Energy Stored in Capacitors and Applications

Capacitors store energy in the electric field between their plates. The mathematical expression for that energy — derivable in three equivalent forms — underpins a host of real-world devices, from camera flashes and pacemaker defibrillators to regenerative-braking systems in electric vehicles. This lesson develops the energy-storage formulas from first principles, explores the geometry of the Q–V graph, and works through quantitative examples drawn from clinical, photographic and automotive applications.

Spec mapping (AQA 7408 §3.7.4 — Energy stored by a capacitor): This lesson covers the three equivalent expressions for capacitor energy, U = ½QV = ½CV² = Q²/(2C), and their derivation from the integral W = ∫V dQ; the geometric interpretation as the area under the Q–V graph; and applications of capacitor energy storage, including defibrillators, camera flash, automotive jump-start capacitor packs, supercapacitors and pulsed-power systems. Charge-discharge dynamics are picked up in Lessons 5 and 7. Refer to the official AQA 7408 specification document for the exact wording.

Synoptic links — where this resurfaces in 7408:

• §3.7.4 Capacitance (Lesson 4 here) — the V = Q/C linear relationship is the algebraic root of the integration.
• §3.5 Electricity — energy supplied by a source W = QV (full charge through full PD) vs energy stored in the capacitor ½QV: the factor of two is the lost-half paradox examined in Lesson 4.
• §3.4.1.7 Energy and power — ½CV² is structurally identical to ½kx² for a spring (electrical compliance ↔ mechanical compliance; voltage ↔ displacement); the energy-as-area-under-graph derivation parallels work done stretching a spring.
• §3.7.5 Magnetic fields / inductors (extension) — inductor energy U = ½LI² is the magnetic analogue; LC oscillator energy oscillates between electric and magnetic forms.
• §3.2.1 Photoelectric effect — flash tubes in cameras and stroboscopes are gas-discharge devices powered by capacitor pulses; the photon flux delivered ties to the §3.2 photon-energy framework.
• §3.4.1.4 Newton's laws / §3.4.1.5 Impulse — defibrillator current pulses deliver well-defined impulses to the heart; the impulse-on-the-thorax framing connects energy storage to clinical physiology (extension reading).

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Three Equivalent Expressions for Stored Energy

When charge Q has been transferred onto a capacitor of capacitance C, the work done by the external source against the rising potential difference V is stored as energy in the electric field of the capacitor. The result is three equivalent expressions:

U = ½QV = ½CV² = Q²/(2C)

All three are equal because Q, V and C are linked by the defining relation C = Q/V. Use the form that matches the data you are given:

Given	Use
Q and V	U = ½QV
C and V	U = ½CV²
Q and C	U = Q²/(2C)

The factor of one-half is the headline result of this lesson — it is the source of much of the cleverness in capacitor circuits and the origin of several A* exam questions.

Derivation by Integration

The work done by an external source in transferring an additional small charge dQ onto a capacitor that already carries charge Q at potential V is:

dW = V dQ

But V = Q/C at this stage in the charging process, so:

dW = (Q/C) dQ

Integrating from Q = 0 to the final charge Q_f:

W = ∫₀^Q_f (Q/C) dQ = (1/C) × [Q²/2]₀^Q_f = Q_f²/(2C)

This is the energy stored — the third of our three forms. Substituting Q_f = CV_f:

W = (CV_f)²/(2C) = ½CV_f² (second form)

Substituting Q_f = C V_f again (i.e. V_f = Q_f/C):

W = ½Q_f × (Q_f/C) = ½Q_f × V_f / 1 = ½Q_f V_f (first form, after recognising V_f = Q_f/C).

The three forms are algebraically identical; any two of {Q, V, C} suffice to compute U.

