123-ME · Capacitance & Inductance

Capacitance & Inductance

Slides 3.1–3.2, 3.4–3.5 · Exercise 7 · Tutorial 6

Every circuit element you've met so far — resistors, sources — reacts to the present. A resistor's voltage depends only on the current flowing through it right now. Capacitors and inductors are different: they have memory. What they do next depends on their whole history, because each one stores energy — a capacitor in an electric field, an inductor in a magnetic field — and that stored energy can't appear or vanish instantaneously. This single idea (energy storage forces continuity) is the through-line of this chapter, and it's also the seed of Chapter 4's RC/RL transients: once you understand *why* $v_C$ and $i_L$ can't jump, the exponential charging curves are just "how fast do they get where they're going." Budget about 25 minutes here — most of it on the two derivations, since everything else (series/parallel rules, energy formulas) falls out of them cleanly.

1. The capacitor: storing charge in a field

A capacitor is two conductors separated by an insulator (a dielectric). Push charge onto one plate and an equal, opposite charge accumulates on the other, with an electric field — and a voltage — built up between them. The defining relationship is a simple proportionality:

q(t) = C\,v(t)

where $C$ , the capacitance in farads (F = coulombs/volt), is a constant set entirely by geometry. For the simplest case — two flat plates of area $A$ separated by distance $d$ , filled with a dielectric of permittivity $\varepsilon = \varepsilon_0 \varepsilon_r$ — geometry gives:

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

More plate area means more room to spread charge for the same field strength (bigger $C$ ); more separation weakens the field the same charge produces, so it takes more voltage to hold that charge (smaller $C$ ). A better dielectric ( $\varepsilon_r$ ) lets the material itself partially cancel the field, again letting more charge pile up per volt.

Series and parallel capacitors combine opposite to resistors, because $C$ scales with area (parallel plates effectively increase area) but $1/C$ scales with separation (series plates effectively increase the gap):

C_{\text{parallel}} = C_1 + C_2 + \cdots \qquad\qquad \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Notation check

Same passive sign convention as every other element in this course: current

i(t)

enters the $+$ terminal of

v(t)

. With that convention,

p(t) = v(t)\,i(t) > 0

means the capacitor is absorbing energy (charging up);

p(t) < 0

means it's releasing stored energy back to the circuit.

2. Deriving $i = C\,dv/dt$

The capacitor's $q = Cv$ relationship is about charge, but circuits are built from currents. The link is the definition of current itself: current is the rate charge flows onto the plate.

Derivation — from

q=Cv

i = C\,dv/dt

Start from the defining relationship and the definition of current:

q(t) = C v(t), \qquad i(t) = \frac{dq(t)}{dt}

Substitute

q(t) = Cv(t)

into the current definition:

i(t) = \frac{d}{dt}\big[C v(t)\big]

Pull $C$ out of the derivative — it's a fixed geometric constant, not a function of time:

i(t) = C\,\frac{dv(t)}{dt}

i(t) = C\,\frac{dv(t)}{dt}

Integrating both sides the other way gives the voltage in terms of the current history — useful when you're handed a current waveform and asked to sketch $v(t)$ (exactly the Exercise 7 and Tutorial 6 problem style):

v(t) = \frac{1}{C}\int_0^t i(\tau)\,d\tau \; + \; v(0)

That $v(0)$ term is not optional bookkeeping — it's the whole reason capacitors have "memory." The present voltage depends on every ampere-second of current that has ever flowed onto the plate, not just the current flowing right now.

3. Why $v(t)$ across a capacitor can't jump

Look at $i = C\,dv/dt$ again. If $v(t)$ had a genuine discontinuity — a vertical jump — then at that instant $dv/dt \to \infty$ , which by the formula demands $i \to \infty$ too. An infinite current would mean an infinite amount of charge moves in zero time, which no real (or ideal) source can supply. A capacitor's voltage is therefore forced to be continuous — it can change quickly, but never in a literal instant. Physically: voltage is a stand-in for how much charge is piled on the plate, and charge is a real, countable quantity — moving a finite amount of it still takes a finite amount of time, even if that time is a microsecond.

The mirror-image statement holds for an inductor, from $v = L\,di/dt$ : if current jumped instantaneously, $di/dt \to \infty$ and so would the voltage spike across the inductor — which is exactly the mechanism behind real inductive "kick" (e.g. a spark when you yank a plug on an energized coil). An inductor's current is therefore continuous, for the mirrored physical reason: current through an inductor is a stand-in for the magnetic flux/momentum it has stored, and that can't unwind in zero time either.

