The Series LCR Circuit & Impedance

Wire a resistor, an inductor and a capacitor in series across an AC supply and measure the three voltages with a good meter: say 40 V across the resistor, 90 V across the inductor, 60 V across the capacitor. Add them: 190 V. Now measure the supply: 50 volts. Every reading is correct, and the audit still refuses to balance — the phasor module left this scandal as a promissory note, and this module cashes it. The resolution is that the three voltages are not teammates pulling in one direction; they are arrows pointing three different ways. The resistor's voltage moves in step with the current; the inductor's runs a quarter turn ahead; the capacitor's a quarter turn behind. Adding them the honest way — tip to tail, as Phasors: The Rotating-Arrow Toolkit taught — produces a resultant whose length is set by a right triangle, not by arithmetic on a line. Out of that triangle falls the chapter's central quantity: the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ , the single ohm-valued number that plays Ohm's-law for a whole AC circuit, and the phase angle $\tan\varphi = (X_L - X_C)/R$ that says whether the circuit as a whole behaves like a coil or like a capacitor. One construction, two formulas, and every series AC problem on every board reduces to drawing a triangle and reading its sides. The frequency at which the triangle collapses to its base — where $X_L$ and $X_C$ annihilate — gets a module of its own next.

Key takeaways

In a series LCR circuit the current is common to all three elements — so the current phasor is the reference arrow, and each voltage hangs off it: $V_R$ in phase, $V_L$ at +90°, $V_C$ at −90°.
Tip-to-tail addition gives $V_0^2 = V_{R0}^2 + (V_{L0} - V_{C0})^2$ — element voltages add like perpendicular legs, never like plain numbers.
Substituting the element laws turns the voltage triangle into the impedance triangle: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ , with $I = V/Z$ and $\tan\varphi = \dfrac{X_L - X_C}{R}$ .
The sign of $X_L - X_C$ is the circuit's personality: positive → inductive (current lags the supply), negative → capacitive (current leads).
$Z$ traces a valley as frequency rises — huge at both extremes, minimum at one special frequency $f_0$ where $X_L = X_C$ ; the current correspondingly peaks there.
Key idea: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ — series AC circuits are solved by one right triangle whose legs are "in-phase ohms" and "quadrature ohms".

The core insight

Because one current threads every element of a series LCR circuit, the element voltages become arrows at fixed angles to that current — in phase for $R$ , ±90° for $L$ and $C$ — and their tip-to-tail sum yields a single Ohm's-law-shaped statement $V = IZ$ with $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and phase angle $\tan\varphi = (X_L - X_C)/R$ : the impedance triangle solves the whole circuit.

Where this lives today

The impedance triangle never retired — it just moved to higher frequencies. The "matching" that audio engineers fuss over between amplifier and speaker, the "50 Ω" printed on every antenna cable and RF connector in a lab, the reason your phone's antenna is precisely the shape it is: all impedance bookkeeping, all descendants of this module's construction. Power engineers run the same triangle at 50 Hz on a national scale — a transmission line is one enormous series RL circuit, and the grid's stability calculations begin from exactly the $Z$ and $\varphi$ defined here. Closer to the syllabus, the triangle is the workhorse of the chapter's remaining modules: the collapse of its vertical leg is Resonance & the Q-Factor; the cosine of its angle prices energy flow in Power in AC Circuits & Power Factor; and drilling the full pipeline — reactances, $Z$ , current, element voltages, personality — until it runs in ninety seconds is the express purpose of the LCR Numericals Ladder archetype. One triangle, four modules of dividends.

The physical picture

The one structural fact about a series circuit — the same fact that ran Resistors in Series & Parallel back in DC — is that every element carries the same current. At any instant, the charge flowing through the resistor is the charge flowing through the coil is the charge flowing onto the capacitor's plate. On AC that shared current is one sinusoid, $i = i_0\sin\omega t$ , and the phasor module's series doctrine follows: draw the current arrow first and hang everything else off it. Now ask what voltage each element demands in order to carry that shared current. The resistor is the easy tenant: $v_R$ rises and falls in step with $i$ — its arrow lies along the current arrow, length $i_0R$ . The inductor charges its toll a quarter cycle early: $v_L$ leads $i$ by 90°, an arrow of length $i_0X_L$ rotated a quarter turn ahead. The capacitor runs a quarter cycle late: $v_C$ lags $i$ by 90°, an arrow of length $i_0X_C$ a quarter turn behind. (Each 90° is a result its own module built from the slope of a sine — AC Through an Inductor and AC Through a Capacitor; here we spend them.) Notice what the geometry has already decided. The inductor's and capacitor's arrows point in exactly opposite directions — one straight up, one straight down, if the current points east. Before any algebra, the two reactive voltages partially cancel: the circuit's net quadrature demand is only the difference $V_L - V_C$ . The supply must furnish the vector sum of all three arrows, and the sum of a horizontal leg and a net vertical leg is the hypotenuse of a right triangle. That triangle is the entire theory of this module.

Figure

The phasor construction — three element voltages hang off one shared current arrow

The flagship derivation: from three arrows to Z

The construction above is the proof; here it is prosecuted step by step, because writing it out is the board answer.

