Most capacitors are in the µF, nF and pF range. I know there are some special ones that go that high, but at the time faraday was around, and the unit was named after him, they didn't have such a thing. Why is the unit so large if we rarely use caps with that high of a value?

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    \$\begingroup\$ For elementary particle physicists, the meter and the second are enormous units. It's all a matter of context. For electronic engineers, mA and uA are common. For electrical engineers, kA and MA are common. \$\endgroup\$ – Alfred Centauri Jun 10 '13 at 23:16
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    \$\begingroup\$ The unit you're talking about wasn't defined as we currently know it until more than a decade past Farady's death. (Source) Units named after people are typically assigned posthumously. \$\endgroup\$ – Warren Young Jun 10 '13 at 23:58
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    \$\begingroup\$ He was giant in his time. We can only hope to posses uF's worth today ;-) Like fE (femto einsteins). \$\endgroup\$ – user6972 Jun 11 '13 at 7:49
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    \$\begingroup\$ ...And you need a unit for those bigger capacitors. If I'm right, they are trying to use "supercapacitors" in electric cars. \$\endgroup\$ – Anonymous Penguin Jun 19 '13 at 19:14

As others have mentioned, 1 farad is 1 coulomb per 1 volt. But the rabbit hole goes deeper -- the question then becomes why is 1 coulomb what it is, and why is 1 volt what it is?

Following this rabbit hole to the bottom will eventually lead us to the 7 base SI units, which are units of measure for the 7 physical attributes of our world: distance, mass, time, electric current, thermodynamic temperature, amount of a substance, and luminar intensity. They're like axioms in mathematics. From here, other units are defined in terms of these. So volt = (kilogram meter meter) / (ampere second second second). Meanwhile coulomb = ampere * second. You'll notice that 1 of a derived unit is expressed in terms of 1's of a base units.

So ultimately, 1 farad is so large because the base units are so large, at least relative to the sizes of electronic components nowadays where we fit billions of transistors onto several square millimeters.


Because it fits in with all the other (SI) units we have. 1 farad is 1 coulomb per volt. It just so happens that 1 coulomb is... a lot of charge.

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    \$\begingroup\$ Let's put it another way; it allows the formula \$f = \frac{1}{2\pi RC}\$ to work with any mysterious conversion factors. \$\endgroup\$ – Kaz Jun 10 '13 at 23:01
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    \$\begingroup\$ It'd be nice to hear why the other SI units (i.e. coloumb) are so big then. Is it the definition of ampere, charge or voltage? \$\endgroup\$ – Macke Jun 11 '13 at 7:14
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    \$\begingroup\$ @Macke 1 coulomb is 1 ampere × 1 second. \$\endgroup\$ – Random832 Jun 11 '13 at 12:44
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    \$\begingroup\$ @Macke: A second is a nice unit for a human-perceivable timescale, but it's huge relative to the amount of time a typical capacitor can supply what would have been a reasonably-measurable amount of current. \$\endgroup\$ – supercat Jun 19 '13 at 22:07

Because 1 Ampere is a unit so large compared to the amount of current we normally use. Because 1 second is a unit so large compared to the audio and rf frequencies we normally use.

If you normally use currents much smaller than 1A, for periods much shorter than 1sec, and don't have a lot of money to waste or a lot of space to waste, you can use capacitors much smaller than 1F.

On the other hand, if you wanted to do electrical power, instead of radio electronics, 1F isn't very big. Here is a recent press release on a 400F capacitor. http://www.engineering.com/ElectronicsDesign/ElectronicsDesignArticles/ArticleID/5290/Is-it-a-battery-No-its-a-Supercap.aspx -- and note that the special feature is that it is no larger than a deck of cards.

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    \$\begingroup\$ 400F with the size of a deck of cards is by no means a large capacitance in a small package. There are capacitors in the kiloFarad range and above, which are much smaller. They, however, operate on very small voltages. \$\endgroup\$ – vsz Jun 11 '13 at 15:19
  • \$\begingroup\$ @vsc Energy stored is proportional to voltage squared, so that is no surprise. \$\endgroup\$ – starblue Jun 11 '13 at 19:40
  • \$\begingroup\$ 1kWh, 300kW, 477kg, 1900x950x455: bombardier.com/en/transportation/sustainability/technology/… \$\endgroup\$ – starblue Jun 11 '13 at 19:45

The SI units for electricity fit in with the SI units for everything else. The relationship becomes clear if you look at the definition of a joule:

$$ J = N\cdot m = W\cdot s $$

Notice that it has both mechanical units you'd naturally consider mechanical (newtons, meters) and electrical units (watts). We can break it down into more basic units:

$$ J = \frac{kg\cdot m^2}{s^2} $$

Or we can expand watts to more basic, but still electrical units:

$$ J = V \cdot A \cdot s $$

And now you have volts and amps, by which the farad can be defined:

$$ F = \frac{A\cdot s}{V}$$

If you analyze this carefully, you will notice that a joule is a watt-second, and a watt is some ratio of current and voltage, but that ratio is undefined. This is why the ampere is an SI base unit, defined as

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10−7 newton per metre of length.

So if you want to blame something for the farad being so large, blame the ampere. Or, blame the other SI base units referenced by its definition, the second, meter, or kilogram (indirectly, by the newton).


It has nothing to do with Faraday. It is a definition.

From Wikipedia:

\$F = \dfrac{A\times s}{V}\$

Manipulated Algebraicly:

\$A=\dfrac{F\times V}{s}\$

And in terms of \$i_c(t)=C\dfrac{\mathrm{d}v}{\mathrm{d}t}\$.

Expressed Algebraicly:

\$I=C\dfrac{\Delta V}{\Delta t}\$


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