Geometric Interpretation — Area Under the Q–V Graph

If you plot Q on the y-axis and V on the x-axis (or vice versa) for an ideal capacitor, you get a straight line through the origin with gradient C (or 1/C, depending on which way round). The area under this straight line between Q = 0 and Q = Q_f is a triangle of base V_f and height Q_f:

Area = ½ × base × height = ½ × V_f × Q_f = ½Q_f V_f = U.

The area-under-graph reading is one of the simplest geometric proofs of the ½-factor: by contrast, the full energy supplied by a constant-voltage source during charging is Q_f V_f (a rectangle of width V_f and height Q_f, not a triangle), so half the supplied energy is dissipated as heat in the charging circuit (the famous "lost-half" paradox).

graph LR
    A["Source supplies<br/>energy QV"] --> B["Capacitor stores<br/>energy ½QV"]
    A --> C["Resistor dissipates<br/>energy ½QV as heat"]

    style A fill:#3b82f6,color:#fff
    style B fill:#27ae60,color:#fff
    style C fill:#e74c3c,color:#fff

This energy bookkeeping holds regardless of R: even in the limit of zero resistance, exactly half the energy supplied by the source ends up dissipated (radiated, in the idealised R = 0 case). This is a direct consequence of the linear V = Q/C relationship and is examined synoptically with conservation of energy.

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Worked Examples

Worked Example 1 — Energy in a Smoothing Capacitor

Question: A 10 000 μF electrolytic capacitor in a DC power supply is charged to 15 V. Calculate the energy stored.

Solution:

U = ½CV² = ½ × 10 000 × 10⁻⁶ × 15² = ½ × 10 000 × 10⁻⁶ × 225 = 1.125 J ≈ 1.1 J

Check with Q²/(2C): Q = CV = 10 000 × 10⁻⁶ × 15 = 0.15 C U = Q²/(2C) = (0.15)² / (2 × 10 000 × 10⁻⁶) = 0.0225 / 0.020 = 1.125 J ✓

A 1-joule capacitor, if discharged through a short-circuit, would deliver a brief peak current of many amperes — enough to weld the contacts of an under-rated switch. This is why power supplies are typically fitted with bleeder resistors to discharge smoothing capacitors safely when power is removed.

Worked Example 2 — Defibrillator Energy and Charge

Question: A medical defibrillator stores 360 J of energy at a charging voltage of 4500 V. Calculate (a) the required capacitance; (b) the charge stored on each plate; (c) the average power delivered if the energy is dumped into the patient over 5.0 ms.

Solution:

(a) U = ½CV² → C = 2U/V² = (2 × 360) / (4500)² = 720 / 2.025 × 10⁷ = 3.56 × 10⁻⁵ F = 35.6 μF

(b) Q = CV = 3.56 × 10⁻⁵ × 4500 = 0.160 C = 160 mC

(c) Average power P = U / Δt = 360 / 5.0 × 10⁻³ = 7.2 × 10⁴ W = 72 kW

The 72 kW peak power is roughly 100 times the continuous power that a mains socket can supply (typical UK socket: ~3 kW). This illustrates the headline strength of capacitor energy storage: not raw stored energy (a battery can store far more), but the ability to release that energy on a very short timescale at high peak power. A car battery and a defibrillator capacitor might store comparable total energies, but only the capacitor can dump it in milliseconds.

Clinical note: Modern biphasic defibrillators (the rectangular waveform variant) typically dose at 120–200 J for adult cardiac arrest; the 360-J figure used here is closer to the older monophasic devices. The energy is delivered to the patient — the energy stored on the capacitor must be somewhat larger to allow for losses in the transformer and electrodes. AQA exam questions tend to use rounded representative figures (~360 J) without further qualification.

Worked Example 3 — Camera Flash Energy and Recharge Time

Question: A camera flash uses a 1500 μF capacitor charged to 300 V. (a) Calculate the energy stored in the capacitor. (b) If the flash takes 2.0 ms to discharge and converts 30% of the stored energy to visible light, calculate the peak visible-light optical power. (c) If the battery supplying the camera can deliver 1.5 W of charging power, estimate the minimum time between flashes.

