Library / Electrical / ELE-10
Electrical
Energy stored in a capacitor
Energy stored in a capacitor charged to a given voltage.
- Ref
- ELE-10
- Rev
- v1
- Field
- Electrical
- Result
- E [mJ]
E in joules, C in farads, V in volts. The calculator takes C in microfarads (µF) and gives E in millijoules (mJ).
Charging a capacitor means pushing charge against the growing voltage across it. The energy stored is half the charge times the voltage, which works out as one half C V squared. Because of the square, voltage matters far more than capacitance.
Calculator
- Stored energy0.1469 J
- Charge stored Q = CV11.75 mC
- Energy at double the voltage587.5 mJ
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Energy scales with V², so raising the working voltage is the most efficient way to store more energy. 10% more voltage gives 21% more energy, while 10% more capacitance gives 10%.
- Charging a capacitor from a fixed supply through a resistor wastes as much energy as is stored, whatever the resistance.
- Stored energy can be released in a very short pulse, which is why capacitors power flashes and pulsed loads.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mJ) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V Voltage | 1 to 400 V | 0.235 | 37,600 | 37,600 | +10% | +21 % |
| C Capacitance | 1 to 10,000 µF | 0.3125 | 3,125 | 3,125 | +10% | +10 % |
Sweep of Voltage
| Voltage (V) | Stored energy (mJ) |
|---|---|
| 1 | 0.235 |
| 2.115 | 1.051 |
| 4.472 | 4.7 |
| 9.457 | 21.02 |
| 20 | 94 |
| 42.29 | 420.4 |
| 89.44 | 1,880 |
| 189.1 | 8,408 |
| 400 | 37,600 |
Sweep one variable
Variables and how each one contributes
Stored energy
What the equation gives you. Everything else feeds it.
Capacitance
Ability of the capacitor to hold charge per volt. Directly proportional: doubling the capacitance at the same voltage doubles the stored energy.
Voltage
Voltage the capacitor is charged to. Energy goes as V²: doubling the voltage stores four times the energy, so voltage is the more powerful lever.
Worked example with the values above
- Convert to SI: C = 470 µF = 4.7×10−4 F.
- Square the voltage: V² = 25² = 625 V².
- E = ½CV² = 0.5 × 4.7×10−4 × 625 = 0.1469 J = 146.9 mJ.
Assumptions and limits
- Ideal capacitor: constant capacitance and no leakage.
- Fully charged to the voltage V.
- This is the energy stored, not the energy drawn from the source while charging, which is twice as much when charging through a resistor from a fixed supply.
Common mistakes
- Forgetting to square the voltage.
- Forgetting the factor ½.
- Using µF directly as farads. 470 µF is 4.7×10−4 F, not 470 F.
Statistics and uncertainty
Capacitance values have wide tolerances (±20% is common for electrolytic capacitors) while the voltage may be set within 1 or 2%. Because V is squared, for independent errors δE/E ≈ √[(δC/C)² + (2δV/V)²]. A ±20% capacitor at a ±2% voltage gives about ±20%, dominated by the capacitance.
Where it is used
- Sizing capacitors for pulsed loads, flashes and backup supplies.
- Safety: energy still stored in high-voltage capacitors after switch-off.
- Comparing capacitors and batteries as energy stores.
Maths you need
- Squares: E depends on V²
- Prefixes: µF to F and mJ to J