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# how meany mah is in 1 uf? Answered

i need to know how meany mah is in 1050 uf and i need to know

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Meany, meany, miney, mah.
Catch a tagger by its tah.
If it hallah, let it gah.

;-)

Seriously though, capacitors and batteries are not exactly the same thing, so there is not an obvious conversion between the units of,

current capacity, or charge,
1 ma*h = (10^-3 A)*(3600 s) = 3.6 A*s = 3.6 C
= 3.6 ampere*seconds = 3.6 coulombs,

used for rating the current capacity of a battery,

versus capacitance, a ratio of charge (on a capacitor) to voltage (measured across the same capacitor). The SI unit for capacitance is the farad, F.

1 F = (1 C)/(1 V)

1 farad = (1 coulomb)/(1 volt)

The actual amount of charge stored on a 1 microfarad capacitor, at ordinary voltages, e.g. 10 volts or 20 volts, the actual amount of charge is small.

Q = C*V = (1 uf)*(10 V) = 10 uC = 10 microcoulombs = 10 uC

Q = C*V = (1 uf)*(20 V) = 10 uC = 20 microcoulombs = 20 uC

A comparison in terms of energy maybe makes more sense, and other answerers here have mentioned this.

The usual naive calculation for energy in a battery is just the current capacity multiplied by the battery voltage, e.g. for a 10 or 20 volt battery with a current capacity of just 1 ma*h,

U = (1 ma*h)*(10 V) = (10*3.6 C)*(10V) = 36 J

U = (1 ma*h)*(20 V) = (10*3.6 C)*(20V) = 72 J

Energy stored in a capacitor depends on the square of the voltage, as the formula, U = 0.5*C*V^2

So, for example, a 1 uF capacitor, charged to a voltage of 10 V, or 20 V, has stored energy:

U = 0.5*C*V^2 = 0.5*(1 uF)*(10 V)^2 = 50 microjoules = 50 uJ

U = 0.5*C*V^2 = 0.5*(1 uF)*(20 V)^2 = 200 microjoules = 200 uJ

So it kind of looks like a 1 ma*h battery beats a 1 uF capacitor, by a factor of roughtly a million = 10^6, in terms of energy storage, at ordinary voltages, 10 or 20V or so.

I mean the numbers might get better at higher voltages, like 100s or 1000s, of volts, since there's that dependence on voltage squared, but you know that might be problematic too, because high voltage things tend to be dangerous.

Running late for the homework I guess ;)

He does not understand condensers...

I didn't know I was doing his homework ! but now it makes sense.

That's why I typically never give the answer to the basic questions like this, but rather try to give the knoledge and understanding required to figure it out.

The short answer is "it matters how many volts the capacitor is charged to"

Here is what you NEED to know first:

Volts are a measure of Voltage.

Amps are a measure of Current.

Ohms are a measure of Resistance.

Watts are a measure of Power.

Joules are a measure of Energy.

Coulombs are a measure of Charge.

Amp-Hours also a measure of Charge.

mAH also a measure of Charge.

--------------------------------------------------------------

Here is how they all relate to each other:

Volts = Amps * Ohms

Watts = Amps * Volts

Joules = Watts * Seconds

Coulombs = Amps * Seconds

3600 * AH = Coulombs

(3600AH in a Coulomb)

Farads = Coulombs / Volts

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Those are the basics. (hopefully I didn't make any typo's) and using basic pre-algebra you should be easily able to rearrange these formulas and substitute variables for the equivalent expressions. For example, The last equation, we can replace Coulombs with the equivalent expression (3600 * AH) and then rearrange for what you want to solve for. You will see that capacitors do not have a definitive mAH rating because the effective mAH depends on the voltage they are charged to.

A 1050uF is a capacitor, that stores charge = electrical Energy = C*V*V/2, the charge must be at some DC voltage..

When we know what the maximum voltage of your capacitor.. We can explain how the exponential discharge curve can be converted to a changing mah that it could deliver to a load...

Look below to see how a capacitor is charged and Dis-charged...