Capacitance Calculator | Q=CV, Energy & RC Circuits
Calculate charge, voltage, energy stored, series/parallel equivalents, and RC time constants for capacitors. Multi-unit support with circuit diagrams.
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Press Enter to calculate · Esc to reset
What Is the Capacitance Calculator | Q=CV, Energy & RC Circuits?
A capacitor stores electrical energy in an electric field between two conductive plates separated by a dielectric (insulating) material. Unlike a battery which stores energy chemically, a capacitor stores and releases energy almost instantaneously, making it essential for filtering, timing, energy buffering, and signal processing in virtually every electronic circuit.
Capacitance measures how much charge a capacitor stores per volt of applied voltage. One farad means one coulomb stored per volt, an enormous amount in practice. Real-world capacitors range from picofarads (pF) in radio circuits to thousands of microfarads (µF) in power supplies, and supercapacitors reach the farad range for energy storage.
The RC time constant (τ = RC) governs how quickly a capacitor charges or discharges through a resistor. This predictable exponential behaviour is the foundation of RC filters, oscillators, debounce circuits, and signal delay lines used everywhere in electronics.
Formula
Core equation, Q = CV
Q = C × V
Solve for any variable:
C = Q / V
V = Q / C
where:
Q = charge stored (coulombs, C)
C = capacitance (farads, F)
V = voltage across capacitor (volts, V)
Energy stored, three equivalent forms
E = ½ × C × V² (from capacitance and voltage)
E = Q² / (2C) (from charge and capacitance)
E = ½ × Q × V (from charge and voltage)
E = energy stored in joules (J)
Capacitors in series
1 / C_eq = 1/C₁ + 1/C₂ + 1/C₃ + …
Result: C_eq is always less than the smallest capacitor
Charge Q is the same on every capacitor
Voltage divides: V_total = V₁ + V₂ + V₃ + …
Capacitors in parallel
C_eq = C₁ + C₂ + C₃ + …
Result: C_eq is always greater than the largest capacitor
Voltage is the same across every capacitor
Charge divides: Q_total = Q₁ + Q₂ + Q₃ + …
RC circuit, time constant
τ = R × C (time constant, seconds)
Charging: V(t) = V₀ × (1 − e^(−t/τ))
Discharging: V(t) = V₀ × e^(−t/τ)
At t = τ : 63.2% of V₀ charged / 36.8% remaining
At t = 5τ: 99.3% settled, fully charged or discharged
R = resistance in ohms (Ω) · C in farads (F) · τ in seconds · V₀ = source voltage
How to Use
- 1Choose a tab: Q = CV to solve for charge, capacitance, or voltage. Energy to compute stored energy in all three forms. Series / Parallel for equivalent capacitance of a network. RC Circuit for time constant and voltage at any time t.
- 2Q = CV tab: Select what to solve for (Q, C, or V), enter the two known values with unit selectors, then press Calculate or hit Enter. Use the Quick Presets to fill a common capacitor value instantly.
- 3Series / Parallel tab: Toggle Series or Parallel, set the unit, fill in 2–6 capacitor values (C1 and C2 required, the rest optional), then click Calculate. A circuit diagram is drawn automatically.
- 4RC Circuit tab: Select Charging or Discharging mode. Enter R, C, and source voltage V₀. Optionally enter a specific time t to see the capacitor voltage at that instant. The full charge/discharge curve is plotted over 5τ.
- 5Reset: Press Esc or click Reset All to clear all inputs and results. Your inputs are saved between sessions automatically.
Example Calculation
Example 1, Q = CV: a 470 µF electrolytic capacitor charged to 16 V.
Given: C = 470 µF = 470 × 10⁻⁶ F, V = 16 V
Q = C × V = 470×10⁻⁶ × 16 = 7.52 mC
E = ½CV² = ½ × 470×10⁻⁶ × 16² = 60.16 mJ
Example 2, Series combination: C1 = 10 µF, C2 = 22 µF, C3 = 47 µF
1/C_eq = 1/10 + 1/22 + 1/47 = 0.1594 µF⁻¹
C_eq = 1 / 0.1594 = 6.27 µF
Check: C_eq (6.27) < smallest input (10 µF) ✓
Example 3, RC circuit: R = 10 kΩ, C = 100 µF, V₀ = 12 V (charging)
τ = R × C = 10,000 × 100×10⁻⁶ = 1.00 s
5τ = 5.00 s (fully charged at 99.3% of V₀)
V at t = τ = 12 × (1 − e⁻¹) = 7.58 V (63.2%)
V at t = 2τ = 12 × (1 − e⁻²) = 10.38 V (86.5%)
V at t = 0.5τ (discharging) = 12 × e⁻⁰·⁵ = 7.28 V
Why the energy formula uses ½, the integration behind it
As a capacitor charges, voltage rises from 0 to V. Each increment of charge dQ is pushed against an ever-increasing voltage, so the work done is not simply Q × V. Integrating: E = ∫₀^Q V dQ = ∫₀^Q (Q/C) dQ = Q²/2C = ½CV². The factor of ½ reflects that the average voltage during charging is V/2, half the final value.
