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Official Formula: E = C x V^2 / 2

Visualization: Joules = Farad x (Voltage x Voltage) x 0,5

My approach: (Voltage x Voltage) x 0,5 -> Voltage-"Constant" to convert Farad into Joules for this Voltage-Class of Capacitors.

After that: VCo x Farad = Joules

Divide the Joules-Value by 3600 to convert into Watt-hours.

Example: 16V 2F Capacitor

Official: 2 x 16^2 / 2 = 2 x 256 /2 = 256J

My approach: 16 x 16 /2 = 128 Voltage-"Constant" for 16V capacitors. = 128 x 2F = 256J

This is easier, if you have to work with Capacitors of same Voltage but different Capacity, as the "Constant" is easier to remember and faster to punch into your calculator.

If i want to calculate Joules from a 16V 2200yf Cap, i just do this: 128 x 0,0022 = 0,2816J

16V 30F Supercap ? (just fiction, i did not look up if it exists) = 128 x 30 = 3840J ( divided by 3600) = 1.0667Wh

In my Case i was scrolling through a Supercap-Datasheet with multiple models and varying Capacity but Voltage was 2,8V on each one.

(2,8 x 2,8 /2 =3,92)

I got 3,92 as the 2,8V Capacitor Voltage-"Constant" and proof-calculated the Wh - Values.

On the 3000F Supercap it was 3,27Wh.

So: 3,92 x 3000 = 11760J ( divided by 3600) = 3,2666Wh

I know it is only 99% accurate in this case, but we just confirmed the Wh-Value.

400F Cap = 3,92 x 400 = 1568J (divided by 3600) = 0,43555Wh

Hope i could help.

Little minigame: Convert Fictional 9V 450F Supercap into Wh-Value !

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ADC BT

Lvl 2
3y ago

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