AC Through an Inductor

Here is a demonstration that has opened this topic for a century. A lamp glows happily in series with a coil of thick copper wire — on a battery. Swap the battery for an AC supply of the same voltage and the lamp dims. Slide an iron core into the coil and it dims further, down to a sulky glow — yet the coil is barely warm. Some opposition is throttling the current without eating any energy, and it answers to frequency, not to the ohmmeter: put the same circuit on a higher-frequency supply and the lamp dims again. That opposition is the inductor's toll on change. Self-Inductance & Inductors built the machinery: a coil threads its own flux, and any change of current induces a back-EMF that fights the change. A battery only suffers this fight once, at switch-on. An alternating current never stops changing — so the fight never ends, and the coil charges a permanent, frequency-priced toll called inductive reactance, $X_L = \omega L$ . Alongside it comes a stranger fact: the current through an inductor runs a quarter-cycle late behind the voltage — the first of the two 90° arrows the phasor module promised. This module builds both results with nothing but the slope of a sine curve, reads the reactance's frequency behaviour off one straight-line graph, and meets the choke coil — the device engineers build when they want to limit an AC current without paying for it in heat. The capacitor's mirror-image story is next; the two of them together make the LCR circuit a triangle problem.

Key takeaways

An ideal inductor on AC passes $i = i_0\sin\omega t$ while its voltage runs $v = \omega L i_0\cos\omega t$ : the voltage leads the current by 90° (the current lags by a quarter cycle).
The peak-to-peak bookkeeping defines inductive reactance $X_L = \omega L = 2\pi f L$ , in ohms: $V_0 = i_0 X_L$ and $V_{\mathrm{rms}} = I_{\mathrm{rms}} X_L$ .
$X_L$ grows in proportion to frequency: the $X_L$ – $f$ graph is a straight line through the origin. DC ( $f = 0$ ) sees an ideal inductor as a plain piece of wire.
Reactance is not resistance: the inductor stores energy in its field for a quarter cycle and hands it all back in the next — averaged over a cycle, an ideal inductor eats nothing.
A choke coil exploits exactly this: it limits AC current almost without wasting power — the full accounting arrives in Power in AC Circuits & Power Factor.
Key idea: $X_L = \omega L$ , with voltage leading current by 90° — the inductor taxes the rate of change of current, so faster oscillations pay a proportionally higher toll.

The core insight

An inductor opposes alternating current with reactance $X_L = \omega L$ — an ohm-valued, frequency-proportional opposition arising from its own back-EMF — and forces the current to lag the applied voltage by a quarter cycle, storing and returning field energy so that its average power consumption is zero.

Where this lives today

Open any phone charger or LED driver and the largest component on the board is usually an inductor — a ferrite-cored choke doing exactly this module's job at 100 kHz instead of 50 Hz. Switch-mode power supplies exploit both ends of the $X_L$ – $f$ line: they chop current fast because a small, cheap inductor presents a big reactance at high frequency ( $X_L = \omega L$ — raise $\omega$ and you can shrink $L$ ), then use chokes to block the fast switching hash from leaking back into the mains. The same logic in reverse explains a classic engineering choice: aircraft power runs at 400 Hz rather than 50 Hz partly so that every onboard transformer and choke can be eight times lighter for the same reactance. Closer to the syllabus: the tube-light ballast humming in a school corridor is a mains-frequency choke limiting the lamp current nearly for free, and the "hum filter" in audio equipment is an inductor blocking 50 Hz precisely because even that low frequency pays a toll if $L$ is large enough. The next module hands the capacitor its mirror-image arrow — reactance falling with frequency, current leading — and once both arrows are on the table, the Series LCR Circuit assembles them into the chapter's central machine.

The physical picture

Start from what a coil is. Self-Inductance & Inductors established the law of the device: when the current through an inductor changes, the changing flux induces an EMF of size $L\,\Delta I/\Delta t$ that opposes the change — the electrical analogue of inertia. A steady current suffers nothing (an ideal coil has no resistance); only change is taxed. Now feed it alternating current, $i = i_0\sin\omega t$ . This current is always changing — that is its job description. Near every zero-crossing it changes fastest; at every crest it pauses for an instant. So the back-EMF, which tracks the rate of change, is largest when the current is zero and vanishes when the current peaks. To keep the current going against this ever-shifting opposition, the supply must apply a voltage that exactly matches the back-EMF instant by instant — biggest at the current's zero-crossings, zero at the current's crests. Read that timing carefully, because it is the module's second headline. The voltage peaks a quarter-cycle before the current does: by the time the current finally crests, the voltage has already fallen to zero and is heading negative. In phasor language (Phasors: The Rotating-Arrow Toolkit), the voltage arrow points 90° ahead of the current arrow. The standard sentence, worth memorizing in both directions: in an inductor, voltage leads current by 90°; the current lags the voltage by 90°.

