Capacitors

AC Circuits · 13 min read

A capacitor is, at heart, two pieces of metal separated by a thin insulator. Connect a voltage across them and positive charge piles up on one plate, an equal negative charge on the other, and an electric field forms in the gap. That charge-store property is what makes the capacitor the second most-used passive component after the resistor — and the foundation for everything AC.

Structure

The canonical “parallel-plate” model is two conductive plates of area $A$ , separated by a distance $d$ , with an insulating material (the dielectric) filling the gap. In real components this gets rolled up, coiled, stacked, or sintered into shapes that fit on a PCB, but the mental model stays the same.

Two conductive plates of area A separated by a dielectric of relative permittivity εr, distance d apart. Connect a voltage V and positive charge +Q piles onto one plate, −Q onto the other — the field between the plates stores energy.

Factors that affect capacitance

For the parallel-plate geometry, three things set $C$ :

C = \varepsilon_0 \, \varepsilon_r \, \frac{A}{d}

Plate area $A$ — more metal facing each other means more room for charge to spread. Double the area, double the capacitance.
Plate separation $d$ — closer plates see a stronger field per volt, so they pull more charge for the same voltage. Halve the gap, double the capacitance.
Dielectric permittivity $\varepsilon_r$ — the material between the plates. Air / vacuum is $\varepsilon_r = 1$ . Paper is ~3.5, mica is ~6, ceramic can run into the thousands. $\varepsilon_0$ is a universal constant ( $8.854 \times 10^{-12}$ F/m).

Three knobs set the capacitance. Double the plate area or halve the gap and C doubles. Swap air for mica and C rises by the relative permittivity εr.

Why do different dielectrics help? The insulating material polarises in the field — its own molecules tilt so their internal charges partly cancel the applied field. A weaker net field for the same charge means you can pack more charge in for the same terminal voltage, which is exactly what bigger C means.

Capacitance is a fixed property of the physical construction — set by geometry (A, d) and material (

\varepsilon_r

), not by the circuit around it. Once you know C, a single formula links it to charge and voltage — which we introduce after the series and parallel rules.

Capacitors in series and parallel

Combine two caps and the equivalent capacitance depends on how they're wired. The rules are the opposite of what resistors do:

\text{Series:} \quad \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots

\text{Parallel:} \quad C_{eq} = C_1 + C_2 + \ldots

Capacitors swap the series/parallel rules you learned for resistors. Series = reciprocal sum (smaller than the smallest cap). Parallel = straight sum (larger than the largest cap).

Series — caps share the same current (and therefore the same charge Q). With identical charge on every capacitor, the voltages distribute inversely with capacitance: $V_k = Q / C_k$ . Adding those voltages gives the total, and dividing by Q gives the reciprocal-sum rule. A practical consequence: series capacitors behave like a single cap smaller than the smallest element. Two 10 µF caps in series give 5 µF.

Parallel — caps share the same terminal voltage V. Each pulls its own charge $Q_k = C_k V$ , and the total charge delivered by the source is the sum of the branches. So the effective C is just the straight sum. Two 10 µF caps in parallel make 20 µF.

The swap with resistors trips people up the first time. The memorable way: in a series cap network the same charge moves through everything but the voltage has to be divided — so more elements in series can hold less charge per volt of total, i.e. smaller C. In parallel, every branch gets its own charge at the same V, so more branches add up to a bigger C.

Resistors sum in series, reciprocate in parallel. Capacitors do the reverse. Series caps: reciprocal sum. Parallel caps: direct sum. The bigger cap always wins in parallel; the smaller cap always wins in series.

The charge-voltage relationship

Apply a voltage $V$ across the plates. Charge $Q$ accumulates in direct proportion. The constant of proportionality is the capacitance:

Q = C V

$C$ is measured in farads (F): one farad stores one coulomb of charge per volt applied. One farad is enormous — real capacitors span picofarads (pF, $10^{-12}$ F) up to a few farads for supercaps, a range of 12 decades. Typical ceramics are pF–nF; typical electrolytics µF–mF. When you see a “100 µF 16 V” cap in a schematic, that's its capacitance and its voltage rating (the maximum it can handle without breakdown).

Rearranging $Q = CV$ gives two equally useful forms:

V = \frac{Q}{C}, \qquad C = \frac{Q}{V}

For a given cap, more charge pushed in means a higher terminal voltage — same relationship as filling a bucket faster raises the water level faster. For a given voltage, a bigger $C$ holds more charge — a bigger bucket holds more water.

