Inductor vs Capacitor: What Is the Difference?

The mirror, stated once

These two components behave as opposites, and the opposition is systematic rather than accidental. Swap voltage for current, electric field for magnetic field, and every statement about one becomes a true statement about the other.

A [capacitor](/glossary/capacitor) resists a change in the voltage across it. Its defining relation is $C = \frac{Q}{\Delta V}$, so changing the voltage means moving charge, and moving charge takes current and therefore takes time. Its stored energy lives in the electric field in the gap between the plates.

An inductor resists a change in the current through it. Its defining relation is $\mathcal{E} = -L\frac{dI}{dt}$, so changing the current produces a back emf that fights the change. Its stored energy lives in the magnetic field its own current creates.

That single sentence pair generates everything else on this page, including the behaviour table two sections down. Rather than memorising four facts about switches, memorise two:

• The voltage across a capacitor cannot jump.
• The current through an inductor cannot jump.

Both follow from the definitions. A jump in capacitor voltage would need an infinite current to move the charge instantly; a jump in inductor current would need an infinite emf. Neither is available, so both quantities are continuous through the instant a switch is thrown, and every entry in the $t = 0$ column below is that continuity written out.

The capacitance and inductance glossary entries define each side; this page is about what the pair does that neither does alone.

Side by side

Two rows repay a second look before the behaviour table.

The energy formulas are not printed in parallel form. The C: E&M sheet gives the capacitor's energy in terms of charge and voltage, $\frac{1}{2}Q\Delta V$, and the inductor's in terms of inductance and current, $\frac{1}{2}LI^2$. Those look like different shapes because they are written in different variables. Substitute $Q = C\Delta V$ into the first and you get $\frac{1}{2}C(\Delta V)^2$, which is the true mirror of $\frac{1}{2}LI^2$. That form is not printed on either the C: E&M sheet or the AP Physics 2 sheet, so you derive it in one line when you want it. Do not go looking for it and conclude the sheet is missing something.

Only the capacitor gets combination rules. The C: E&M sheet prints $\frac{1}{C_{\mathrm{eq},s}} = \sum_i \frac{1}{C_i}$ and $C_{\mathrm{eq},p} = \sum_i C_i$, the reverse of the resistor rules. There is no equivalent-inductance line anywhere on the sheet, and the CED's Topic 13.5 learning objective is worded to describe a circuit containing a combination of resistors and a single inductor. So the course never asks you to combine inductors.

The t = 0 and t = infinity table, cell by cell

Every cell in this table is stated somewhere in the CED, and every one of them is easy to invert under time pressure. Take uncharged capacitor and zero initial current inductor, which is the standard setup, and assume the switch closes at $t = 0$.

Every row except the last two is reversed between the columns. Here is where each comes from.

Capacitor at $t = 0$: a wire. Essential knowledge 11.8.B.3.i states that immediately after being placed in a circuit, an uncharged capacitor acts like a wire, and charge can easily flow to or from the plates. It has no charge yet, so by $C = Q/\Delta V$ it has no voltage across it, and a component with no voltage across it is indistinguishable from a piece of wire.

Capacitor as $t \to \infty$: a break. Essential knowledge 11.8.B.3.iv states that after a long time, a charging capacitor approaches a state of being fully charged, reaching a maximum potential difference at which there is zero current in the circuit branch in which the capacitor is located. Zero current through a branch is what a break means.

Inductor at $t = 0$: a break. Essential knowledge 13.5.A.4.i states that when a switch is initially closed or opened in a circuit containing an inductor, the induced emf will be equal in magnitude and opposite in direction to the applied potential difference across the branch containing the inductor. The back emf exactly cancels the applied voltage, so no current starts, and the whole applied voltage appears across the inductor.

Inductor as $t \to \infty$: a wire. Essential knowledge 13.5.A.4.iii states that after a time much greater than the time constant of the circuit, an inductor will behave as a conducting wire with zero resistance. The current has stopped changing, so $\frac{dI}{dt} = 0$, so the back emf is zero, so there is nothing across it.

The two energy rows are the exception that proves the rule, and they trip people who have learned the table as "everything is opposite". Both components start with no stored energy and end with the maximum they will hold, because both are storing something that had to be built up. What differs is what they hold at each moment: the capacitor's energy tracks its voltage, which grows, and the inductor's tracks its current, which also grows. Opposite variables, same direction of travel.

