Chapters/Ch 6· intermediate

Capacitors and Inductors

Everything for Capacitors and Inductors in one place — 6 topics, 10 key formulas, a live simulator and a 10-question quiz.

01Capacitor: Structure & v-i Relationship

A capacitor is two conducting plates separated by a non-conducting dielectric. When voltage is applied, positive charge accumulates on one plate and equal negative charge on the other, storing energy in the electric field between them.

Capacitance is the ratio of stored charge to applied voltage:

$C = \frac{Q}{V}$

Unit: farad (F), where $1 F = 1 C/V$ . One farad is huge — practical values are in µF (microfarad, $1 0^{- 6}$ ), nF ( $1 0^{- 9}$ ), and pF ( $1 0^{- 12}$ ).

Parallel-plate formula:

$C = ε \frac{A}{d} = ε_{0} ε_{r} \frac{A}{d}$

where $A$ is the plate area, $d$ is the plate separation, $ε_{0} = 8.854 \times 1 0^{- 12}$ F/m is the permittivity of free space, and $ε_{r}$ is the relative permittivity of the dielectric (1 for air; ~2.5 for polyester film; ~7 for mica; ~10 000 for some ceramics).

The fundamental v-i relationship:

$i = C \frac{d v}{d t}$

Current flows only while voltage is *changing*. Three immediate consequences:

DC steady state: $\frac{d v}{d t} = 0$ , so $i = 0$ . A capacitor behaves as an open circuit to DC.
Voltage cannot change instantaneously: a step change in $v$ would require infinite current. So $v_{C} (0^{+}) = v_{C} (0^{-})$ across any switching event.
AC reactance: in a sinusoidal AC circuit, the higher the frequency, the more current flows for the same voltage — capacitors *pass* high frequencies and *block* DC.

Integrating the v-i relationship gives the voltage in terms of past current:

$v (t) = \frac{1}{C} \int_{- \infty}^{t} i (τ) d τ = v (0) + \frac{1}{C} \int_{0}^{t} i (τ) d τ$

The capacitor 'remembers' all the charge that's ever flowed into it (until it leaks away).

Solved examples

The voltage across a 50 µF capacitor varies as

v (t) = 20 cos (1000 t)

V. Find

i (t)

$i = C \frac{d v}{d t}$
$d v / d t = - 20 \times 1000 sin (1000 t) = - 20000 sin (1000 t)$ V/s
$i = 50 \times 1 0^{- 6} \times (- 20000 sin (1000 t))$

Answer

i (t) = - 1 sin (1000 t)

A =

- sin (1000 t)

Common pitfalls

Trying to apply Ohm's law $v = i R$ to a capacitor. Capacitors don't have resistance in this sense — their v-i is *differential*.
Forgetting capacitor voltage is continuous across a switching instant. The current can jump; the voltage cannot.

02Energy Stored in a Capacitor

The instantaneous power delivered to a capacitor is $p = v \cdot i = v \cdot C d v / d t$ . Integrating from zero voltage to some final $V$ gives the energy stored in the electric field:

$W = \frac{1}{2} C V^{2}$

Equivalent forms:

$W = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$

Energy is in joules. Notable facts:

A capacitor stores energy when charging (energy goes from source into the field) and returns it when discharging — unlike a resistor, which dissipates it as heat.
Energy scales as $V^{2}$ . Doubling the voltage *quadruples* the stored energy. This is why high-voltage capacitor banks (defibrillators, photoflash, industrial pulsed power) can be dangerous even at moderate capacitance.
A 1 F supercapacitor charged to 5 V stores 12.5 J — comparable to a small battery, deliverable in milliseconds.

Always treat any large or unknown capacitor as charged until you've shorted it through a resistor. Even 'discharged' capacitors can self-recharge from dielectric absorption.

Solved examples

Energy stored in a 470 µF capacitor charged to 25 V.

$W = \frac{1}{2} C V^{2}$
$W = 0.5 \times 470 \times 1 0^{- 6} \times 2 5^{2}$
$W = 0.5 \times 470 \times 1 0^{- 6} \times 625$

Answer

W \approx 0.147

J = 147 mJ

Common pitfalls

Using $W = \frac{1}{2} C V$ (missing the square). The energy is quadratic in voltage.
Forgetting that large capacitors can deliver dangerous current bursts even if their voltage doesn't look high.

