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Make a Capacitor with stuff you already have (how it works+calculations)

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if you use a resister to lower the voltage out of the capacitor will it have a higher f or amp hour?

There are many theoretically oriented people who want to be more exact than necessary. The person asking wants simply to replace the battery with charged capacitor. The answer is:

1F=0,2777mAh/V

so: if you have 1,2V battery with 0,27mAh capacity, than you can replace it with the capacitor of 1F charged to let's say 2V. Your circuit should cope with it. After consuming 0,27mAh from the capacitor it will get discharged down to 1V which should be also ok for your circuit.

So:

* check the minimum and maximum voltage for your circuit. Take the difference between those two,

* multiply it by 0,2777 per each needed mAh

and you will get needed capacity in F.

And remember to start with fully charged capacitor up to the maximum possible voltage.

sorry one correction:

*check the minimum and maximum voltage, take the difference

* divde 0,277 (not multiply) by the min/max voltage difference

you will get number of F.

How many kilderkins in a MeV?

2 Kilderkins = 1 Barrel

1 1/2 Barrels = 1 Hogshead

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

You might ask the question, "How many donkeys are equal to one school bus... approximately?", and then wonder why no one knows what the monkey you're talking about.

There is some thing unstated, some thing donkeys and school buses have in common, some thing which will make the question make sense. Is it that donkeys and school buses are both heavy? Is it that donkeys and school buses can both be used to transport people? Knowing the quality which is common to both donkeys and school busses, the thing

that you care about, is important.Regarding milliampere*hours and farads, I suspect the thing you are interested in is energy storage.

For this you should know batteries and capacitors store energy in somewhat different ways. For a battery, the energy stored is approximately its voltage multiplied by its current*time capacity, e.g. volts times ampere*hours.

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.

If you say, that the charged capacitor C, with voltage V has a charge on it, Q=CV, you know how many Couloumbs of charge are stored. Now, if 1 Couloub flows past a point in a second, we say a current of 1A is flowing. Likewise 0.001 C/S = 1mA.

Steve

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

charge(coulombs). Farads are a unit ofcapacitance, which is charge per voltage. Get it right.L