The qualification that the table hides

"A capacitor is a wire at $t = 0$" and "an inductor is a break at $t = 0$" are shortcuts, and they are only true for the standard starting conditions. Learn the two continuity rules instead, because they never fail.

A pre-charged capacitor is not a wire at $t = 0$. Its voltage cannot jump, so at the instant the switch closes it still has whatever voltage it had a moment before. A capacitor charged to $6.0 \ \mathrm{V}$ and then switched into a new circuit behaves, at that instant, like a $6.0 \ \mathrm{V}$ battery, not like a wire. Only an uncharged capacitor starts at zero volts, which is what makes it look like a wire.

An inductor already carrying current is not a break at $t = 0$. Its current cannot jump, so at the instant the switch is thrown it keeps carrying whatever current it was carrying. Open the switch on a circuit where an inductor was carrying $2.0 \ \mathrm{A}$ and the inductor will keep pushing $2.0 \ \mathrm{A}$ for an instant, through whatever path is available, generating whatever voltage that takes. Only an inductor starting from zero current looks like a break.

The CED covers the second case explicitly. Essential knowledge 13.5.A.3.iv states that for an inductor with an initial current, the time constant represents the time required for the current in the inductor to reach approximately 37 percent of its initial value, which is the decay case rather than the rise case. And 13.5.A.1 notes that a resistor will dissipate energy that was stored in an inductor as the current changes: the energy in that magnetic field has to go somewhere, and it goes into the resistor.

So the safest formulation, and the one that answers every version of the question, is:

• Whatever the capacitor's voltage was an instant before the switch, it is that same value an instant after.
• Whatever the inductor's current was an instant before the switch, it is that same value an instant after.

Everything else in the circuit rearranges around those two fixed points.

The case that separates them: the same circuit twice

Build one circuit, put a capacitor in the slot, then put an inductor in the same slot, and the numbers swap ends.

Take a $12 \ \mathrm{V}$ battery, a $200 \ \Omega$ resistor in series, and then a $400 \ \Omega$ resistor in parallel with the component under test.

The two columns contain the same four numbers in reversed order. Worked example three derives every one of them.

That is what "exact opposites" means in practice, and it is also the reason this pairing is worth a page rather than two glossary entries. If you can do the capacitor version of a circuit, you can do the inductor version by reading your own answer backwards in time, and vice versa. On a timed exam that is worth more than any individual formula.

The energy each one holds, and where it sits

Both components store energy and neither dissipates any. That is what separates them jointly from the resistor, and it is why both appear in the same conversation about circuits that do something over time.

The capacitor's energy is in the electric field. Separating charge onto two plates takes work against the attraction of the charge already there, and that work is recoverable. The C: E&M sheet prints $U_C = \frac{1}{2}Q\Delta V$, and the energy stored in a capacitor entry gives the alternative forms. The factor of one half is there because the voltage rose from zero to its final value as the charge went on, so the average voltage during the process was half the final one.

The inductor's energy is in the magnetic field. Essential knowledge 13.4.A.2 states that inductors store energy in the magnetic field that is generated by current in the inductor, with the relevant equation $U_L = \frac{1}{2}LI^2$. The energy stored in an inductor entry covers the form itself. The same factor of one half appears for the same reason: the current rose from zero.

The CED then says something about that energy which has no capacitor equivalent in the course. Essential knowledge 13.4.A.2.i states that the energy stored in the magnetic field generated by an inductor in which current is flowing can be dissipated through a resistor or used to charge a capacitor. That second clause is the LC circuit, Topic 13.6, where the two components trade energy back and forth: the capacitor discharges into the inductor, building a magnetic field, which then collapses and recharges the capacitor the other way round. The sheet prints its frequency as $\omega_{LC} = \frac{1}{\sqrt{LC}}$.

So the two components are not only opposites; they are complementary. An LC circuit oscillates precisely because one of them resists a change in voltage while the other resists a change in current, and neither can stop the other from happening. That is the same mathematics as a mass on a spring, with charge playing the part of displacement and current the part of velocity.

When it costs a mark

Reversing the behaviour table. The error that turns a whole multi-part question wrong at the first step. If you remember nothing else, remember that an uncharged capacitor starts conducting freely and stops, while an inductor starts blocking and stops. Both eventually settle; they just settle to opposite states.

Assuming zero current through a capacitor means zero current everywhere. At steady state the capacitor branch carries nothing, but the rest of the circuit usually carries plenty. Redraw the circuit with that branch deleted and solve what is left.

