Capacitors

Capacitors

A capacitor stores energy in an electric field between two conducting plates separated by an insulator (dielectric). Capacitors are essential components in electronics for timing, filtering, and energy storage.

Capacitance

Capacitance, C, is the charge stored per unit potential difference:

C = Q/V

where:

• C = capacitance (farads, F)
• Q = charge stored (coulombs, C)
• V = potential difference (volts, V)

1 farad is very large; typical values are μF (10⁻⁶), nF (10⁻⁹), or pF (10⁻¹²).

Energy Stored in a Capacitor

The energy stored equals the area under the V-Q graph (which is a straight line through the origin):

E = ½QV = ½CV² = Q²/(2C)

Derivation

When a small charge dQ is added at voltage V: dW = V dQ = (Q/C) dQ

Total work: W = ∫₀ᵠ (Q/C) dQ = Q²/(2C) = ½CV²

Worked Example

A 470 μF capacitor is charged to 12 V. Find the charge stored and the energy stored.

Q = CV = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C = 5.64 mC

E = ½CV² = ½ × 470 × 10⁻⁶ × 12² = 0.0338 J = 33.8 mJ

Capacitors in Parallel

Capacitors in parallel share the same voltage. Total charge = sum of individual charges:

C_total = C₁ + C₂ + C₃ + ...

(Opposite rule to resistors in parallel.)

Capacitors in Series

Capacitors in series store the same charge. Total voltage = sum of individual voltages:

1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + ...

For two capacitors: C_total = C₁C₂/(C₁ + C₂)

Charging a Capacitor Through a Resistor

When a capacitor charges through a resistor from a supply voltage V₀:

Q = Q₀(1 − e^(−t/RC)) where Q₀ = CV₀

V = V₀(1 − e^(−t/RC))

I = I₀ e^(−t/RC) where I₀ = V₀/R

Discharging a Capacitor Through a Resistor

Q = Q₀ e^(−t/RC)

V = V₀ e^(−t/RC)

I = I₀ e^(−t/RC)

All quantities decay exponentially during discharge.

The Time Constant

τ = RC (units: seconds)

where R is in ohms and C in farads.

The time constant is the time for the charge (or voltage, or current) to fall to 1/e ≈ 37% of its initial value during discharge, or to reach 63% (= 1 − 1/e) of its final value during charging.

After 5 time constants (5RC), the capacitor is considered fully charged or discharged (> 99%).

Time	Fraction remaining (discharge)
1τ	0.368 (37%)
2τ	0.135 (14%)
3τ	0.050 (5%)
5τ	0.007 (0.7%)

Worked Example — Discharge

A 220 μF capacitor is charged to 9.0 V and then discharged through a 47 kΩ resistor. Find the voltage after 15 s.

τ = RC = 47 × 10³ × 220 × 10⁻⁶ = 10.34 s

V = V₀ e^(−t/RC) = 9.0 × e^(−15/10.34) = 9.0 × e^(−1.451) = 9.0 × 0.234 = 2.11 V

Determining the Time Constant Experimentally

Method 1 — Direct: Plot V against t during discharge. Read off the time when V = V₀/e = 0.37V₀.

Method 2 — Logarithmic: Take natural logs of V = V₀ e^(−t/RC):

ln V = ln V₀ − t/(RC)

Plot ln V against t: straight line with gradient = −1/(RC) and y-intercept = ln V₀.

Factors Affecting Capacitance

For a parallel plate capacitor:

C = ε₀εᵣA/d

where:

• ε₀ = permittivity of free space
• εᵣ = relative permittivity of the dielectric
• A = area of overlap of the plates (m²)
• d = separation of the plates (m)

Increasing capacitance: larger plates, smaller separation, higher permittivity dielectric.

Dielectrics

A dielectric is an insulating material placed between the plates. It increases capacitance by a factor of εᵣ (relative permittivity).

The dielectric works by polarisation: the electric field between the plates causes a slight displacement of charge in the dielectric molecules, creating an opposing field that partially cancels the applied field, allowing more charge to be stored at the same voltage.

Applications

• Smoothing circuits: Capacitors smooth the output of rectified AC in power supplies
• Timing circuits: The RC time constant controls delays in 555 timers and similar circuits
• Coupling and decoupling: Block DC while passing AC signals
• Camera flash: Rapid discharge of stored energy
• Defibrillators: Store and rapidly release energy to restart the heart (typically 200–360 J)