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Circuits, electronics and noise
Energy dissipated during capacitor charge sharing
Energy lost when two charged capacitors are joined in parallel and settle.
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- ELN-08
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- v1
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- Circuits, electronics and noise
- Result
- \Delta E [J]
Joining two charged capacitors loses energy even though charge is conserved. The energy lost is ΔE = ½[C₁C₂/(C₁ + C₂)](V₁ − V₂)², equal to the initial energy less the final energy. It depends only on the voltage difference and the series-equivalent capacitance, and it goes into heat in the connection or radiation. It is zero when V₁ = V₂.
Energy lost when two charged capacitors are joined in parallel and settle. The loss goes to heat or radiation in the connection and is zero if the voltages are equal.
Calculator
- Initial stored energy, ½C₁V₁² + ½C₂V₂²8.1×10−5 J
- Final stored energy, ½(C₁ + C₂)V_f²2.7×10−5 J
- Share of the initial energy that is lost0.6667
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.
- Joining the two capacitors dissipates 5.4×10−5 J, 66.7 % of the 8.1×10−5 J that was stored.
- With equal starting voltages nothing is lost, and with one capacitor at zero volts the loss is C₂/(C₁ + C₂) of the energy of the charged one.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V2 Initial voltage on C₂ | −50 to 50 V | 6.667e-7 | 0.002321 | 0.00232 | +2 V | −39.5 % |
| V1 Initial voltage on C₁ | −50 to 50 V | 0 | 0.001667 | 0.001667 | +2 V | +49.4 % |
| C1 Capacitance 1 | 1e-9 to 0.001 F | 4.049e-8 | 1.614e-4 | 1.613e-4 | +10% | +6.45 % |
| C2 Capacitance 2 | 1e-9 to 0.001 F | 4.048e-8 | 8.084e-5 | 8.08e-5 | +10% | +3.13 % |
Sweep of Initial voltage on C₁
| Initial voltage on C₁ (V) | Energy dissipated (J) |
|---|---|
| −50 | 0.001667 |
| −37.5 | 9.375e-4 |
| −25 | 4.167e-4 |
| −12.5 | 1.042e-4 |
| 0 | 0 |
| 12.5 | 1.042e-4 |
| 25 | 4.167e-4 |
| 37.5 | 9.375e-4 |
| 50 | 0.001667 |
Sweep one variable
Variables and how each one contributes
Energy dissipated
What the equation gives you. Everything else feeds it.
Capacitance 1
The capacitance of capacitor 1, in F. A larger C₁ raises the series capacitance and so the energy lost. Raising it by 10 % (to 2.2×10−6) moves the result by +6.45 %, from 5.4×10−5 to 5.748×10−5 J.
Initial voltage on C₁
The initial voltage across capacitor 1, in V. Raises the loss with the square of the voltage difference. Raising it by 10 % (to 9.9) moves the result by +21 %, from 5.4×10−5 to 6.534×10−5 J.
Capacitance 2
The capacitance of capacitor 2, in F. A larger C₂ raises the series capacitance and so the energy lost. Raising it by 10 % (to 4.4×10−6) moves the result by +3.13 %, from 5.4×10−5 to 5.569×10−5 J.
Initial voltage on C₂
The initial voltage across capacitor 2, in V. Raises or lowers the loss with the square of the voltage difference. Raising it by 10 % (to 0) moves the result by +0 %, from 5.4×10−5 to 5.4×10−5 J.
Worked example with the values above
- Series-equivalent capacitance: C₁C₂/(C₁ + C₂) = 1.3333×10−6 F.
- Voltage difference squared: (9 − 0)² = 81 V².
- ΔE = ½ × 1.3333×10−6 × 81 = 5.4×10−5 J.
Assumptions and limits
- Ideal linear capacitors and inductors with constant values.
- Leakage current, series resistance and saturation are neglected.
Common mistakes
- Assuming no energy is lost because the charge is conserved.
- Forgetting to square the voltage difference.
- Using the parallel capacitance in place of the series-equivalent one.
Statistics and uncertainty
Real capacitors have tolerances of 5 % to 20 % and change with voltage, temperature and age, so results scale with those errors. Inductor values fall when the core saturates.
Where it is used
- Timing, smoothing and switched-capacitor circuits.
- Energy storage in inductors and capacitors for converters and pulse circuits.
Maths you need
- Capacitance: C = Q/V
- Charge conservation: charge on isolated nodes does not change
- Energy in a capacitor or inductor: ½CV² and ½LI²