03Series and Parallel Capacitors

Parallel capacitors. When connected in parallel, capacitors share the same voltage but store separate charges. Equivalent capacitance:

$C_{eq} = C_{1} + C_{2} + \dots + C_{n}$

Plate areas effectively add — bigger total area for the same plate separation gives a bigger capacitor.

Series capacitors. When in series, the same charge $Q$ accumulates on each (charge has nowhere else to go), but the voltage splits inversely with capacitance. Equivalent:

$\frac{1}{C _{eq}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots + \frac{1}{C _{n}}$

For exactly two:

$C_{eq} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$

Series equivalent is *smaller* than any individual capacitor — opposite to resistors.

Energy split in series: same $Q$ on all → smaller $C$ stores *more* energy ( $W = Q^{2} /2 C$ ).

Energy split in parallel: same $V$ across all → larger $C$ stores *more* energy ( $W = \frac{1}{2} C V^{2}$ ).

Why use series of capacitors? Two 200 V capacitors in series can handle 400 V across the pair (with a balancing resistor network in practice). This is how high-voltage capacitor banks are assembled from lower-rated units.

Solved examples

C_{1} = 6

µF and

C_{2} = 3

µF in series. Find

C_{eq}

$\frac{1}{C _{eq}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6}$
$C_{eq} = \frac{6}{3} = 2$ µF

Answer

C_{eq} = 2

µF

Common pitfalls

Adding series capacitors directly. Series of capacitors uses the *reciprocal* formula, like parallel resistors.
Forgetting that the smaller capacitor in a series pair takes the larger voltage. Critical for voltage-rating choices.

04Capacitor Classification, Labelling, and Standard Values

Major types by dielectric:

Type	Dielectric	Range	Voltage	Typical use
Ceramic (MLCC, disc)	Class 1 NP0 / Class 2 X7R	1 pF – 100 µF	6.3 V – 1 kV	Decoupling, RF, timing
Aluminum electrolytic	$Al_{2} O_{3}$ (polarised)	0.1 µF – 1 F	6.3 – 450 V	Bulk smoothing
Tantalum	$Ta_{2} O_{5}$ (polarised)	0.1 – 1000 µF	2 – 50 V	Compact filtering
Film (polyester / polypropylene / PTFE)	Plastic film	1 nF – 100 µF	50 V – 2 kV	Audio, snubbers, precision
Mica	Natural/silvered mica	1 pF – 10 nF	100 V – kV	HF, high-precision
Supercapacitor	Electric double layer	0.1 – 3000 F	2.5 – 5 V	Energy buffering

Polarised types (electrolytic, tantalum) must be connected with correct polarity — reverse voltage destroys the dielectric and the capacitor can rupture.

Labelling — three-digit codes used on small ceramics and films give the value in picofarads: first two digits are significant figures, third is the power of ten multiplier.

Code	Value
104	$10 \times 1 0^{4}$ pF = 100,000 pF = 100 nF = 0.1 µF
473	$47 \times 1 0^{3}$ pF = 47 nF
220	$22 \times 1 0^{0}$ pF = 22 pF
2A104K	$100 V, 0.1 μ F, \pm 10$

Tolerance letter $(o f t e n f o ll o w s t h eco d e) : J = \pm 5$

Standard E-series values — capacitors come in geometrically-spaced 'preferred values' so the tolerance bands tile the range. E12 base values per decade: 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82. Common stock values you'll actually see in a parts drawer: 0.1, 0.22, 0.47, 1, 2.2, 4.7, 10, 22, 47, 100, 220, 470 µF, and similar across nF and pF.

Common pitfalls

Connecting a polarised capacitor backwards. The negative leg is marked with a stripe (electrolytic) or a bar (tantalum). Reverse voltage = damage.
$T re a t in g t h e p r in t e d v a l u e a se x a c t . A^{'} 100 μ F^{'} \pm 20$

05Charging Phase of a Capacitor (RC Circuit)

Connect an uncharged capacitor $C$ through a resistor $R$ to a DC source $V_{s}$ by closing a switch at $t = 0$ . The capacitor starts at 0 V (it can't change voltage instantly), so the entire $V_{s}$ momentarily appears across $R$ , giving an initial current $V_{s} / R$ . As charge accumulates on the capacitor, its voltage rises, the voltage across $R$ falls, and current decreases — until eventually $v_{C} = V_{s}$ and current stops.

Governing differential equation (from KVL):

$RC \frac{d v}{d t} + v = V_{s}$

Solution for the charging response:

$v_{C} (t) = V_{s} (1 - e^{- t / RC})$

$i (t) = \frac{V _{s}}{R} e^{- t / RC}$

The time constant $τ = RC$ is the natural timescale of the circuit. R in ohms, C in farads, $τ$ in seconds.

