What is the formula for energy stored in a capacitor?

The standard formula is E = ½ × C × V², where C is capacitance in farads and V is voltage in volts. Two equivalent forms are E = Q² / (2C) and E = ½ × Q × V, where Q = C × V is the stored charge. All three give the same result and all three are shown in the working section.

Why is the energy formula ½CV² and not CV²?

When you charge a capacitor, the voltage across it rises from 0 to V as charge builds up. Early charge flows against almost no voltage (little work done); later charge flows against the full voltage (more work done). Integrating P = V × dQ/dt across the whole charging process gives exactly half the energy you would get if the full voltage were present throughout — hence the ½ factor.

How do I calculate total capacitance for capacitors in parallel?

In parallel all capacitors share the same terminal voltage. Total capacitance simply adds: C_total = C1 + C2 + ... + Cn. Because each capacitor charges to V, total stored energy is E = ½ × C_total × V². This is always greater than any individual capacitor alone.

How do I calculate total capacitance for capacitors in series?

In series the same charge Q flows through every capacitor. The reciprocals add: 1/C_total = 1/C1 + 1/C2 + ... + 1/Cn. The result is always smaller than the smallest individual capacitor. Each capacitor shares a different fraction of the total voltage, proportional to its own charge-to-capacitance ratio.

What happens to the energy when a capacitor discharges through a resistor?

The voltage decays exponentially: Vt = V₀ × e^(−t/RC), where RC is the time constant τ. The energy at any moment is E(t) = ½ × C × Vt². The energy that has left the capacitor is converted to heat in the resistor. After 5τ less than 0.7% of the original energy remains — the conventional "fully discharged" threshold.

What is the RC time constant and why does it matter?

τ = R × C (in seconds when R is in ohms and C in farads). It sets the timescale of charging and discharging. After one τ the capacitor has discharged to about 36.8% of its original voltage (lost about 86.5% of its energy). Engineers use τ to size filter capacitors, set oscillator frequencies, and limit inrush current in power supplies.

Capacitor Energy Calculator

The capacitor energy calculator covers the three quantities engineers and students reach for most often: the energy stored in a charged capacitor, the equivalent capacitance of a series or parallel network, and the voltage and power profile as a capacitor discharges through a resistive load. Everything runs in your browser — no data is uploaded or stored.

How it works

Energy stored (Tab 1)

A capacitor stores energy in the electric field between its plates. The governing formula is:

E = ½ × C × V²

where C is capacitance in farads and V is the voltage across the plates. Because charge Q = C × V, two equivalent forms are useful:

E = Q² / (2C) and E = ½ × Q × V

All three forms are shown in the working section so you can cross-check. Results are also expressed in watt-hours (Wh) and kilowatt-hours (kWh) for comparison with battery-sized storage.

Series and parallel networks (Tab 2)

When capacitors are combined:

Parallel — all share the same voltage; capacitances add directly: C_total = C1 + C2 + … + Cn. Total energy is ½ × C_total × V².
Series — all carry the same charge Q; reciprocals of capacitances add: 1/C_total = 1/C1 + 1/C2 + … + 1/Cn. Each capacitor develops a different share of the total voltage.

The per-capacitor breakdown table shows the charge and energy held by each individual capacitor after solving the network.

RC discharge and power (Tab 3)

When a charged capacitor is discharged through a resistor R, the voltage decays as:

Vt = V₀ × e^(−t/RC)

The energy remaining at time t is:

E(t) = ½ × C × Vt²

The peak current occurs at t = 0 when the full voltage appears across R: I_peak = V₀ / R. Average power over the elapsed interval equals the energy dissipated divided by the time: P_avg = (E₀ − E(t)) / t.

Worked example

A 1 000 µF capacitor charged to 400 V (typical in a camera flash or power-factor correction bank):

Stored energy: E = 0.5 × 1 000×10⁻⁶ × 400² = 80 J
Equivalent to 80 J ÷ 3 600 = 0.0222 Wh

If that capacitor discharges into a 100 Ω load:

Time constant τ = 100 Ω × 1 000 µF = 0.1 s
After 100 ms (one τ) the voltage has dropped to 400 × e⁻¹ ≈ 147 V
Remaining energy: 0.5 × 1 000×10⁻⁶ × 147² ≈ 10.8 J — about 13.5% of the original
Average power over that 100 ms: (80 − 10.8) / 0.1 = 692 W

Capacitance	Voltage	Energy
100 µF	12 V	7.2 mJ
470 µF	5 V	5.875 mJ
1 000 µF	400 V	80 J
10 mF	2.7 V	36.45 mJ

Formula note

The factor of ½ in E = ½CV² arises because charging is not done at constant voltage. As the first element of charge dQ flows onto an empty plate it encounters zero opposing voltage; the last element dQ flows against the full voltage V. Integrating V × dQ from 0 to Q = CV gives E = ½CV². This is analogous to the ½mv² kinetic-energy formula and for the same mathematical reason: both involve integrating a linearly increasing force (or voltage) over a displacement (or charge).