Energy is stored in an electric field.

A capacitor has conducting electrodes separated by an insulating dielectric. Applying a voltage causes charge separation and stores energy in the electric field.

Capacitance describes charge stored per volt, measured in farads (F). Microfarads (µF) are commonly used because one farad is a large unit for many ordinary electronic circuits.

Two electrodes, separated charge

Separated charge and electric field in a capacitorTwo conducting electrodes carry equal and opposite charges, plus Q and minus Q. An insulating dielectric separates them. Field arrows point from the positive electrode to the negative electrode; they do not show electrons crossing the dielectric.+Q−Q+−+−+−Insulating dielectricField points from + to −
  1. Positive electrode (+Q): a deficit of electrons relative to its uncharged state.
  2. Negative electrode (−Q): an equal excess of electrons in this ideal, initially neutral example.
  3. Dielectric: insulation between the conductors. The separated charges create an electric field; the whole capacitor can still have zero net charge.
Simplified parallel-plate view, not a drawing of a particular component. Edge effects and the dielectric’s microscopic polarisation are omitted. The arrows show electric-field direction, not a path for charge through the insulation.

Q = C × V describes charge separation. Here, Q is the magnitude of charge on one electrode, C is capacitance and V is the magnitude of voltage between them. It does not mean Q extra coulombs have been added to the capacitor as a whole.

For 1,000 µF at 24 V, Q = 0.001 × 24 = 0.024 C: +0.024 C on one electrode and −0.024 C on the other in this ideal example. Their sum is zero, yet energy is stored in the field.

Charging and supplying a load.

For an ideal capacitor, current flows while its voltage changes. With steady DC voltage, ideal capacitor current falls to zero; real devices can still have leakage.

“Current flows” refers to the connected circuit. In the ideal model, charge does not conduct across the insulating gap: changes in the electric field connect the electrical behaviour of the two sides. This is why a capacitor can respond to AC even though it blocks a steady DC current after charging.

Its opposition to sinusoidal AC depends on frequency and capacitance. That is capacitive reactance, part of impedance; the related distinction between energy stored and returned, and power used matters in AC circuits.

Why charging and discharging take time.

A resistor limits how quickly charge can move into or out of a capacitor. In a simple RC circuit, resistance R and capacitance C together set the rate of change. The voltage changes quickly at first, then more slowly.

The time constant: τ = R × C

The Greek letter τ (tau) means the circuit’s time constant, measured in seconds. With R = 1,000 Ω and C = 0.001 F, τ = 1 second.

After one time constant, charging has reached about 63.2% of its final voltage. In the separate discharge case, about 36.8% of the initial voltage remains.

Charging from 0 V towards 24 V

Charging through a resistorAn ideal capacitor starts at zero volts and charges towards 24 volts. At one second it reaches about 15.2 volts, or 63.2 percent of the final voltage. Both curves use a 1 kilohm resistor and 1000 microfarad capacitor, giving a one-second time constant.Capacitor voltage (V)0122401234515.2 V at 1 sTime (s)1 τ = 1 s

Discharging from an initial 24 V

Discharging through a resistorAn ideal capacitor starts at 24 volts and discharges towards zero. At one second about 8.8 volts remains, or 36.8 percent of its starting voltage. Both curves use a 1 kilohm resistor and 1000 microfarad capacitor, giving a one-second time constant.Capacitor voltage (V)012240123458.8 V at 1 sTime (s)1 τ = 1 s
Calculated ideal RC curves with the same axes. Each graph starts at its own t = 0. Both use R = 1 kΩ and C = 1,000 µF; the charging case has a constant 24 V source, while the discharging case has no source and only the resistor as its load. Leakage, ESR and other circuit paths are omitted. These are mathematical models, not discharge instructions or safe waiting times.

One time constant does not mean “finished”. After five time constants, these ideal curves are about 99.3% charged or have about 0.7% of their initial voltage remaining. A real installation may have other paths, stored energy and faults; elapsed time alone does not establish that equipment is safe.

Voltage matters as well as capacitance.

An ideal 1,000 µF capacitor at 24 V

1,000 µF = 0.001 F. Stored energy is ½ × C × V², so ½ × 0.001 × 24² = 0.288 joules.

Doubling the voltage gives four times the stored energy for the same capacitance. The voltage rating is a limit for the component, not a target that every circuit must reach.

Smoothing supplies and supporting a drive’s DC link.

In a power supply, a capacitor can support the load between rectified pulses. In a VFD, DC-link capacitors help manage the intermediate DC supply. Capacitors are also used in filtering and timing circuits.

A capacitor does not make an unlimited backup supply. Its voltage changes as energy is transferred, and its size and circuit conditions affect how long it can support a load.

Real components have limits.

Capacitors have resistance, temperature limits and ageing behaviour. Some types are polarised, while others are not. Matching capacitance alone is not enough to select a replacement.

Stored energy may remain after external power is removed. A stopped or disconnected drive is not evidence that its internal capacitors are discharged.

A closer look

Equivalent series resistance (ESR) models part of a real capacitor’s losses. Ripple current through those losses produces heat. In a simple resistor-capacitor charging circuit, the time constant is R × C; this is a mathematical model, not a safe waiting-time rule for equipment.

For the ideal charging graph, Vc(t) = Vs[1 − e^(−t/RC)] with initial capacitor voltage zero; for the discharging graph, Vc(t) = V₀e^(−t/RC). Here Vs is the fixed charging-source voltage, V₀ is the discharge starting voltage, t is elapsed time, and e is approximately 2.718. Current follows i = C dVc/dt: the steeper the voltage curve, the larger the current. In the ideal charging example it starts at Vs/R = 24 mA, then falls exponentially. Charge and voltage approach their final values asymptotically, not at a sudden cutoff.

Check your understanding

If voltage doubles on the same ideal capacitor, does stored energy double?

Reveal the explanation

It quadruples, because stored energy is proportional to voltage squared: E = ½CV².

Further reading.

Optional reading to explore this lesson in more detail.

  • TDK: capacitor construction and behaviour

    Begin with the construction diagram, then the three basic-properties sections. They explain charging and AC behaviour visually; the frequency relationship is useful further reading.

  • OpenStax: capacitors and capacitance

    The opening definitions distinguish the equal and opposite electrode charges from net charge, then introduce C = Q/V. The later geometry calculations are optional.

  • OpenStax: RC circuits

    Read the charging and discharging sections for the exponential equations and time constant. The derivation is optional; the voltage and current plots help explain why change slows over time.