Applying Q = CV in circuits

With both the combination rules and the formula in hand, you can find the charge on every capacitor in a battery-driven network. The method is always the same: collapse combinations to a single $C_{eq}$ , find total $Q$ from $Q = C_{eq} \, V$ , then work backwards to individual charges and voltages.

Single capacitor

The simplest case: one capacitor and one battery. When the circuit reaches steady state the capacitor voltage equals the battery voltage exactly, so $Q = C \times V$ is applied directly.

The battery pumps charge onto the plates until V_C = 9 V. Q = CV = 100 µF × 9 V = 900 µC on each plate.

Two capacitors in parallel

Parallel caps share the same voltage — each is connected directly across the battery. First find $C_{eq} = C_1 + C_2$ , then find the total charge $Q_{total} = C_{eq} \, V$ . Each individual charge follows $Q_k = C_k \, V$ since both caps see the same voltage.

Both caps charge to 9 V. C_eq = 47 + 100 = 147 µF. Q₁ = 47 µF × 9 V = 423 µC; Q₂ = 100 µF × 9 V = 900 µC; Q_total = 1.32 mC.

Combination: series and parallel

When series and parallel elements mix, reduce in stages: collapse the parallel group first, then treat the result as a single cap in series with the remaining element.

Reduce the parallel pair first (400 µF), then apply the series formula with C₁. C_eq = 80 µF. The same charge Q = 960 µC flows through C₁ and splits between C₂ and C₃.

After finding $Q = 960\,\mu\text{C}$ , note that this is the charge on $C_1$ (series elements carry the same Q). The voltage across the parallel pair is $V_{23} = Q / C_{23} = 960 / 400 = 2.4\,\text{V}$ , and $V_1 = Q / C_1 = 960 / 100 = 9.6\,\text{V}$ — which sums to 12 V as expected.

The systematic approach: (1) reduce parallel groups — direct sum; (2) reduce series groups — reciprocal sum; (3) apply

Q = C_{eq} \, V

for total charge; (4) back-substitute to find individual charges and voltages using

Q = C_k V_k

Charging and discharging

Everything above describes a capacitor sitting in steady state. The interesting behaviour happens during the transition: the moment a switch closes and the capacitor first sees a voltage source, or the moment the source is removed and the cap dumps its charge through whatever path is available.

Put a resistor $R$ in series with a capacitor $C$ and a DC source $V_s$ , then close the switch. The voltage across a capacitor can't jump instantaneously — that would require infinite current — so $v_C$ ramps up from zero toward $V_s$ along an exponential curve:

v_C(t) = V_s \left( 1 - e^{-t/\tau} \right)

The whole shape of that curve is governed by a single number, the time constant:

\tau = R \cdot C

With $R$ in ohms and $C$ in farads, $\tau$ comes out in seconds. A 10 kΩ resistor charging a 100 µF cap gives $\tau = 1\,\text{s}$ ; a 1 MΩ × 1 nF combination gives $\tau = 1\,\text{ms}$ . Pick R and C, you've picked your timing.

The 63 % rule. At

t = \tau

, a charging capacitor has reached

V_s (1 - 1/e) \approx 0.632 \, V_s

— just shy of two-thirds of the way to the target. That number is universal: it doesn't depend on R, C, or V_s.

Engineers treat the cap as fully charged after about five time constants. At $t = 5\tau$ the cap has reached roughly 99.3 % of $V_s$ — close enough to call it done.

The current through the same circuit follows the mirror curve. At the instant the switch closes, the cap looks like a short to a sudden change; current jumps to $V_s/R$ and then decays as $v_C$ climbs and the voltage left across R shrinks:

i(t) = \frac{V_s}{R} \, e^{-t/\tau}

Charging shape — v_C rises along $V_s(1 - e^{-t/\tau})$ ; current decays as v_C catches up. The full interactive playground lives on the Simulate page.

Discharging is the symmetric story. Disconnect the source and short the cap through R; v_C decays exponentially from its starting value $V_0$ toward zero with the same time constant:

v_C(t) = V_0 \, e^{-t/\tau}

Same $\tau$ , same five-tau rule for “effectively empty.” The current reverses direction (out of the cap instead of in) and decays alongside the voltage.

Discharging shape — v_C decays along $V_0\,e^{-t/\tau}$ ; current reverses out of the cap. Same τ controls both halves.

Why this matters. Anything that charges or discharges a capacitor — power-supply soft-start, RC lowpass filters, RC oscillators, sample-and-hold circuits, switch-debounce networks — runs on the same exponential. Head to the Simulate page for the full interactive plotter with sliders, animation, and a scrub cursor.