Capacitor Calculator
Four calculators in one. Combine capacitors in series or parallel, solve charge, voltage, capacitance and stored energy from any two of them, or enter a resistor and a capacitor to get the RC time constant, the voltage at any moment and the full charging or discharging curve. The working is shown for each. Free, no sign-up.
Add a voltage to see the charge, voltage and energy of each capacitor.
Results
6.667 µF
Equivalent capacitance
2
Capacitors
80 µC
Total charge
480 µJ
Total energy
The working
- 1 ÷ (1/10 µF + 1/20 µF) = 6.667 µF
In series every capacitor holds the same charge, so the smallest capacitance takes the largest voltage. Check each capacitor's voltage rating against its share.
Each capacitor with this voltage
| Capacitor | Capacitance | Voltage across | Charge | Energy |
|---|---|---|---|---|
| C1 | 10 µF | 8 V | 80 µC | 320 µJ |
| C2 | 20 µF | 4 V | 80 µC | 160 µJ |
How to Use the Capacitor Calculator
- 1. Choose what to calculate. Series and parallel combine several capacitors, Charge and energy solves Q = CV and E = ½CV², and RC charge and discharge models a resistor charging a capacitor.
- 2. Enter the values with their units. Capacitance runs from pF to F, voltage from µV to MV, charge from pC to C and energy from nJ to kJ, so a 4.7 µF part and a 25 V rating go in as printed.
- 3. Read the results. The unknown quantities appear with the SI prefix that suits them, together with the formula and your numbers.
- 4. Use the drawing. The series and parallel modes sketch the bank, the charge mode plots Q against V, and the RC mode plots the voltage curve with the time constant marked.
The Formulas
| Quantity | Formula | Notes |
|---|---|---|
| Capacitors in parallel | C = C₁ + C₂ + C₃ + … | Plates in effect add area |
| Capacitors in series | 1/C = 1/C₁ + 1/C₂ + 1/C₃ + … | Always less than the smallest |
| Charge | Q = C × V | Coulombs = farads × volts |
| Stored energy | E = ½CV² = ½QV = Q² ÷ 2C | Joules |
| RC time constant | τ = R × C | Seconds, with R in ohms and C in farads |
| Charging | v(t) = V × (1 − e^(−t/τ)) | Rises toward the source voltage V |
| Discharging | v(t) = V × e^(−t/τ) | Falls from V toward zero |
Capacitors combine the opposite way to resistors. Parallel capacitors add because each one provides extra plate area to hold charge at the same voltage. In series the same charge sits on every capacitor, the voltage divides between them, and the total capacitance falls below that of the smallest one — two equal capacitors in series give half the value of one.
Worked Examples
- Series, 12 V: 10 µF and 20 µF in series give 1 ÷ (1/10 µF + 1/20 µF) = 6.667 µF. The charge is 6.667 µF × 12 V = 80 µC on each, so the voltages are 80 µC ÷ 10 µF = 8 V and 80 µC ÷ 20 µF = 4 V. The smaller capacitor carries the larger voltage.
- Parallel, 12 V: 1 µF, 2.2 µF and 4.7 µF in parallel total 7.9 µF, holding 94.8 µC and 568.8 µJ at 12 V.
- Charge and energy: a 100 µF capacitor at 12 V holds Q = 100 µF × 12 V = 1.2 mC and E = ½ × 100 µF × (12 V)² = 7.2 mJ.
- RC circuit: with 10 kΩ and 100 µF the time constant is 10,000 Ω × 0.0001 F = 1 s. Two seconds after a 5 V step the capacitor is at 5 × (1 − e⁻²) = 4.323 V, 86.5% of the source, and the current has fallen to 67.67 µA from its starting 0.5 mA.
Reading an RC Curve
After one time constant a charging capacitor has reached 63.2% of the source voltage, and a discharging one has fallen to 36.8%. After two it is at 86.5%, after three 95.0%, after four 98.2% and after five 99.3%. The rule of thumb that a capacitor is "fully charged after 5τ" really means within 1% — it never reaches the source voltage exactly. The time to reach any fraction f of the change is t = −τ × ln(1 − f), which gives 0.693τ for the halfway point and 2.303τ for 90%.
The same RC pair is also a first-order filter. As a low-pass filter its −3 dB corner is at fc = 1 ÷ (2πRC), which is 0.159 Hz for the 1 s example above. The calculator shows it beside the time constant.
Real capacitors differ from the ideal
Electrolytic capacitors often have a tolerance of ±20%, and their capacitance changes with temperature, voltage and age. Every capacitor also has a voltage rating that must exceed the voltage it will see — in a series bank, check each one against its own share. Large capacitors can hold a dangerous charge for long after power is removed, so discharge them before handling.
Frequently Asked Questions
- How do you add capacitors in parallel and in series?
- In parallel, add the capacitances: 1 µF + 2.2 µF + 4.7 µF = 7.9 µF. In series, add the reciprocals and invert: for 10 µF and 20 µF, 1/C = 0.1 + 0.05 = 0.15 per µF, so C = 6.667 µF.
- What is the formula for the energy stored in a capacitor?
- E = ½CV², in joules with C in farads and V in volts. Equivalent forms are ½QV and Q² ÷ 2C. Doubling the voltage quadruples the energy, which is why high-voltage capacitor banks are hazardous.
- What is the RC time constant?
- τ = R × C, the time in seconds for a capacitor charging through a resistor to reach 63.2% of the source voltage, or for a discharging one to fall to 36.8% of its starting value. Ten kilohms with 100 µF gives 1 s.
- How long does a capacitor take to charge fully?
- In theory it never finishes. After 5τ it is within 0.7% of the source voltage, which is usually treated as fully charged; after 3τ it is at 95%. A larger R or C gives a longer time.
- Why does a series pair give less capacitance?
- The two capacitors together act like one capacitor whose plates are further apart, because the charge on the inner plates only moves between them. Greater spacing means less capacitance, and the voltage divides across the pair.
- What is the unit farad?
- One farad stores one coulomb of charge per volt. It is a very large unit, so practical capacitors are marked in microfarads (µF), nanofarads (nF) and picofarads (pF); supercapacitors reach hundreds of farads.