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Circuits, electronics and noise
Final voltage after capacitor charge sharing
Common voltage after two charged capacitors are disconnected from their sources and joined in parallel.
- Ref
- ELN-07
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- Vf [V]
When two charged capacitors are isolated and then joined in parallel, the total charge C₁V₁ + C₂V₂ is conserved and spreads over the total capacitance C₁ + C₂. The final common voltage is the capacitance-weighted average of the starting voltages. A negative starting voltage means the capacitor is connected with opposite polarity.
Common voltage after two charged capacitors are disconnected from their sources and joined in parallel. Charge is conserved, and the voltages are signed in the final polarity.
Calculator
- Total charge, C₁V₁ + C₂V₂1.8×10−5 C
- Total capacitance, C₁ + C₂6×10−6 F
- Change in the voltage of capacitor 1−6 V
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 2×10−6 F at 9 V to 4×10−6 F at 0 V leaves both at 3 V, with 1.8×10−5 C of charge.
- The larger capacitor dominates: with a much bigger C₂ the final voltage stays close to V₂.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (V) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V2 Initial voltage on C₂ | −50 to 50 V | −30.33 | 36.33 | 66.67 | +2 V | +44.4 % |
| V1 Initial voltage on C₁ | −50 to 50 V | −16.67 | 16.67 | 33.33 | +2 V | +22.2 % |
| C2 Capacitance 2 | 1e-9 to 0.001 F | 0.01796 | 8.996 | 8.978 | +10% | −6.25 % |
| C1 Capacitance 1 | 1e-9 to 0.001 F | 0.002249 | 8.964 | 8.962 | +10% | +6.45 % |
Sweep of Capacitance 1
| Capacitance 1 (F) | Final common voltage (V) |
|---|---|
| 1e-9 | 0.002249 |
| 5.623e-9 | 0.01263 |
| 3.162e-8 | 0.07059 |
| 1.778e-7 | 0.3831 |
| 1e-6 | 1.8 |
| 5.623e-6 | 5.259 |
| 3.162e-5 | 7.989 |
| 1.778e-4 | 8.802 |
| 0.001 | 8.964 |
Sweep one variable
Variables and how each one contributes
Final common voltage
What the equation gives you. Everything else feeds it.
Capacitance 1
The capacitance of capacitor 1, in F. A larger C₁ pulls the final voltage toward V₁. Raising it by 10 % (to 2.2×10−6) moves the result by +6.45 %, from 3 to 3.194 V.
Initial voltage on C₁
The initial voltage across capacitor 1, in V, signed in the final polarity. Moves the final voltage in proportion to C₁/(C₁ + C₂). Raising it by 10 % (to 9.9) moves the result by +10 %, from 3 to 3.3 V.
Capacitance 2
The capacitance of capacitor 2, in F. A larger C₂ pulls the final voltage toward V₂. Raising it by 10 % (to 4.4×10−6) moves the result by −6.25 %, from 3 to 2.813 V.
Initial voltage on C₂
The initial voltage across capacitor 2, in V, signed in the final polarity. Moves the final voltage in proportion to C₂/(C₁ + C₂). Raising it by 10 % (to 0) moves the result by +0 %, from 3 to 3 V.
Worked example with the values above
- Total charge: C₁V₁ + C₂V₂ = 2×10−6 × 9 + 4×10−6 × 0 = 1.8×10−5 C.
- Total capacitance: 6×10−6 F.
- V_f = 1.8×10−5 / 6×10−6 = 3 V.
Assumptions and limits
- Ideal linear capacitors and inductors with constant values.
- Leakage current, series resistance and saturation are neglected.
Common mistakes
- Adding capacitances in series as if they were in parallel.
- Forgetting the sign of a voltage when two capacitors are connected with opposite polarity.
- Assuming no energy is lost when two charged capacitors at different voltages are joined.
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