4. The inductor: storing energy in a magnetic field

An inductor is (usually) a coil of wire. Current flowing through it sets up a magnetic field that links the coil's own turns; if that current changes, the changing field induces a voltage across the coil that opposes the change (Faraday/Lenz). That opposition is exactly what the defining relationship encodes:

v(t) = L\,\frac{di(t)}{dt} \qquad\qquad i(t) = \frac{1}{L}\int_0^t v(\tau)\,d\tau + i(0)

$L$ , the inductance in henries (H = V·s/A), plays the same structural role for current that $C$ plays for voltage — and note the relationship is the exact algebraic mirror of the capacitor's: swap $v \leftrightarrow i$ and $C \leftrightarrow L$ and one equation becomes the other. That symmetry is why series/parallel combination also mirrors resistors directly (unlike capacitance):

L_{\text{series}} = L_1 + L_2 + \cdots \qquad\qquad \frac{1}{L_{\text{parallel}}} = \frac{1}{L_1} + \frac{1}{L_2} + \cdots

Capacitor ↔ inductor dictionary

Every capacitor fact has an inductor mirror:

q=Cv \leftrightarrow \lambda = Li

(flux linkage),

i=C\,dv/dt \leftrightarrow v=L\,di/dt

, "

v

can't jump"

\leftrightarrow

i

can't jump", series/parallel rules swap, and (next section)

E=\tfrac12Cv^2 \leftrightarrow E=\tfrac12Li^2

. If you memorize the capacitor derivations, you get the inductor ones for free by swapping symbols.

5. Stored energy: $E=\tfrac12Cv^2$ and $E=\tfrac12Li^2$

Both elements are lossless — no resistor inside them to dissipate energy as heat. All the energy delivered while charging a capacitor (or ramping up current in an inductor) is recoverable later; it's stored, not spent. We find how much by integrating instantaneous power $p(t)=v(t)i(t)$ over the time it took to establish that voltage/current.

Derivation — capacitor energy

E=\tfrac12Cv^2

Instantaneous power absorbed by the capacitor (passive sign convention), then substitute

i = C\,dv/dt

p(t) = v(t)\,i(t) = v(t)\left(C\frac{dv}{dt}\right) = Cv\frac{dv}{dt}

Energy is the time-integral of power. Charging from

v=0

(uncharged) up to some voltage

v

, change the integration variable from

t

v

itself (this is the standard trick —

dt

and one factor of

dv/dt

cancel):

E = \int_0^t p\,d\tau = \int_0^t Cv\frac{dv}{d\tau}\,d\tau = \int_0^{v} Cv'\,dv'

Integrate the simple power-of-

v

integral:

E = C\left[\frac{v'^2}{2}\right]_0^{v} = \frac12 C v^2

E = \frac12 C v^2

Run the identical steps with $v=L\,di/dt$ in place of $i=C\,dv/dt$ (swap $C\to L$ , $v\to i$ throughout) and the inductor's energy falls out the same way:

E = \frac12 L i^2

Both are always $\ge 0$ — you cannot extract more energy from either element than you put in, which is exactly what "lossless energy storage" (as opposed to a source) means.

6. The water-flow analogy

Circuits are invisible, which makes intuition hard to build. A classic trick is to map every electrical quantity onto a plumbing system you can actually picture: voltage is water pressure, current is water flow rate. Under that mapping, every element in this course gets a mechanical twin — including the two you just met.

Walk the loop and every "can't jump" fact becomes physically obvious:

Pump ↔ voltage source — pushes flow around the loop by maintaining a pressure difference, exactly like a source maintains a voltage.
Water wheel ↔ resistor — flow through it converts pressure drop into wasted energy (turning the wheel, which we imagine as friction/heat) — irreversible, like $I^2R$ .
Elastic diaphragm reservoir ↔ capacitor — water flowing in stretches the diaphragm, building pressure. You cannot instantly inflate/deflate a stretchy diaphragm to a new shape — it has to physically stretch, which takes time proportional to the flow rate. That's why pressure (voltage) can't jump: the diaphragm's shape (charge) is a physical, continuous quantity.
Heavy paddle wheel ↔ inductor — a massive wheel has rotational inertia: you can't instantly start or stop it spinning, because that would need infinite torque. Flow rate (current) through it is exactly this rotational speed — it resists sudden changes for the same reason a flywheel does. That's why current can't jump.

Why the analogy earns its keep

It's not just cute — it converts two abstract calculus facts (

dv/dt\to\infty

needs

i\to\infty

;

di/dt\to\infty

needs

v\to\infty

) into two mechanical facts you already believe: stretchy things can't teleport into a new shape, and heavy spinning things can't teleport into a new speed.

7. Interactive: watch a capacitor charge from constant current

Exercise 7's problems (P3.6, P3.7) all hinge on the same picture: a constant current $i$ into a capacitor produces a voltage that ramps up linearly, not exponentially (that only shows up once a resistor is in the loop, in Chapter 4). Drag the sliders below and watch $v(t)=\frac{1}{C}\int i\,dt + v(0)$ trace out live, or flip to the inductor to see the mirrored $i(t) = \frac1L\int v\,dt + i(0)$ relationship — and notice that whichever quantity is highlighted in bold on each plot (v for the capacitor, i for the inductor) is always the smooth, unbroken curve, while the source you're driving it with can be any shape you like, including a hard step.