Figure

AC through an ideal inductor — the voltage peaks a quarter cycle before the current

From the slope of a sine to X_L = ωL

Turn the story into the working formula. The coil's law relates voltage to the slope of the current's graph: $v = L \times$ (slope of the $i$ – $t$ tangent) — that is Self-Inductance's $\varepsilon = -L\,\Delta I/\Delta t$ read from the supply's side, in the tangent-slope language this track uses wherever an instantaneous rate is needed. And the slope curve of a sine is a fact you already own from the oscillations chapter: SHM Kinematics established that when a position runs as $A\sin\omega t$ , its rate runs as $\omega A\cos\omega t$ — the slope curve of a sine is a cosine, scaled by $\omega$ . Apply it to the current $i = i_0\sin\omega t$ :

v = L \times (\text{slope of } i) = \omega L\, i_0 \cos\omega t \qquad (1)

where: $i_0$ is the peak current, $\omega = 2\pi f$ the angular frequency, and $L$ the inductance in henry. Two headlines sit in this one line. The shape: the voltage is a cosine while the current is a sine — and a cosine is a sine pushed a quarter-cycle ahead, which is the 90° lead, now in symbols. The size: the voltage's peak is the current's peak multiplied by $\omega L$ . That multiplier plays exactly the role $R$ plays in Ohm's law, so it gets an ohm-flavoured name:

X_L = \omega L = 2\pi f L, \qquad V_0 = i_0 X_L, \qquad V_{\mathrm{rms}} = I_{\mathrm{rms}} X_L \qquad (2)

where: $X_L$ is the inductive reactance, in ohms — check the units: henry × (rad/s) = (V·s/A)(1/s) = V/A = Ω ✓. The peak (or rms) values obey an Ohm's-law-shaped relation, and only the peak values: because of the 90° offset, $v \neq i X_L$ instant by instant — at the moment the current peaks, the voltage is zero. Reactance relates amplitudes, never simultaneous values. An anchor pair for 50 Hz mains ( $\omega = 314$ rad/s): a 0.1 H coil offers $X_L = 31.4\ \Omega$ ; a 25.5 mH coil offers $8.0\ \Omega$ .

Derivation AWhy the current lags by exactly a quarter cycle

The goal is the 90° phase relation of equation (1), pinned down so firmly that you can reproduce the argument, not just the slogan. The strategy is to read the slope of the current's graph at the four landmark instants of one cycle and check that the voltage — which must equal $L$ times that slope — traces a cosine. The thing to watch: lag and lead are about which curve reaches its landmarks first, and it is the current that arrives late.

Step 1. Take $i = i_0\sin\omega t$ and walk its graph. At $t = 0$ the current is zero but climbing at its steepest — maximum positive slope. A quarter period later the current crests — slope zero. Another quarter on, it plunges through zero — maximum negative slope. At three-quarters, it bottoms out — slope zero again.

Step 2. The coil's voltage is $L$ times that slope, so it reads: maximum at $t = 0$ , zero at $T/4$ , most negative at $T/2$ , zero at $3T/4$ . Plot those four values and the curve through them is $\cos\omega t$ — exactly equation (1), $v = \omega L i_0\cos\omega t$ .

Step 3. Compare arrival times. The voltage hits its peak at $t = 0$ ; the current does not crest until $t = T/4$ . Every landmark of the current happens one quarter-period after the same landmark of the voltage: the current lags by $T/4$ , i.e. by $90°$ . In the identity language: $\cos\omega t = \sin(\omega t + 90°)$ , so $v = V_0\sin(\omega t + 90°)$ against $i = i_0 \sin\omega t$ — the voltage's phase is $90°$ ahead. Why the physics wants it this way: the coil only "notices" change. The supply must shout loudest (maximum $v$ ) where the current changes fastest (its zero-crossings) and can fall silent where the current momentarily stops changing (its crests). A voltage tied to the current's slope is automatically a quarter cycle ahead of the current itself.

