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Capacitor energy calculator

A charged capacitor is a small tank of energy. Type its capacitance and the voltage it's charged to, and this gives the energy in joules, the charge in coulombs, and how much of it your circuit can actually use before the voltage falls too low.

µ or u = micro: 4700µ = 0.0047 F.

Usually the supply that filled it.

Where your circuit stops working. A red LED needs about 2 V.

The load, for the time constant. Leave empty to skip.

Energy stored, E = ½ × C × V²

0.19 J
Charge, Q = C × V
42.3 mC
Usable down to 2 V
0.181 J
Stranded below it
9.4 mJ
Energy in watt-hours
52.9 µWh
Time constant, τ = R × C
2.21 s
Current at the start
19.1 mA
Voltage rating to buy
16 V or more

The voltage rating leaves room for spikes: about half as much again as the voltage, so an electrolytic capacitor lasts.

The formulas

E = ½ × C × V² and Q = C × V

C is in farads, so turn microfarads into farads first: 4700 µF is 0.0047 F. The voltage is squared, so it matters most: charge the same capacitor to twice the voltage and it holds four times the energy.

A circuit that stops working below some voltage V2 can only use the energy above it:

Eusable = ½ × C × (V1² − V2²)

A worked example

A 4700 µF capacitor charged to 9 V, running a red LED through 470 Ω:

E = ½ × 0.0047 F × 9² = ½ × 0.0047 × 81 ≈ 0.19 J

It holds Q = 0.0047 × 9 ≈ 0.042 C. The LED stops at about 2 V, so ½ × 0.0047 × (81 − 4) ≈ 0.18 J is usable; the last 9.4 mJ stays stranded. Through 470 Ω the time constant is 470 × 0.0047 ≈ 2.2 s, so the LED fades over a few seconds. The resistor turns much of that energy into heat on the way.

Free lesson

Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Energy in a capacitor” is lesson 5 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 4 before it first.

Or start from the very first lesson, no account needed

The ideas behind it

Questions people ask

How do you calculate the energy stored in a capacitor?

E = ½ × C × V², in joules, with C in farads and V in volts. 4700 µF is 0.0047 F, so at 9 V it holds ½ × 0.0047 × 81 ≈ 0.19 J.

What is the charge on a capacitor?

Q = C × V, in coulombs. 4700 µF at 9 V holds 0.0047 × 9 ≈ 0.042 C. The energy can also be written E = ½ × Q × V.

Why does doubling the voltage give four times the energy?

The voltage is squared in E = ½ × C × V². Twice the voltage stores twice the charge, and each bit of it went in at twice the push, so 2 × 2 = 4 times the joules.

How much of the energy can my circuit use?

Only what lies between the starting voltage and the lowest voltage the circuit works at: ½ × C × (V1² − V2²). An LED stops at about 2 V, so 4700 µF from 9 V gives up about 0.18 J of its 0.19 J.

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