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Electronics & RFIEC 62368-1 • NFPA 70E • IEEE 18

Capacitor Stored Energy

Enter capacitance, voltage and a discharge resistor to get joules, charge, peak current and safe discharge time.

E = ½ C V²  •  Q = C V  •  t = τ ln(V / 50)
Calculated Result
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Step-by-step

  1. Enter valid values to begin.

Ideal capacitor charged to V, discharged through an ideal resistor (no ESR, leakage or dielectric absorption). Hazard thresholds (under 1 J low, 1–20 J caution, over 20 J high) are a conservative rule of thumb only, not a standard; IEC 62368-1 defines energy source classes and its own limits. 50 V is used as a commonly applied safe-touch level for the discharge target — check your applicable standard.

IEC 62368-1 • NFPA 70E • IEEE 18 • Stored Energy Safety

Capacitor Stored Energy: How Much Can It Actually Hurt You?

Core Engineering Principles

Energy in a capacitor is E = ½CV², and the square on the voltage is what bites. Double the voltage and you store four times the energy, which is why a 470 µF capacitor on a 400 V inverter DC bus holds far more than a 4700 µF part on a 24 V rail. We see the same thing in photoflash units: a few hundred microfarads charged to a few hundred volts, and the stored joules dump into a tube in microseconds. A bus capacitor shorted by a screwdriver is limited only by loop resistance, so expect a crater in the blade and a flash.

Treat “off” as a claim you must verify. Power-factor-correction and DC-bus capacitors rely on a bleeder resistor, often a few hundred kΩ, so the decay time constant can run to minutes, and a failed or open bleeder leaves the full charge sitting there. Dielectric absorption makes it worse: a capacitor shorted briefly can recover part of its voltage once the short is removed, so a “dead” cap can come back and bite. Before service we discharge through a resistor, measure, then leave a shorting strap on.

E = ½ × C × V²  •  Q = C × V  •  Ctotal = N × C (parallel)
Ipeak = V / R  •  τ = R × C  •  t = τ × ln(V / 50)  •  Ppeak = V² / R

NEC & Standard References

IEC 62368-1 (audio/video, information and communication technology equipment) classifies energy sources and sets the safeguards, including decay of stored charge after the mains is removed; use its classes, not this page’s thresholds. NFPA 70E requires establishing an electrically safe work condition and verifying absence of voltage, and treats stored energy as a hazard that must be released or restrained. IEEE Std 18 covers shunt power capacitors, including their internal discharge resistors and discharge time requirements. Check the adopted edition.
Worked Example: Inverter DC Bus Before a Service Call
Given: one 470 µF capacitor, 400 V, discharge resistor 100 Ω.
1. E = ½ × 470 × 10⁻⁶ × 400² = 37.6 J (37,600 mWs, about 0.0104 Wh).
2. Q = 470 × 10⁻⁶ × 400 = 188 mC.
3. Peak current V / R = 400 / 100 = 4 A; peak power V² / R = 1600 W, so use a pulse-rated resistor.
4. τ = 100 × 470 µF = 47 ms.
5. Time to fall to 50 V = τ × ln(400 / 50) = 47 × ln 8 = 97.7 ms; 5τ = 235 ms.
6. The 37.6 J sits above the 20 J rule-of-thumb line, so the page flags it as high hazard. The 100 Ω resistor works in about a tenth of a second; the unit’s bleeder may take minutes.
Safety & Installation Rules
  • Never short with a screwdriver. It welds and sprays molten metal; use a resistor on insulated leads.
  • A “discharged” cap can recharge. Dielectric absorption brings the voltage back after a short, so re-measure after a few minutes.
  • Series banks drift. Without balancing resistors, leakage mismatch overloads one capacitor, so check each cell separately.
  • Ripple current heats the part. ESR times ripple squared shortens electrolytic life.
  • Derate voltage. Run capacitors well below rating; this page’s thresholds are a rule of thumb, not a code limit.