No, that is not correct.

. Convertir coulomb en ampère-heure.

But what we see is that power decreases over the duration of the experiment. Of course, we can make a very good guess at what voltage is being implied here - 5V, the standard for phones and many many other consumer devices which use Lithium ion based batteries.

Without complex electronics, its impossible to "replace" one, because the capacitor voltage drops as soon as you start to pull charge from it, so it won't be 1.2 V....If we say that we do have perfect electronics, 100% efficient, then since Q=CV , C=Q/V and we have 2.6 C/ 1.2 V = 2F Steve.

Now, if your capacitor is losing voltage, it’s also losing power rapidly (P=V^2/R) and current, so it’s hard to compare capacitors with batteries when supplying circuits with power. http://en.wikipedia.org/wiki/Battery_%28electricity%29#Battery_capacity_and_discharging, http://en.wikipedia.org/wiki/Capacitor#Energy_storage. If you wanted to store the iPhone 7 plus 2900mAh equivalent you'd need a minimum of 4176 F in capacitance - assuming perfect voltage regulation and energy transfer as above.

* theoretical battery of 1000mAh and 1,2V: at the beginning has 1,2V at output, and keeps that 1,2V at output for as long as you do not reach 1000mAh (for instance discharging it with 1000mA in 1 hour, or 1mA in 1000 hours).

Then you could work out the power it delivers at each point in time by multiplying voltage and current (P=IV).

But your Watts calculations are not important as I show below.3) I don't think Watts can be added as you have done. Actually, 99,486,918.8 kilderkins = 1 furlong^3, Well technically none - a kilderkin is an old English measure of volume (beer or Ale) 2 Kilderkins = 1 Barrel 1 1/2 Barrels = 1 Hogshead, 2.999999999999999999999999999999999999999999999999999999999999999999, Without complex electronics, its impossible to "replace" one, because the capacitor voltage drops as soon as you start to pull charge from it, so it won't be 1.2 V....If we say that we do have perfect electronics, 100% efficient, then since Q=CV , C=Q/V and we have 2.6 C/ 1.2 V = 2FSteve. 6 décimales Ampere-hours are a unit of charge (coulombs). Trouvez le produit FARAD pour votre modèle de voiture. Statement or question?As a statement it is true, as we need the capacitor’s internal resistance to calculate the Amps.If it is a question & the leads are shorted together with a very thick lump of metal (i.e. Gigafarad (GF)

1 décimales Définition Classé sous : Physique. Farad. Catalogues en ligne. Décifarad (dF) Not: does mAh have an equivalent value in farads, which, if I dare be so bold to venture, I would say yes, I think an estimate could be calculated.   And so with our above assumptions: At 5V, to store the energy equivalent of 1mAh , you need a 1.44F capacitor. After that it immediately dies and the output is 0V.

* check the minimum and maximum voltage for your circuit. You could argue that, as you're calculating the Watts for 1 second then that is equal to energy (measured in Joules), so you are adding energy not power, BUT...4) Why do you assume the voltage rises by 1 Volt each second?Farads x Volts = Amps x seconds20 x Volts = 1 x seconds (assuming 1 amp constant current)or 20 x Volts = seconds (voltage rise is constant with constant current)to rise 1 Volt takes 20 seconds (not 1 second)to rise 2 Volts takes 20 x 2 = 40 secondsto rise 16 Volts takes 20 x 16 = 320 seconds 5) The energy stored in the capacitor =(Farads x Volts x Volts) / 2 Joules =(20 x 16 x 16)/2 Joules = 5120 JoulesWatts x seconds = Joules or seconds =Joules / Watts seconds = 5120 / 136 seconds = ~ 38 seconds (~ means about)In ideal conditions a 20F capacitor could provide 136 Watts (your answer) for 38 seconds.In 3) above the heater uses energy from the mains electricity supply so does not run out.

(Farad * Volts * Volts)/2 Joules.

Regarding milliampere*hours and farads, I suspect the thing you are interested in is energy storage. 3 years ago. In electronics there is nothing like ideal theoretical circuit and some shortcuts in thinking are perfectly justifieble and ... ...long story short: the above question makes sense. Hectofarad (hF) Isn't the Farad scale a similar kind of measure?

