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Lessons · Engineering · energy in a capacitor

A capacitor stores energy, and the energy goes as the square of the volts

A capacitor C charged to V holds charge Q = C V and energy E = ½ C V².

Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.

In beta. This lesson was written for Hone and has not yet been checked by a licensed engineer. Practice material, not professional advice. What that means.

What it is for

A drive's DC bus -- direct current -- capacitors are still charged ten minutes after the power is off, and the energy in them is enough to weld a screwdriver. The number on the warning label came from this line.

How to think about it

Convert microfarads to farads first: 470 µF is 470e-6 F. Then a half, times C, times the volts squared. Doubling the voltage quadruples the energy, which is why high-voltage capacitors are the dangerous ones.

Worked example

E = ½ C V²; also Q = C V
Energy and charge. Both need C in farads.
A 470 µF capacitor charged to 24 V
The numbers on the can and the meter.
E = 0.5 × 470e-6 × 24² = 0.5 × 470e-6 × 576 = 0.1354 J
About an eighth of a joule.
At 48 V: 0.5 × 470e-6 × 48² = 0.5414 J, four times as much
Twice the volts, four times the energy. The square is the whole warning.

Your turn

1000 µF at 10 V. Write the energy.

E = 0.5 × 1000e-6 × ² = 0.05 J

The trap

Leaving C in microfarads, or forgetting the half. 470 × 576 is 270,720 of nothing. Convert to farads first, and remember the half is not decoration.

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