Constant-source charging widget

Element:

Charging current

i

(mA): 2.0 Capacitance

C

(μF): 100 Initial value

v(0)

: 0.0

Adjust the sliders to see

v(t)

trace out.

8. Self-check

Self-check 1

A capacitor with $v(0)=0$ is charged by a constant current of $2$ mA. After $5$ ms its voltage is $4$ V. What is $C$ ?

Self-check 2

Which statement correctly explains why an inductor's current cannot change instantaneously?

Self-check 3

Two $10\ \mu\text{F}$ capacitors, both initially uncharged, are connected in series across a $50$ V source. What is the voltage across each?

9. Practice problems

Problem 1 — Charging time (based on P3.6)

A $10{,}000\ \mu\text{F}$ capacitor, initially uncharged, is charged by a constant current source of $100\ \mu\text{A}$ . How long does it take to reach $10$ V?

Method 1 — direct integration. With $v(0)=0$ and constant $i$ : $v(t) = \frac{1}{C}\int_0^t i\,d\tau = \frac{it}{C} \;\;\Rightarrow\;\; t = \frac{vC}{i} = \frac{(10)(10{,}000\times10^{-6})}{100\times10^{-6}} = \frac{0.1}{10^{-4}} = 1\text{ s}$

Method 2 — cross-check via charge. Total charge needed is $Q = Cv = (10{,}000\times10^{-6})(10) = 0.1$ C. At a constant $100\ \mu\text{A} = 10^{-4}$ A, the time to deliver that charge is $t = \frac{Q}{i} = \frac{0.1}{10^{-4}} = 1\text{ s}$ Both methods agree: $t = 1$ s.

Problem 2 — Power direction and energy (based on P3.7)

A $50\ \mu\text{F}$ capacitor carries a constant current $i(t) = 3$ mA (passive sign convention), with $v(0) = -20$ V. Find the power at $t=0$ and at $t=1$ s, and state whether the capacitor is absorbing or delivering energy at each instant.

Method 1 — direct $v(t)$ , then $p=vi$ . $v(t) = \frac{1}{C}\int_0^t i\,d\tau + v(0) = \frac{(3\times10^{-3})t}{50\times10^{-6}} - 20 = 60t - 20\text{ V}$ At $t=0$ : $v=-20$ V, so $p = vi = (-20)(3\times10^{-3}) = -60$ mW — negative, so the capacitor is delivering energy (it's still "discharged past zero," releasing what little charge of the opposite sign it held). At $t=1$ s: $v = 60-20 = 40$ V, so $p = (40)(3\times10^{-3}) = 120$ mW — positive, absorbing energy (now genuinely charging up).

Method 2 — cross-check via stored energy change. Energy at $t=0$ : $E_0 = \tfrac12 Cv(0)^2 = \tfrac12(50\times10^{-6})(400) = 10$ mJ. Energy at $t=1$ s: $E_1 = \tfrac12(50\times10^{-6})(40)^2 = 40$ mJ. Since $E$ is falling near $t=0$ (moving from $v=-20$ V toward $v=0$ , magnitude of stored energy is decreasing) the instantaneous power there must be negative, and since $E$ is rising by $t=1$ s the power there must be positive — matching Method 1's signs without ever computing $p$ directly.

$p(0) = -60$ mW (delivering); $p(1\text{ s}) = +120$ mW (absorbing).

Problem 3 — Exponential inductor current (based on P3.50)

A $300$ mH inductor carries $i(t) = 5e^{-200t}$ A. Find the voltage $v(t)$ , and the energy stored at $t=0$ and as $t\to\infty$ .

Method 1 — direct differentiation. $v(t) = L\frac{di}{dt} = (0.3)\frac{d}{dt}\left[5e^{-200t}\right] = (0.3)(5)(-200)e^{-200t} = -300e^{-200t}\text{ V}$ Energy: $E(t) = \tfrac12 Li^2 = \tfrac12(0.3)(5e^{-200t})^2 = 3.75\,e^{-400t}$ J. At $t=0$ : $E(0) = 3.75$ J. As $t\to\infty$ : $E\to 0$ J (all the stored energy is eventually delivered back into the circuit — an ideal inductor doesn't dissipate any of it).

Method 2 — cross-check via power integral. Instantaneous power is $p(t) = v(t)i(t) = (-300e^{-200t})(5e^{-200t}) = -1500\,e^{-400t}$ W (negative throughout — the inductor is releasing energy the whole time, consistent with decaying current). Total energy released from $t=0$ to $\infty$ should equal the initial stored energy: $-\int_0^\infty p(t)\,dt = \int_0^\infty 1500\,e^{-400t}\,dt = 1500\left[\frac{-e^{-400t}}{400}\right]_0^\infty = \frac{1500}{400} = 3.75\text{ J}$ This matches $E(0)=3.75$ J from Method 1 exactly, confirming both the voltage expression and the energy accounting.

$v(t) = -300e^{-200t}$ V; $E(0) = 3.75$ J, $E(\infty) = 0$ J.