Behavioural milestones:

Time	$v_{C}$	$i$
$0^{+}$	0	$V_{s} / R$ (peak)
$1 τ$	63.2% of $V_{s}$	36.8% of peak
$2 τ$	86.5%	13.5%
$3 τ$	95.0%	5.0%
$5 τ$	$99.3$	0.7%

Practical rule: capacitor is considered $f u ll yc ha r g e d a f t er 5 τ$ .

Solved examples

A 100 µF capacitor charges through a 10 kΩ resistor from a 12 V source. Find τ and the voltage at

t = 1

$τ = RC = 10000 \times 100 \times 1 0^{- 6} = 1$ s
$v_{C} (1) = 12 (1 - e^{- 1}) = 12 \times 0.632$

Answer

τ = 1

v_{C} (1) \approx 7.59

Common pitfalls

Treating the initial current as something other than $V_{s} / R$ . At $t = 0^{+}$ , the capacitor is a short circuit (zero voltage), so the full source drives R alone.
$F or g e tt in g t ha t 5 τ i s a r u l eo f t h u mb . I nhi g h - p rec i s i o n se tt in g s (osc i ll a t or t imin g) yo u ma y n ee d 10 τ or m ore .$

06Discharging Phase of a Capacitor

Take a capacitor previously charged to $V_{0}$ and remove the source — close a switch that connects only $R$ and $C$ in a loop. The stored charge on the capacitor now drives current *backward* through the resistor, gradually returning the capacitor to zero.

Governing differential equation:

$RC \frac{d v}{d t} + v = 0$

Solution:

$v_{C} (t) = V_{0} e^{- t / RC}$

$i (t) = - \frac{V _{0}}{R} e^{- t / RC}$

The negative sign on $i$ shows the current reverses direction relative to charging — it now flows *out* of the capacitor's positive plate.

Milestones:

Time	$v_{C}$ (% of $V_{0}$ )
$1 τ$	36.8%
$2 τ$	13.5%
$3 τ$	5.0%
$4 τ$	1.8%
$5 τ$	$0.7$

Energy bookkeeping during discharge: every joule of stored capacitor energy ends up dissipated as heat in the resistor. The total heat = $\frac{1}{2} C V_{0}^{2}$ , regardless of $R$ . (Smaller $R$ = faster discharge but same total heat.)

Solved examples

A 100 µF capacitor charged to 12 V discharges through 10 kΩ. Find the current at

t = 0^{+}

and the voltage at

t = 2

Initial current $i (0^{+}) = - V_{0} / R = - 12/10 000 = - 1.2$ mA
τ = $RC = 1$ s, so $t = 2$ s = 2τ
$v_{C} (2) = 12 \times e^{- 2} = 12 \times 0.1353$

Answer

i (0^{+}) = - 1.2

mA;

v_{C} (2) \approx 1.62

Common pitfalls

Reversing the sign of the discharge current. By convention, current during discharge is *opposite* to the charging direction.
Computing total energy dissipated using $i^{2} R \cdot t$ . The current decays exponentially — you'd have to integrate. Use $\frac{1}{2} C V_{0}^{2}$ for the total directly.

§Key formulas

i = C \frac{d v}{d t}

Capacitor i-v · A

v = \frac{1}{C} \int i d t + v (0)

Capacitor voltage · V

w = \frac{1}{2} C v^{2}

Capacitor energy · J

1/ C_{e q} = 1/ C_{1} + 1/ C_{2} + \dots

Series capacitors · F

C_{e q} = C_{1} + C_{2} + \dots

Parallel capacitors · F

v = L \frac{d i}{d t}

Inductor v-i · V

i = \frac{1}{L} \int v d t + i (0)

Inductor current · A

w = \frac{1}{2} L i^{2}

Inductor energy · J

L_{e q} = L_{1} + L_{2} + \dots

Series inductors · H

1/ L_{e q} = 1/ L_{1} + 1/ L_{2} + \dots

Parallel inductors · H

Open in cheat sheet

Capacitors and Inductors

01Capacitor: Structure & v-i Relationship

02Energy Stored in a Capacitor

03Series and Parallel Capacitors

04Capacitor Classification, Labelling, and Standard Values

05Charging Phase of a Capacitor (RC Circuit)

06Discharging Phase of a Capacitor

§Key formulas

Simulator & graphs

Chapter quiz