0v-2.4v 10F).

pourquoi le Sars-CoV-2 est-il si infectieux ? Nanofarad (nF) 2005, l'année mondiale de la Physique démarre sur Futura-Sciences, Révision du Bac : révisez votre physique en quiz, Physique : chronologie des grandes étapes, Prix Nobel de physique 2008 : la physique des particules à l'honneur, Lancement national de l'Année mondiale de la physique, Dossier : les évolutions marquantes de la physique, Lire la suite : Définition | Candela | Futura Sciences. 3.7v 2000mAh), you would have to design a special circuit.

I’ll try to use water as an analogy: “1 Farad = how many mAh?” is a bit like saying “if I have a tank that fills to a height of 1m when I put in 1L of water, how many mL per second will I be able to leak and maintain this for exactly 1 hour.

Likewise 0.001 C/S = 1mA.Steve.

Take the difference between those two, * multiply it by 0,2777 per each needed mAh. I think we've got to make some assumptions here about what exactly is being asked.

In practice: the battery is not ideal, the voltage starts at 1,4V or similiar, drops a bit during lifetime of the battery, also the circuit is never ideal and usually how many mA it drains depends on input voltage. My apologies. Reply Well first of all, what is a power meter?

A multimeter is able to measure various parameters that make up power, but it doesn't actually measure power, but if you were able to monitor the current and voltage, you could surely ascertain how power (or whatever other measurements you were interested in) by doing some basic math with those values... My power meter is able to display/log current, voltage, Ah, Wh and surely anything else needed with some basic conversion/math... As such, by connecting the Capacitor to a load, surely I could obtain the mAh, along with any other applicable metrics...On top of that, all portable power sources (that is to say, batteries) that I am aware of have a voltage that varies with their capacity, or rather, their state of charge, so why is measuring the capacity of these in any way different from measuring a Capacitor?

But they are both measures of capacitance, and I'm pretty sure you could measure how long a capacitor can supply a given load, which means you could find out how many mA it can supply in an hour, then you could look at how many F it is and tell how many mAh= F. I know they arent the same, I was asking for a close number of about how much. Is it that donkeys and school buses can both be used to transport people? Centifarad (cF) The farad is the SI unit of capacitance, equal to coulombs/volt. http://en.wikipedia.org/wiki/Battery_%28electricity%29#Battery_capacity_and_discharging For a capacitor, stored energy is equal to one half the square of the voltage multiplied by the capacitance of the capacitor, in farads.http://en.wikipedia.org/wiki/Capacitor#Energy_storage Anyway you should meditate on these facts and formulas, and then come back when you've got a question that makes sense. No, The FARAD is the ONLY measurement of capacitance. approximately? If you do use a regulator, you will be fine, except you are wasting a lot of power at the beginning of the discharge (Imagine a capacitor charged to 10V with a 3V regulator: that’s initially 7V across your regulator and 70% of your power wasted on the regulator).

You could connect your charged capacitor to a small load say (10K ohm) and measure both the current and voltage periodically, until the capacitor is drained to 0V. ~R. Exafarad (EF)

It also depends on both + - plates and their surrounding medium inside the battery. Unité légale de capacité électrique (symbole F). I have one sitting here that I built. Kilofarad (kF)

volts times ampere*hours. Now, if 1 Couloub flows past a point in a second, we say a current of 1A is flowing. Térafarad (TF) One amp (1,000 mA) represents a rate of electron flow of 1 coulomb of electrons per second, so a 1-farad capacitor can hold 1 amp-second of electrons at 1 volt. if you use a resister to lower the voltage out of the capacitor will it have a higher f or amp hour? Here is something that I found to be useful : > A capacitor's storage potential, or capacitance, is measured in units called farads. Then the question also becomes "it depends" because capacitors don't hold constant voltage, but we're looking at a theoretical maximum here. There is some thing unstated, some thing donkeys and school buses have in common, some thing which will make the question make sense. After consuming 0,27mAh from the capacitor it will get discharged down to 1V which should be also ok for your circuit. h ou Ah) est une unité de charge électrique.C'est la quantité d'électricité traversant une section d'un conducteur parcouru par un courant d'intensité de 1 ampère pendant 1 heure. Femtofarad (fF)

Coulomb par Volt