Power & Energy
Energy stored in a capacitor
A capacitor charged to a voltage V stores energy in its electric field: E = ½ × C × V², in joules. The voltage counts double, because it's squared: charge the same capacitor to twice the voltage and it holds four times the energy. A 4700 µF capacitor at 9 V holds about 0.19 J.
E = ½ × C × V²
Example: 4700 µF at 9 V.
Where does the half come from? As charge flows in, the voltage builds up from zero. The first bit of charge goes in at almost 0 V and the last at the full V, so on average it's pushed in at half the final voltage. With Q = C × V, that's E = ½ × Q × V.
Live experiment
Watch the energy tank drain
A 9 V battery fills a capacitor while it lights Pip through 470 Ω. Open the switch and the capacitor runs him alone, fading. The tank shows the charge and ½CV², and at the end where the joules went: Pip, the resistor's heat, and what's left stranded.
“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.
Not all of it is usable
A capacitor's voltage falls as it empties, and most circuits stop working below some voltage. The energy you can actually use lies between the start and that limit:
A red LED needs about 2 V, so from 9 V it can take ½ × 0.0047 × (81 − 4) ≈ 0.18 J; the rest stays stranded. And the resistor in series with it turns much of that into heat on the way.
Charging costs energy too: filling a capacitor from a fixed voltage through a resistor wastes as much in the resistor as ends up stored, whatever the resistor's value.
How long it lasts
The resistance sets how fast the energy leaves; the capacitance sets how much there was. Their product, τ = R × C, is the time scale: 470 Ω × 4700 µF ≈ 2.2 s. Because energy goes with V², it drains twice as fast as the voltage falls.
Capacitors vs batteries
| Store | Energy |
|---|---|
| 4700 µF at 9 V | 0.19 J |
| Camera flash, 330 µF at 300 V | about 15 J |
| Supercapacitor, 10 F at 2.7 V | about 36 J |
| AA alkaline cell | several thousand joules |
A capacitor holds far less than a battery, but it can deliver its energy in a thousandth of a second. A camera flash charges a capacitor for a second or two and dumps it at once. Some electric buses run on supercapacitors, recharged in seconds at each stop.
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.
“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 neededWork it out
Practice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: 0.19 J. ½ × 0.0047 × 81 is 0.19 J. Small, but enough for a few seconds of fading glow, even though R1 turns most of it into heat on the way.
- Why not “0.04 J”? 0.0047 × 9 ≈ 0.04 is the charge, Q = C × V, in coulombs. Energy is ½ × C × V², which comes to about 0.19 J.
- Why not “9 J”? 9 J would need a far bigger capacitor. ½ × 0.0047 × 81 is only about 0.19 J.
- Why not “42 J”? That's over two hundred times too much. With 4700 µF as 0.0047 F, ½ × 0.0047 × 81 is only about 0.19 J.
Show the answer
Answer: Four times the energy. Energy goes as V²: twice the volts, four times the joules.
- Why not “Twice the energy”? Twice the voltage would double the charge, Q = C × V, but energy goes with V², so it's four times.
- Why not “The same energy”? Higher voltage always stores more energy in the same capacitor.
- Why not “Half the energy”? More voltage never stores less energy in the same capacitor.
Show the answer
Answer: ½ × C × V². Half C V squared.
- Why not “V ÷ C”? V ÷ C isn't a meaningful quantity here. Energy grows with both C and V.
- Why not “C ÷ V²”? Energy grows with voltage, so V can't be on the bottom. It's ½ × C × V².
- Why not “C × V”? C × V is the charge, Q, in coulombs, not the energy.
“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.
Questions people ask
What is the formula for energy stored in a capacitor?
E = ½ × C × V², in joules, with C in farads and V in volts. A 4700 µF capacitor at 9 V holds ½ × 0.0047 × 81 ≈ 0.19 J. It can also be written E = ½ × Q × V, since Q = C × V.
Why is there a half in E = ½CV²?
Because the voltage builds up as you fill it. The first charge goes in at almost 0 V and the last at the full V, so on average each coulomb is pushed in at half the final voltage: E = Q × V ÷ 2.
How does a capacitor compare with a battery?
A battery stores far more energy: an AA cell holds several thousand joules, while even a 10 F supercapacitor at 2.7 V holds about 36 J. But a capacitor can deliver or take its energy in a fraction of a second, which is why camera flashes use one.
How much of a capacitor's energy can I actually use?
Only what lies between its starting voltage and the lowest voltage your circuit works at: ½ × C × (V1² − V2²). An LED stops around 2 V, so a 4700 µF capacitor from 9 V gives up about 0.18 J of its 0.19 J; the rest stays stranded.
Related topics
- How does a capacitor work?Two plates and a gap: how a capacitor stores charge and energy, and why it blocks steady DC.
- Capacitor discharge explainedFast at first, then slower: the 37% rule, the formula, and how long a capacitor takes to empty.
- Electrical power explained: P = V × IP = V × I: how fast energy turns into heat, and why a resistor has a power rating.
- RC time constant explainedτ = R × C: how long a capacitor takes to charge through a resistor, and the 63% rule.