can't be directly converted. An amp is current, a kWh is energy. However, "10 amps at 240 volts for 1 month" is energy. Google will happily convert the units, if you ask it "(10 amps * 240 volts * 1 month) in kwh": 1,753 kWh. Coming from the other direction, since kWh is energy, it doesn't ...


9

Voltage is the difference in energy between two points in an electric field, expressed per unit of charge. A Volt is a Joule per Coulomb: \$V = J/C\$. The voltage between two points tells us how much energy each electron will gain or lose when it moves between those two points. The separation of opposite charges stores energy. If we separate charges such ...


9

Your maths is wrong. 100 Watts generated by solar power at 12V. Your 8.33 amps is the correct calculation. The mistake you have made is assuming that this current stays the same value when added to the 110V source. You cannot just connect the systems. To feed the 12V (DC) into the 110V (AC) you need to convert them to the same thing - usually the 110V AC. ...


9

Your converter can't do magic. It converts energy from one form (current at 12V DC) to another form (current at 110V AC). It can't produce energy, to the contrary: like any physical device it is imperfect, so it will loose energy. A typical figure might be 20% loss, or 80% efficiency. Let's assume you draw 1A of current at the 110V AC side, which amounts to ...


9

Sometimes the power company will charge you. The issue isn't that the power companies need to make more "power", rather poor power factors mean that various equipment which transfers the power from the station to your house has to be bulkier, and thus more expensive. In the past traditionally this was only a major concern for industrial customers who had ...


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