Step 4. Read the energy ledger in the same graph. From $t = 0$ to $T/4$ , current rises: the coil's field builds, absorbing energy ( $\tfrac{1}{2}Li^2$ grows — Energy in an Inductor & the Magnetic Field). From $T/4$ to $T/2$ , current falls: the field collapses and pumps every joule back into the circuit. Two deposits and two full refunds per cycle: the average power drawn by an ideal inductor is zero. Verified: numerically, with $i_0 = 1$ A, $\omega = 314$ rad/s, $L = 0.1$ H — slope of $\sin(314t)$ at $t = 0$ computed from a small interval gives $313.9\ \mathrm{s^{-1}} \approx \omega$ ✓; $v$ at $t = 0$ is $31.4$ V $= \omega L i_0$ ✓; at the current's crest ( $t = 5$ ms) the computed slope is $0.0$ and $v = 0$ ✓; product $vi$ integrated over a full cycle sums to zero to numerical precision ✓.

Reactance grows with frequency

Equation (2) is a straight line through the origin: double the frequency, double the toll. This graph is a stated syllabus demand (ISC asks for "graph of $X_L$ vs $f$ " explicitly) and a favourite MCQ skeleton everywhere else.

Figure

X_L versus frequency — a straight line through the origin, steeper for larger L

The two ends of the line are the exam's favourite limits. At DC ( $f = 0$ ): no change, no back-EMF, no reactance — an ideal inductor is a short circuit, a plain piece of wire. This is why the opening demo's lamp burned bright on the battery. (A real coil retains its small winding resistance, the only thing then limiting the current — which is exactly why connecting a transformer or fan coil designed for AC straight to DC can burn it out.) At high frequency: the toll grows without bound; to a fast enough signal, the coil is effectively an open switch. Inductors are therefore low-pass gatekeepers: they wave DC and slow wiggles through and block fast ones. That gatekeeping, purchased without heat, is a product. A choke coil is an inductor built to limit AC current in place of a resistor: in series with a fluorescent tube (the classic "ballast"), it holds the tube's current down while dissipating almost nothing, where a resistor doing the same job would run hot and bill you for it. The claim "almost nothing" is the quarter-cycle refund ledger of the derivation — energy parked in the field comes back, every cycle. ISC flags the choke "analytical only," and the analysis is a one-liner after the next two modules: the full arithmetic of why its power consumption is zero (the famous $\cos\varphi = 0$ , "wattless current") is owned by Power in AC Circuits & Power Factor.

Limiting cases

case	result	meaning
$f \to 0$ (DC)	$X_L = 0$	ideal coil = plain wire; only winding resistance survives
$f$ doubled	$X_L$ doubled	straight-line law; current halves for the same supply
$\omega \to \infty$	$X_L \to \infty$	coil = open switch to fast signals
iron core inserted	$L$ up ⇒ $X_L$ up	current drops — the dimming-lamp demo's second act
$L \to 0$	$X_L \to 0$	no coil, no toll — circuit reverts to its resistive self
average power, ideal coil	$0$	field energy borrowed and fully returned each cycle

Common trapAn inductor resists current

It doesn't — it resists change of current, and the difference is worth marks in at least three recurring questions. First: on DC, an ideal inductor offers no opposition at all ( $X_L = 0$ ); students who think "coil = opposition" predict a dim lamp on the battery and lose the plot of the classic demo. Second: reactance is not resistance — an ideal inductor limits an AC current to $I = V/X_L$ while dissipating zero average power, because every joule parked in its magnetic field during one quarter cycle is repaid in the next. A resistor limiting the same current would convert the lot to heat. Third, the subtlest: $V = IX_L$ relates amplitudes only. At the instant the current through an inductor is maximum, the voltage across it is zero — dividing simultaneous readings of $v$ and $i$ gives anything between $0$ and $\infty$ , not $X_L$ . Reactance is peak-to-peak bookkeeping on curves that refuse to peak together.

Worked examples

Board level — a coil on the mains (CBSE/ISC pattern, 3 marks)

A 0.5 H inductor of negligible resistance is connected to a 230 V, 50 Hz supply. (a) Find its reactance and the rms and peak currents. (b) What happens to the rms current if the frequency is doubled to 100 Hz?