Assuming zero voltage across an inductor means zero current through it. At steady state the inductor has no voltage across it and carries the maximum current. Redraw it as a plain wire and solve.

Using $\frac{1}{2}C(\Delta V)^2$ and then also using $\frac{1}{2}Q\Delta V$ on the same capacitor and adding them. They are the same energy written two ways, not two contributions.

Treating $L_{\mathrm{sol}} = \frac{\mu_{\mathrm{core}}N^2A}{\ell}$ with $N$ as turns per length. The sheet's symbol list defines $N$ as the number of loops and $n$ as the number of loops per unit length, and the inductance formula uses $N$, squared, with the length $\ell$ appearing separately in the denominator. The solenoid field formula on the same page uses $n$. Two different symbols, two different equations, one page apart.

Combining inductors in series or parallel. There is no printed rule and the course does not ask for it: the Topic 13.5 learning objective describes a circuit containing a combination of resistors and a single inductor. If a problem seems to want two inductors combined, re-read it.

Bringing inductors into an AP Physics 2 answer. The words inductance and inductor do not appear anywhere in the AP Physics 2 CED, and $L$ is not in the symbol list on its equation sheet. Capacitors are on that exam; inductors are not.

What they have in common

The contrast is total but it is not a contrast between an important component and a trivial one, and four properties are genuinely shared.

• Neither dissipates energy. An ideal capacitor and an ideal inductor both store and return; only the resistor turns electrical energy into thermal energy. In an RC or LR circuit, every joule that leaves the battery and does not end up in the storage element ended up in the resistor.
• Both make the circuit take time. Put either with a resistor and you get an exponential approach to a steady state, with a characteristic time constant. The constants are built in opposite ways, which is the subject of RC vs LR circuits.
• Both are defined by geometry and material. $C = \frac{\kappa\varepsilon_0 A}{d}$ depends on plate area, separation and the dielectric between the plates. $L_{\mathrm{sol}} = \frac{\mu_{\mathrm{core}}N^2A}{\ell}$ depends on turns, area, length and the permeability of the core. Neither depends on the voltage or current you apply, which is a point exam questions test by changing the applied voltage and asking whether $C$ or $L$ changed. It did not.
• Both reach a steady state that looks like a plain conductor or a plain gap. After a long time, every reactive component in a DC circuit has become either a wire or a break, and the circuit is a resistor network again. That is why steady-state questions are usually the easy part of an RC or LR problem.

Those shared properties are also why the two get confused in the first place. Both are the non-resistor in a circuit that changes over time, both have a Greek-lettered time constant, and both appear in the same unit sequence. The way to keep them apart is not to memorise which is which, but to hold on to the one asymmetry that generates the rest: voltage is the capacitor's continuous variable, current is the inductor's.

Where these sit in the courses

Capacitors are Topic 10.3 in Unit 10, Conductors and Capacitors, which the CED weights at 10 to 15 percent of the AP Physics C: Electricity and Magnetism multiple-choice section over roughly 8 to 16 class periods. Topic 10.4 adds dielectrics, and they return in Topic 11.8 as part of RC circuits.

Inductors are Topic 13.4 in Unit 13, Electromagnetic Induction, weighted at 10 to 20 percent over roughly 10 to 20 class periods, and they return immediately in Topic 13.5 as LR circuits and Topic 13.6 as LC circuits.

The three-unit gap between them is worth noticing, because it explains why students meet them as unrelated objects. The capacitor arrives early, out of electrostatics, as two plates holding charge. The inductor arrives at the very end, out of induction, as a coil fighting a change in its own current. Nothing in the course sequence puts them side by side until Topic 13.6 does, by which point most of the confusion has set in.

In AP Physics 2 the split is starker still. That course has capacitors, at Topic 10.6 and again at Topic 11.8, and its equation sheet prints $C = Q/\Delta V$, $C = \kappa\varepsilon_0 A/d$, $E_C = \frac{Q}{\kappa\varepsilon_0 A}$, $U_C = \frac{1}{2}Q\Delta V$, the two combination rules and $\tau = R_{\mathrm{eq}}C_{\mathrm{eq}}$. It has no inductors at all: no inductance topic, no $L$ in its symbol list, no energy formula, no LR circuit. So this comparison only exists inside AP Physics C: Electricity and Magnetism.