### Conversion of electrical units (Capacitance) into energy and power units

**Symbols
used in this notes are:**

**F
– Farad; C – Coulomb; V – Volt; A – Ampere; J –Joules; s – second; W – Watt;
g – gram; h – hour**

** **

**Correlation
between electrical (capacitance) and energy and power units**

**F
= C/V**

**W
= A×V**

**C
= A×s**

**J
= W×s (Energy unit)**

**F
=(A×s)/ V**

**So,
F×(V^2) = A×s×V = W×s = J
Or, F = (W×s) / V^2 = J / V^2
**

__Note__: Faraday's Constant is different from Farad. (Not confused, though both represented by symbol 'F'.) 1 Faraday = 96485 C 1 Faraday = (96485 C /3600 s) = 26.801 A·h

** **

**Capacitance
is expressed in, F**

**Energy
is expressed in, Wh**

**Power
is expressed in, W**

** **

**Specific
Capacitance is expressed as: F/kg (or) F/g**

**Specific
Energy is expressed as: Wh/kg (or) Wh/g**

**Specific
Power is expressed as: W/kg (or) W/g**

** **

**Assume
a supercapacitor has a Specific Capacitance of 2000 F/g and discharges at a
potential difference of 1 V in 900 s, then Specific Energy and Specific Power
is related to its Specific Capacitance as:**

** **

**Formula:**

**Specific
energy = 0.5 × Specific Capacitance × (dV)^2**

** **

**Here,
Specific Capacitance = 2000 F/g and dV = 1 V**

**=
0.5 × 2000 F/g × (1 V)^2**

**=
1000 F/g V^2 = 1000 J/g **

**=
1000 W×s /g**

**=
{1000 W×s /g} / 3600 = 0.2778 Wh/g **

**=
0.2778 Wh/g × 1000 = 277.8 Wh/kg**

**So,
Specific Energy = 277.8 Wh/kg**

** **

**Specific
Power = Specific Energy / time (in hour)**

**Discharge
time = 900 s/3600 = 0.25 h**

**So,
Specific Power = (277.8 Wh/kg) / (0.25 h) = 1111.2 W/kg **

** **

** **

**By,**

**Dr. M Kanagasabapathy**

**Associate Professor **

**Department of Chemistry**

**Rajapalayam Rajus’ College**

**Madurai Kamaraj University**

**Rajapalayam (TN) 626117 (IN)**

** **

** **

**Also check this
related page:**