How do you calculate the energy stored in a capacitor?

E = ½ × C × V², where C is capacitance in farads and V is voltage in volts. The result is in joules. Doubling the voltage quadruples the stored energy.

How much energy does a 1F supercapacitor hold at 2.7V?

E = ½ × 1 × 2.7² = 3.645 joules. A typical AA battery holds about 10,800 joules, so a 1F supercapacitor stores about 0.034% of a AA battery. Supercapacitors excel at high power density (fast charge/discharge), not energy density.

What is the charge stored in a capacitor?

Q = C × V, where Q is charge in coulombs, C is capacitance in farads, and V is voltage in volts. A 100µF capacitor charged to 12V holds Q = 100×10⁻⁶ × 12 = 1.2 mC.

Capacitor Energy & Charge Calculator

Calculate energy stored and charge in a capacitor, or find the capacitance needed to store a target energy.

Capacitor Parameters

Capacitance

Voltage

Results

Energy (E)7.200 mJ

Charge (Q)1.200 mC

AA battery equivalent (1 AA ≈ 10 800 J)1/1.5M of a AA battery

Capacitor Energy Storage

A capacitor stores energy in its electric field. The energy depends on both the capacitance and the square of the voltage — doubling the voltage quadruples the stored energy, while doubling capacitance only doubles it.

This makes voltage the dominant parameter for energy storage. Supercapacitors (also called ultracapacitors or EDLC) exploit extremely high capacitance (1–3000 F) to store significant amounts of energy, bridging the gap between conventional capacitors and batteries.

Stored Energy

E = ½ × C × V²

Stored Charge

Q = C × V

Required Capacitance

C = 2E / V²

Key Points

Energy scales with V² — higher voltage means far more energy
A 400 V, 1000 µF camera flash capacitor stores 80 J
A defibrillator capacitor stores ~200–400 J at ~2000 V
Supercapacitors (1–3000 F) can start car engines
Capacitors discharge instantly; batteries discharge slowly
Never exceed the capacitor's rated voltage — risk of explosion

Applications

Camera flash units
UPS hold-up capacitors
Rail gun and pulsed power systems
Supercapacitor energy storage and backup
Power supply bulk capacitors (ripple filtering)

Formula Reference

Capacitor energy and charge

Energy:       E = ½ × C × V²   (Joules)
Charge:       Q = C × V         (Coulombs)
Also:         E = Q²/(2C) = ½ × Q × V

Discharge current (into R):
I(t) = (V₀/R) × e^(–t/RC)
Peak power: P_peak = V²/R   (at t=0)

Capacitor bank (series):   1/C_total = 1/C1 + 1/C2 + ...
Capacitor bank (parallel): C_total = C1 + C2 + ...

Energy Reference Table

Capacitor	Capacitance	Voltage	Energy	Application
Ceramic 100nF	100nF	50V	125µJ	Bypass, decoupling
Film 10µF	10µF	100V	50mJ	Motor start, audio
Electrolytic 1000µF	1000µF	25V	312mJ	Power supply filter
Supercap 1F	1F	2.7V	3.6J	Backup power, IoT
Supercap 100F	100F	2.7V	364J	Energy harvest, UPS
Camera flash	100–1000µF	300V	4.5–45J	Strobe, defibrillator

More Examples

Camera flash capacitor (330µF, 300V)

E = ½ × 330e-6 × 300² = 14.85J. Flash duration ≈ 1ms → average power = 14,850W! Discharge resistor 10Ω: I_peak = 300/10 = 30A, τ = 330µF×10Ω = 3.3ms ✓

Energy: 14.85J · Peak current: 30A · τ = 3.3ms

Supercap backup for IoT node (1F, 2.5V)

System draws 10mA at 2.5V. E_available = ½×1×(2.5²–1.8²) = 1.49J (down to 1.8V cutoff). Runtime = E/(V_avg×I) = 1.49/(2.15×0.010) = 69s of backup power.

Backup runtime: ≈ 69 seconds at 10mA

Design tip

NEVER touch a charged capacitor > 50V — lethal current possible even from small caps.
Always discharge through a resistor: R = V/I_safe (I_safe < 10mA for safety).
Supercapacitors: pair in series for higher voltage (add balancing resistors 1–10kΩ across each).
Electrolytic polarity: reverse voltage > 1V causes rapid failure and possible venting.

Did you know? A 1 F supercapacitor charged to 2.7 V stores 3.645 J — enough to power a 1 W GPS tracker for 3.6 seconds. Flash photography units use large electrolytic capacitors (330–1000 µF, 300–400 V) that store 15–80 J — discharged in microseconds to produce the intense light burst.