Understanding Capacitance | Q=CV, Energy & RC Circuits
Capacitors in everyday electronics
Capacitors are among the most ubiquitous passive components in electronics. Understanding how to calculate capacitance, energy storage, and RC timing is essential for circuit design, component selection, and fault diagnosis across every discipline of electrical engineering.
| Application | How the capacitor is used | Typical value |
|---|---|---|
| Power supply filter | Smooth ripple after rectification | 1,000–10,000 µF electrolytic |
| Decoupling / bypass | Block DC, pass AC noise to ground | 100 nF ceramic beside each IC |
| RC oscillator / 555 timer | Set timing intervals via τ = RC | µs–seconds range |
| Audio crossover filter | Frequency-dependent voltage divider | nF–µF depending on crossover point |
| Flash camera strobe | Store and release energy in milliseconds | 200–1,000 µJ |
| DRAM memory cell | Charge represents stored bit (1 or 0) | fF range per cell |
| Supercapacitor store | Bridge power gaps in EVs and UPS systems | 1–3,000 F at 2.5–2.7 V |
| RF tuning circuit | Tune resonant frequency with inductor (LC) | 10–1,000 pF |
Capacitor types and their characteristics
The dielectric material and construction determine a capacitor's value, voltage rating, frequency behaviour, and stability. Choosing the wrong type is a common cause of circuit malfunction even when the capacitance value is correct:
| Type | Range | Strength | Limitation |
|---|---|---|---|
| Ceramic (MLCC) | 1 pF – 100 µF | Tiny, cheap, wide frequency range | Capacitance drops with applied voltage (Class II) |
| Electrolytic | 1 µF – 100,000 µF | High capacitance, low cost | Polarised; limited frequency; ESR matters |
| Tantalum | 0.1 µF – 2,200 µF | Stable, compact | Fail short under voltage spikes; expensive |
| Film (polyester) | 1 nF – 100 µF | Stable, non-polarised, low loss | Larger than ceramic for same value |
| Mica | 1 pF – 10 nF | Very stable, low loss at RF | Expensive, limited values |
| Supercapacitor | 0.1 F – 3,000 F | Huge capacitance, long cycle life | Low voltage rating (2.5–2.7 V per cell) |
Series vs. parallel, when to use each
Choosing between series and parallel combinations depends on what the circuit requires:
- ›Series → higher voltage rating. Each capacitor handles a fraction of the total voltage. Use when no single capacitor withstands the supply voltage. Add balancing resistors if capacitor values are mismatched, as smaller capacitors will see higher voltage.
- ›Parallel → larger capacitance. Values simply add. Use when the needed value is unavailable as a single part, or when spreading ripple current across multiple smaller capacitors improves reliability.
- ›Mixed (decoupling): a large electrolytic (100–1,000 µF) in parallel with a ceramic (100 nF) is the classic power supply decoupling combination, the electrolytic handles bulk low-frequency storage; the ceramic handles high-frequency transients thanks to its lower ESR.
RC time constant and filter design
An RC circuit acts as a frequency-dependent voltage divider. The cutoff frequency f_c determines where the circuit transitions from passing to attenuating signals:
f_c = 1 / (2π × R × C)
Low-pass filter: passes signals below f_c, attenuates above
High-pass filter: passes signals above f_c, attenuates below
Example: R = 10 kΩ, C = 100 nF
f_c = 1 / (2π × 10,000 × 100×10⁻⁹) ≈ 159 Hz
See the Ohm's Law Calculator for resistance analysis and the Wave Calculator for LC resonance frequency in tuned circuits.
Energy storage, why voltage rating matters
The energy formula E = ½CV² highlights a critical design point: energy scales with the square of voltage. Doubling the voltage quadruples the stored energy, and quadruples the energy released in a failure event:
- ›A 1,000 µF cap at 50 V stores 1.25 J, enough to cause a painful shock or destroy sensitive components
- ›The same capacitor at 16 V stores only 0.128 J, a factor of ~10× less energy
- ›In defibrillators, a capacitor charged to ~2,000 V stores ~140 J and releases it in milliseconds through the patient
- ›In switched-mode power supplies, check the ripple current rating (RMS current), not just the capacitance value
Common mistakes when working with capacitors
- ›Unit confusion, mixing µF, nF, and pF without conversion. Always convert to farads before applying Q = CV. This calculator handles unit conversion automatically.
- ›Ignoring ESR, electrolytic capacitors have significant equivalent series resistance that causes voltage drop and heat under high-frequency or pulsed loads.