Solution:

(a) U = ½CV² = ½ × 1500 × 10⁻⁶ × 300² = ½ × 1500 × 10⁻⁶ × 9.0 × 10⁴ = 67.5 J

(b) Visible-light energy delivered = 0.30 × 67.5 = 20.25 J Peak visible-light power ≈ 20.25 / 2.0 × 10⁻³ = 1.0 × 10⁴ W = 10 kW

(c) Minimum recharge time = U / P_battery = 67.5 / 1.5 = 45 s

In practice, professional flash units use slightly less energy (~40–80 J) and high-voltage rapid-recharge circuitry to cut the recharge interval to ~5 s. The 45-s figure here is consistent with cheap point-and-shoot cameras of the early digital era.

Worked Example 4 — Supercapacitor Energy Density

Question: A 100 F commercial supercapacitor is rated at 2.7 V maximum. (a) Calculate the maximum energy it can store. (b) If the cell weighs 250 g, calculate its energy density in Wh kg⁻¹. (c) Compare with a typical lithium-ion cell at ~200 Wh kg⁻¹.

Solution:

(a) U_max = ½CV² = ½ × 100 × (2.7)² = ½ × 100 × 7.29 = 364.5 J ≈ 360 J

(b) Energy in watt-hours: U = 364.5 / 3600 = 0.1013 Wh. Energy density = 0.1013 / 0.250 = 0.405 Wh kg⁻¹

(c) Lithium-ion is ~200 Wh kg⁻¹, so the supercapacitor has roughly 500× lower energy density by mass — it stores a fraction of the energy per kilogram of an equivalent battery.

However, supercapacitors can charge and discharge in seconds rather than hours, can deliver peak power densities of ~10 kW/kg (vs ~1 kW/kg for lithium-ion), and survive ~10⁶ charge-discharge cycles (vs ~10³ for lithium). The trade-off — low energy density, high power density — makes them ideal for regenerative braking (where energy must be recovered in seconds during deceleration) and load-levelling in hybrid drivetrains, but unsuitable as a primary energy store for an EV's main propulsion battery.

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Application Case Study — The Automated External Defibrillator

A modern Automated External Defibrillator (AED) is the clearest practical example of capacitor energy storage in daily life. The device is now standard in airports, shopping centres, gymnasiums and railway stations.

Operating principle. When a victim of cardiac arrest is found in ventricular fibrillation, a controlled electric pulse of well-defined energy and waveform delivered through the chest can depolarise the heart muscle and allow the natural sinoatrial pacemaker to restart. The energy must be:

1. Large enough to depolarise the cardiac myocytes (typical 120–360 J at the electrodes).
2. Delivered fast enough to produce a brief, intense current pulse (peak current ~30 A; pulse duration 4–10 ms).
3. Controlled in waveform — modern biphasic waveforms reverse polarity midway through the pulse, reducing the energy required and the risk of post-shock myocardial damage.

A 9-V battery alone cannot deliver this — its internal resistance limits peak current to a fraction of an amp. The solution is to charge a capacitor of ~30–100 μF to ~2–5 kV using a DC–DC converter, then dump the stored energy through the patient electrodes via a thyristor switch.

Energy bookkeeping.

Stage	Energy / J	Time / s
Battery → capacitor (charging)	~50 (lost in converter) + ~360 (stored)	~10–20
Capacitor → patient (discharging)	~360 stored → ~200–300 delivered (rest in cables, ~10% in electrodes, ~10% in transformer)	~0.005

The peak power delivered to the patient (~72 kW from Worked Example 2) is about 24 000 times the average power drawn from the battery during charging (~3 W). The capacitor acts as a time-compression element: it pulls energy in over tens of seconds and releases it over milliseconds.

Safety considerations. A charged 360-J capacitor at 4500 V is a serious hazard: