# How can you calculate (or convert) the Wh of a capacitor whose energy is given in Farads

When trying to compare the energy in a battery to the energy in a capacitor, the units don't match up. How can one compare a battery whose Ah are 10 and Voltage is 3 (for a total of 30 Wh) to a capacitor whose Farads is X and voltage is Y?

To begin, let's call things with their names (no offense). What is measured in Farads is capacitance $C$. What is measured in $Ah$ is the charge that can be stored in a battery or a capacitor.

From the definition of capacitance, the charge on the walls of a capacitor with capacitance $C$ and potential difference $V$ is $$q = CV$$ so you obtain the value you're interested in. However, if $C$ is in Farad and $V$ is in volts, $q$ will be measured in Coulombs. $$\text{Ampere} = \frac{\text{Coulomb}}{\text{Second}}$$ so

$$1\text{ Ampere}\cdot\text{hour} = \text{1 Coulomb}\cdot\frac{\text{hour}}{\text{second}} = 3600 \text{ Coulombs}$$ So, in your notation, the stored charge is $XY/3600$.

• Can you tell me why in this data sheet:maxwell.com/products/ultracapacitors/docs/… – Jack Dagmy Jan 31 '13 at 20:31
• Your calculations doesn't work. – Jack Dagmy Jan 31 '13 at 20:31
• I don't know, here we are talking about a physical device, which can be imperfect, have leakage currents etc, I don't know how they calculate it. You should ask an experimental physicist or someone expert in electronics. – Bzazz Jan 31 '13 at 20:37

Energy in a capacitor is $CV^2/2 = QV/2$ because its voltage starts at 0 when uncharged (unlike a battery, where the voltage is more or less constant). See note 8 in the Maxwell doc you referenced, which uses this formula and then converts from joules to watt-hours in the same fashion correctly described by Bzazz's answer.

V=J/C

30Wh x 60min x 60sec = 10800Ws(or J) / 3v = 3600C

• Answers can be more useful if not limited to formulae or equations without any indication about the meaning of symbols. – GiorgioP Jan 16 '19 at 5:59