Step 1. Reactance: $X_L = 2\pi f L = 2\pi \times 50 \times 0.5 = 314 \times 0.5 = 157\ \Omega$ .

Step 2. rms current: $I_{\mathrm{rms}} = V_{\mathrm{rms}}/X_L = 230/157 = \boxed{1.46\ \mathrm{A}}$ .

Step 3. Peak current: $i_0 = \sqrt{2}\,I_{\mathrm{rms}} = 1.414 \times 1.46 = \boxed{2.07\ \mathrm{A}}$ . (The 230 V label is rms — AC Signals: Peak, Average & RMS — so the current answer comes out in rms and needs $\sqrt{2}$ for the peak.)

Step 4. Part (b): $X_L \propto f$ , so doubling $f$ doubles $X_L$ to $314\ \Omega$ and halves the current: $I = 230/314 = \boxed{0.73\ \mathrm{A}}$ . One proportionality sentence earns the mark faster than recomputing from scratch.

Competitive level — reading the wave, auditing the energy (JEE-Main pattern, ~2.5 minutes)

A current $i = 2.0\sin(314\,t)$ A flows through an ideal 0.5 H inductor. (a) Write the expression for the voltage across it. (b) What does an AC voltmeter across the inductor read? (c) At $t = 5.0$ ms, what is the instantaneous power delivered to the inductor, and what is the average power over a full cycle?

Step 1. Part (a): amplitude first — $V_0 = \omega L i_0 = 314 \times 0.5 \times 2.0 = 314$ V. Phase: voltage leads by $90°$ . So $v = \boxed{314\sin(314\,t + \pi/2)\ \mathrm{V}} = 314\cos(314\,t)$ V.

Step 2. Part (b): the voltmeter reads rms: $V_{\mathrm{rms}} = 314/\sqrt{2} = \boxed{222\ \mathrm{V}}$ .

Step 3. Part (c): at $t = 5.0$ ms, $\omega t = 314 \times 0.005 = 1.57$ rad $= 90°$ : the current is at its crest ( $i = 2.0$ A) and the voltage is at zero ( $v = 314\cos 90° = 0$ ). Instantaneous power $p = vi = \boxed{0}$ — the field is momentarily full, neither charging nor discharging.

Step 4. Average power over a cycle: $\boxed{0}$ , by the quarter-cycle refund ledger — the positive $vi$ lobes (field charging) exactly cancel the negative ones (field returning energy). No ideal inductor ever shows up on an energy bill. Timing note: 2.5 minutes. The speed move in (a) is refusing to differentiate anything — amplitude is $\omega L i_0$ , phase is "+90°," done. Part (c) is a trap for autopilot $P = VI$ reflexes: check the instant before you multiply; at a current crest the inductor's voltage is zero, always.

Exam traps

Who leads whom. In an inductor, voltage leads current by 90° (current lags). The classic mnemonic CIVIL holds the whole chapter: in a Capacitor, I before V; V before I in an L. Misassigning the lag flips signs through every LCR problem downstream.

$\omega$ versus $f$ . $X_L = \omega L = 2\pi f L$ . Fed " $50$ Hz" you must multiply by $2\pi$ ; fed " $\sin(314\,t)$ " you already hold $\omega$ . Writing $X_L = fL$ is the single most common reactance slip, and setters bait it by mixing both notations in one paper.

Reactance ≠ resistance. $X_L$ limits current but dissipates nothing; do not feed it into $P = I^2R$ as if it were an $R$ . Power questions about pure inductors have the answer zero — stated with the energy-refund reason for full marks.

Amplitude relation, not instant relation. $V_0 = i_0X_L$ and $V_{\mathrm{rms}} = I_{\mathrm{rms}}X_L$ ; never divide simultaneous instantaneous values. If a question hands you $v$ and $i$ at one instant, it is testing exactly this.

DC questions in AC clothing. "An ideal inductor is connected to a battery…" — current is limited only by whatever resistance exists; $X_L$ does not apply ( $f = 0$ ). Real-coil versions: the winding resistance takes over, and the coil may overheat.

