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Capacitors store energy in an electric field between their plates, and inductors store it in a magnetic field around their winding. Both can release that energy quickly, which makes them essential for smoothing, filtering, timing, and pulsed-power circuits. The energy depends on the square of voltage (capacitor) or current (inductor), so doubling either quadruples the stored energy.
| Quantity | Capacitor | Inductor |
|---|---|---|
| Energy | E = ½CV² | E = ½LI² |
| Solve for value | C = 2E/V² | L = 2E/I² |
| Solve for V or I | V = √(2E/C) | I = √(2E/L) |
| Field type | Electric | Magnetic |
Energy is in joules when C is in farads, V in volts, L in henries, and I in amps.
The two storage formulas are straightforward:
Ecapacitor = ½CV2 Einductor = ½LI2
Both depend on the square of the electrical quantity, which is the part worth internalising. Doubling the voltage on a capacitor quadruples its stored energy; halving it cuts the energy to a quarter.
A 470 µF bus capacitor charged to 400 V holds ½ × 470×10−6 × 4002 = 37.6 J. For comparison, a few joules delivered across the chest is enough to be lethal, and this capacitor will hold its charge for minutes or longer after the supply is disconnected if no bleeder resistor is fitted.
This is exactly why drives and power supplies carry discharge-time warnings, why bleeder resistors are mandatory in commercial designs, and why the only safe procedure is to measure the bus voltage before touching anything — never to assume that time elapsed since switch-off is sufficient.
A 2 mH inductor carrying 20 A stores only 0.4 J, which sounds harmless beside the capacitor. The hazard is different: that energy is released in microseconds when the current path opens, and v = L di/dt means a small energy delivered extremely quickly produces a very high voltage. It is a component-destruction and arcing hazard rather than a shock hazard.
It cuts both ways usefully. A capacitor bank operated at 80% of rated voltage holds only 64% of its full energy — relevant when sizing hold-up capacitance. Conversely, raising a DC-link voltage modestly buys a disproportionate increase in ride-through time, which is often cheaper than adding capacitance.
E = ½CV², where C is in farads and V in volts, giving energy in joules. Doubling the voltage quadruples the stored energy.
E = ½LI², where L is in henries and I in amps. The energy is held in the magnetic field and depends on the square of the current.
Because the energy is the integral of instantaneous power as the voltage (or current) builds up from zero — that integration of a linear ramp gives the ½ factor.
Capacitors (especially supercapacitors) generally store far more energy per volume than inductors, which is why energy storage usually uses capacitors or batteries rather than inductors.
It can be — a large capacitor at high voltage (like a camera flash or power-supply bulk cap) stores enough energy to give a serious shock or burn. Always discharge before handling.
The inductor tries to maintain its current, producing a large voltage spike (the inductive kick). A freewheeling diode or snubber safely absorbs this energy.
Ideal ones do not, but real capacitors leak slowly and real inductors have winding resistance, so the energy gradually dissipates as heat.
Through a suitable resistor (not a direct short, which can damage it or arc). The resistor limits the current while the energy dissipates as heat.
The joule (J). One joule is one watt for one second; the calculator also shows mJ, µJ, and nJ for small values.
The inductor stores energy from the input during one part of the cycle and releases it to the output during the next — the ½LI² per cycle times frequency sets the power handled.
Yes — it can also be written E = ½QV = Q²/2C, since charge Q = CV. All three forms give the same energy.
Because both the stored charge (or flux) and the driving voltage (or current) rise together, so their product — the energy — grows as the square.
E = ½CV², in joules when capacitance is in farads and voltage in volts. A 470 µF capacitor at 400 V stores 37.6 J.
E = ½LI², in joules when inductance is in henries and current in amperes. A 2 mH inductor carrying 20 A stores 0.4 J.
Because it retains its charge. A 470 µF capacitor at 400 V holds 37.6 J, far more than enough to be lethal, and without a bleeder resistor it can stay charged for minutes. Always measure the bus voltage before working on the circuit.
A resistor permanently across a capacitor that discharges it safely once the supply is removed. It is sized to bring the voltage down within a stated time while dissipating acceptably little power in normal operation.
Because energy depends on the square of voltage. The same relationship means a modest increase in DC-link voltage gives a disproportionately large increase in hold-up energy.
They are dangerous in different ways. The capacitor holds far more energy and presents a genuine electric-shock hazard. The inductor holds less but releases it in microseconds, producing very high voltages that destroy semiconductors and cause arcing.
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