- ›Polarity errors, electrolytic and tantalum capacitors must be installed with correct polarity. Reverse voltage causes failure or explosion.
- ›Voltage derating ceramics, Class II ceramic capacitors (X5R, X7R) lose 30–70% of their rated capacitance when operated near their voltage rating. Always check the capacitance vs voltage curve in the datasheet.
- ›Assuming 5τ = safe to touch, at 5τ, 0.67% of charge remains. For high-voltage applications this residual can still be dangerous. Discharge through a bleed resistor and verify with a meter.
Frequently Asked Questions
What is the difference between capacitance, charge, and voltage in Q = CV?
- ›Capacitance (C), a fixed property set by the capacitor's construction: plate area, plate separation, and dielectric material. It does not change with circuit conditions (except voltage-dependent ceramics).
- ›Charge (Q), the actual electrical charge accumulated on the plates when voltage is applied. Q changes as the capacitor charges or discharges.
- ›Voltage (V), the potential difference across the plates. For a given capacitor, Q and V always move proportionally: more voltage → proportionally more charge.
- ›Use the Q = CV tab to solve for any one of the three when the other two are known.
How do I calculate capacitors in series?
- ›Use the reciprocal formula: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + …
- ›Then C_eq = 1 ÷ (sum of reciprocals)
- ›For just two capacitors: C_eq = (C₁ × C₂) / (C₁ + C₂)
- ›The result is always smaller than the smallest individual capacitor
- ›Voltage distributes inversely to capacitance, the smallest capacitor gets the highest voltage
- ›Use the Series / Parallel tab to compute this for up to 6 capacitors with a circuit diagram
What is the RC time constant and what does it mean physically?
- ›τ (tau) = R × C, expressed in seconds
- ›It is the time for a charging capacitor to reach 63.2% of the supply voltage
- ›Or for a discharging capacitor to fall to 36.8% of its initial voltage
- ›After 5τ, the capacitor is 99.3% settled, treated as fully charged or discharged
- ›Larger R slows charging (more resistance limits current); larger C slows it too (more charge needed)
- ›Same τ can be achieved with different R and C pairs, e.g. 10 kΩ / 100 µF = 100 kΩ / 10 µF = 1 s
Why is the energy formula E = ½CV² and not CV²?
- ›As a capacitor charges, the voltage across it rises from 0 to V, not a constant value
- ›Each small increment of charge dQ is pushed against an ever-increasing voltage
- ›Integrating this work: E = ∫₀^Q V dQ = ∫₀^Q (Q/C) dQ = Q²/2C = ½CV²
- ›The average voltage during charging is V/2, so energy = charge × average voltage = Q × V/2 = ½QV
- ›A battery at constant voltage V delivers energy = QV; a capacitor delivers ½QV from the same charge
When should I use series vs. parallel capacitors?
- ›Series: need higher voltage rating, each capacitor shares the total voltage. Add balancing resistors if values are mismatched.
- ›Parallel: need larger capacitance, values add directly. Also reduces equivalent ESR, improving high-frequency performance.
- ›Decoupling combination: large electrolytic (100 µF) in parallel with ceramic (100 nF), each handles a different frequency range.
- ›Precise small values: series combinations create values smaller than any available part (e.g. two 4.7 pF in series → 2.35 pF).
What are the capacitance units and how do I convert between them?
- ›1 F (farad) = 1,000 mF (millifarads)
- ›1 mF = 1,000 µF (microfarads), power supply capacitors
- ›1 µF = 1,000 nF (nanofarads), film capacitors, decoupling
- ›1 nF = 1,000 pF (picofarads), RF and precision circuits
- ›To use 100 µF in the formula: 100 × 10⁻⁶ = 0.0001 F
- ›This calculator converts automatically, just pick the unit from the dropdown
Can I use this calculator for supercapacitors?
- ›Yes, the Q = CV and Energy tabs work for any capacitance value including supercapacitors
- ›Enter capacitance in farads (F) and select the F unit from the dropdown
- ›Supercapacitors typically operate at 2.5–2.7 V per cell; series stacking increases voltage at the cost of capacitance
- ›Example: 100 F supercapacitor at 2.5 V → E = ½ × 100 × 2.5² = 312.5 J = 0.087 Wh
- ›Also consider self-discharge rate and internal resistance (ESR) for complete energy storage design
How do I find the voltage across each capacitor in a series circuit?
- ›In series, all capacitors share the same charge Q
- ›Step 1, Find Q: use the Series tab to get C_eq, then Q = C_eq × V_total
- ›Step 2, For each capacitor: V_n = Q / C_n
- ›Step 3, Verify: V₁ + V₂ + V₃ + … should equal V_total
- ›Smaller capacitors get higher voltages, they must be rated to withstand their individual share
- ›Add a resistor in parallel with each capacitor to balance voltage distribution when capacitors are mismatched
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