Boards and depth. CBSE/NEET: pure-inductor circuit with phasor diagram, $X_L = \omega L$ , the frequency proportionality — this module's spine. ISC: adds the explicit $X_L$ – $f$ graph and the phasor-representation language ("phase lag and phase lead" is syllabus wording), plus the choke coil analytically — the wattless-current arithmetic lands in the power module. JEE-Main/NEET: expression-writing ( $v$ from $i$ and vice versa), instantaneous-vs-average power MCQs. JEE-Advanced: inductors inside LCR and LC circuits (later modules) rather than alone. AP Physics 2 has no AC on its syllabus at all, and IB (2023 onward) dropped AC circuits entirely — this module is CBSE/ISC/JEE/NEET territory.

Origins · 1 minMihajlo Pupin and how this was discovered

Mihajlo Pupin

1858–1935 · Idvor, Serbia → Columbia University, New York

Pupin arrived at Castle Garden in 1874 as a fifteen-year-old Serbian shepherd with five cents and no English, and ended up the man who made $X_L = \omega L$ pay. By 1900, long-distance telephone calls died of distortion within a few hundred kilometres of cable. Pupin's patented cure was gloriously counterintuitive: add more inductance — deliberately, in lumps, as "loading coils" spaced down the line like beads on a string, tuned so the cable carried voice frequencies faithfully instead of smearing them. AT&T bought the patent, planted loading coils by the hundred thousand (the 1915 New York–San Francisco line used them by the tens of thousands alongside the new vacuum-tube repeaters), and the shepherd boy became a Columbia professor, a Pulitzer-winning memoirist (From Immigrant to Inventor, 1924), and the teacher of a generation that included Millikan and Armstrong. One honest asterisk: Oliver Heaviside had published the theory of inductive loading years earlier and never patented a thing — the priority fight embittered him for life. Pupin's gift was the engineer's half: turning the coil's frequency toll from an enemy into a product line.

History

1831

— Faraday's induction experiments reveal that changing current, not current, is what a coil fights

1832

— Joseph Henry describes self-induction: a coil opposes changes in its own current

1887

— Heaviside's cable theory shows deliberate inductance can cure a telephone line's distortion — a proposal without a patent

1900

— Pupin patents the loading coil; AT&T licenses it and "pupinization" spreads down the world's telephone trunks

1915

— the first transcontinental telephone line opens, its inductance engineered coil by coil from New York to San Francisco

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.4) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, a.c. through an inductor · Pupin, From Immigrant to Inventor (1924), for the loading-coil story.

Provenance. All results standard. A verifier should recompute: $2\pi \times 50 \times 0.5 = 157.1\ \Omega$ , $230/157 = 1.46$ A, $\sqrt{2}\times 1.46 = 2.07$ A; doubled frequency $314\ \Omega$ , $0.73$ A; $V_0 = 314 \times 0.5 \times 2.0 = 314$ V, rms $222$ V; $\omega t$ at $5$ ms $= 1.57$ rad $= 90°$ , $vi = 0$ there; unit audit H·s⁻¹ = Ω; anchors $0.1\ \mathrm{H} \leftrightarrow 31.4\ \Omega$ and $25.5\ \mathrm{mH} \leftrightarrow 8.0\ \Omega$ at 50 Hz.

Prerequisites. Phasors: The Rotating-Arrow Toolkit (the 90° language) · Self-Inductance & Inductors (the back-EMF law this module rides). Leads to. AC Through a Capacitor (the mirror-image arrow, next) · The Series LCR Circuit & Impedance (both arrows assembled) · Power in AC Circuits & Power Factor (the choke's zero-power claim cashed) · Alternating-Current Circuits & Phasors (UG deep dive).

If the inductor has no resistance, what is physically stopping the current from becoming infinite?

Its own back-EMF. At every instant the coil generates an EMF proportional to how fast the current is changing, pointed against the change — and on AC the current is perpetually changing, so the opposition never rests. The supply voltage is spent entirely on out-shouting this back-EMF (there is no resistance to spend it on), and that contest is what caps the current at $V_0/\omega L$ . Notice the two knobs in the cap: a bigger $L$ shouts louder per unit change, and a bigger $\omega$ forces faster change — either raises the toll. On DC the contest ends: nothing changes, the back-EMF dies, and only actual resistance (ideally none) remains. That is why "no resistance" and "finite AC current" coexist without contradiction.

How can the voltage across the inductor be at its peak when the current through it is zero? Doesn't zero current mean nothing is happening?

Zero current is when the most is happening — the current is sprinting through zero at its maximum rate of change, and rate of change is the only thing an inductor charges for. Freeze the wave at a zero-crossing: in the next millisecond the current will change by more than in any other millisecond of the cycle, so the flux is changing fastest, the back-EMF is largest, and the supply must apply its biggest voltage right now to keep the schedule. Conversely at the current's crest the wave is momentarily flat — no change, no back-EMF, zero volts, even though the current itself is maximal. Divorce "big current" from "big voltage" in your head for this chapter; in reactive elements they happen a quarter cycle apart, on purpose.

Where does the energy go during the quarter cycles when the inductor is absorbing power?

Into the magnetic field, on deposit. While the current grows from zero to its crest, the coil's stored energy grows from zero to $\tfrac{1}{2}Li_0^2$ (Energy in an Inductor & the Magnetic Field owns that formula) — that quarter-cycle's worth of supply energy is parked in the field around the windings. During the following quarter, the current falls, the field collapses, and the coil drives energy back into the circuit — the collapsing flux acts as a source, exactly the break-spark physics from Self-Inductance tamed into a smooth rhythm. Deposit, refund, deposit, refund, 100 times a second on 50 Hz mains. Nothing is lost anywhere (no resistance, no heat), which is why the average power is exactly zero and why utilities need the separate bookkeeping you'll meet as "wattless current."

Is X_L = ωL valid at every instant, like Ohm's law?

No — and this is the sharpest conceptual difference from a resistor. For a resistor, $v = iR$ holds moment by moment; the two waves are locked in phase and their ratio is always $R$ . For an inductor, $X_L$ relates only the amplitudes (and hence the rms values): $V_0 = i_0X_L$ . The instantaneous ratio $v/i$ swings through every value from zero (at current crests) to infinite (at current zeros) during each cycle, because the waves are 90° apart. So use $X_L$ for sizes and the "+90°" rule for timing, and never write $v(t) = i(t)X_L$ . The object that restores a genuinely Ohm's-law-shaped statement — one number carrying both size and timing for a whole circuit — is the impedance, assembled two modules from now.

Why does an iron core make the lamp dimmer in the demonstration?

The core multiplies the inductance. Iron's magnetization amplifies the flux through the coil enormously for the same current (the ferromagnetism module's story), and inductance is flux-per-ampere — so sliding the core in raises $L$ by a large factor, typically hundreds for a good core. Bigger $L$ means bigger $X_L = \omega L$ , a smaller current, a dimmer lamp — a live graph of the reactance formula with your hand as the frequency… actually as the $L$ -axis. It is also a preview of a design rule: every mains-frequency choke, ballast and transformer you will meet carries an iron or ferrite core, because winding enough turns to get the needed $L$ from air alone would take impractical kilometres of copper.

My book solves this with derivatives — di/dt and L di/dt. Do I need calculus for this module?

No. The one calculus-flavoured fact used here is that the slope-curve of $\sin\omega t$ is $\omega\cos\omega t$ — and you already own it as an oscillations-chapter result (the velocity of an SHM oscillator peaks at the centre and vanishes at the extremes, with amplitude scaled by $\omega$ ). Everything else is reading landmarks off a graph: slope biggest at the zero-crossings, zero at the crests, therefore the voltage curve is the current curve's slope story, a quarter cycle ahead. If you do know differentiation, the derivation compresses to two lines ( $v = L\,di/dt$ with $i = i_0\sin\omega t$ gives $v = \omega Li_0\cos\omega t$ ) and boards accept it happily — CBSE and ISC both present it that way. But nothing in this chapter's exams requires manipulating a derivative you couldn't replace with the slope-of-tangent sentence.

What happens if I connect a real inductor — one with some winding resistance — to the AC supply?

You get a two-personality component: the same physical coil is simultaneously a resistance $R$ (the copper) and a reactance $X_L$ (the geometry), and the two oppositions do not simply add, because their voltages are 90° out of step. Handling that honestly — adding them like perpendicular legs of a triangle — is precisely the machinery of the Series LCR module ( $Z = \sqrt{R^2 + X_L^2}$ for this case). For now, two serviceable limits: if $X_L \gg R$ (a good choke at mains frequency), the coil behaves nearly ideally and the module's results apply; if you put the coil on DC, $X_L$ vanishes entirely and $R$ alone limits the current — often ruinously, which is why AC-designed coils can burn out on DC.