Derivation ABuilding the impedance triangle from the shared current

The goal is the two central formulas of the chapter — $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and $\tan\varphi = (X_L - X_C)/R$ — derived from nothing but the three elements' phase rules and one tip-to-tail addition. The strategy: express each element's voltage amplitude in terms of the one shared current, place the three arrows at their fixed angles, add the perpendicular legs by Pythagoras, and then divide the whole triangle by the current to convert volts into ohms. The thing to watch is that only amplitudes and angles enter — no instant-by-instant addition is ever attempted.

Step 1. Let the shared current be $i = i_0\sin\omega t$ . The element laws give the three voltage amplitudes: $V_{R0} = i_0R$ , $V_{L0} = i_0X_L$ , $V_{C0} = i_0X_C$ , with phases $0$ , $+90°$ , $-90°$ relative to the current — each rule inherited by name from its own module.

Step 2. Add the arrows tip to tail (the operation Phasors: The Rotating-Arrow Toolkit legitimized: arrows rotating at the same $\omega$ keep fixed mutual angles, so they add like ordinary vectors). $V_L$ and $V_C$ lie on the same line, oppositely directed: their sum is a single vertical arrow of length $|V_{L0} - V_{C0}|$ , pointing up if the inductor wins, down if the capacitor does. This vertical arrow is perpendicular to $V_R$ .

Step 3. The supply voltage is the resultant of the horizontal leg $V_{R0}$ and the vertical leg $V_{L0} - V_{C0}$ . Pythagoras:

V_0^2 = V_{R0}^2 + (V_{L0} - V_{C0})^2

This is the audit that fixes the opening scandal — element voltages combine as perpendicular legs. Check it on the promissory numbers: $V_R = 40$ , $V_L = 90$ , $V_C = 60$ → net vertical $30$ , so $V = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50$ V ✓. The meters were honest; the addition was wrong.

Step 4. Substitute the element laws and factor out the shared $i_0$ :

V_0 = i_0\sqrt{R^2 + (X_L - X_C)^2} \equiv i_0 Z, \qquad Z = \sqrt{R^2 + (X_L - X_C)^2}

Why dividing by $i_0$ is the decisive move: it converts the voltage triangle into a triangle of pure ohms — legs $R$ and $X_L - X_C$ , hypotenuse $Z$ — a property of the circuit alone, independent of how hard it is driven. This impedance triangle is the object you redraw in every problem.

Step 5. Read the angle. The supply arrow makes an angle $\varphi$ with the current arrow, fixed by the triangle's legs:

\tan\varphi = \frac{V_{L0} - V_{C0}}{V_{R0}} = \frac{X_L - X_C}{R}

If $X_L > X_C$ the vertical leg points up: the supply voltage leads the current — the circuit is net inductive. If $X_C > X_L$ it points down: the current leads — net capacitive. The full circuit obeys $V_0 = i_0Z$ and $V_{\mathrm{rms}} = I_{\mathrm{rms}}Z$ : Ohm's law regained, at the price of one angle. Verified: with the anchor set $R = 30\ \Omega$ , $X_L = 70\ \Omega$ , $X_C = 30\ \Omega$ : $Z = \sqrt{30^2 + 40^2} = 50\ \Omega$ ✓; $\tan\varphi = 40/30 = 4/3$ , $\varphi = 53.1°$ ✓; on a 200 V supply $I = 4$ A, and the element voltages $120$ , $280$ , $120$ V recombine as $\sqrt{120^2 + 160^2} = 200$ V ✓ — the books balance.

Two habits worth building immediately. First, always compute $X_L - X_C$ before squaring — the difference, with its sign, is the circuit's personality, and carelessly adding the reactances is the single most expensive slip in the chapter. Second, note what the triangle says about sizes: $Z$ can never be smaller than $R$ , and individual element voltages can tower far above the supply — in the verified anchor, a 200 V supply drives 280 V across the inductor, with no contradiction, because arrows at 90° do not compete along the same line.

Who leads: the circuit's two personalities

The sign of $X_L - X_C$ splits every series LCR circuit into one of two regimes, and — since $X_L$ rises with frequency while $X_C$ falls — the same circuit visits both as the frequency sweeps.

Net inductive ( $X_L > X_C$ , high frequencies): the inductor's toll dominates; the vertical leg points up; the supply voltage leads the current by $\varphi$ . The circuit behaves like a resistor in series with a leftover inductor.

Net capacitive ( $X_C > X_L$ , low frequencies): the capacitor dominates; the current leads the supply voltage. The circuit is a resistor plus a leftover capacitor. Between the regimes sits the crossover $X_L = X_C$ : the vertical leg vanishes, the triangle collapses to its base, $Z$ bottoms out at $R$ exactly, and $\varphi = 0$ — the circuit momentarily impersonates a plain resistor. The frequency where this happens, $f_0$ , is fixed by the components alone, and the current there is the largest the circuit can ever deliver, $I = V/R$ . The physics of that special point — why it deserves the name resonance, how sharp the peak is, what the sharpness buys — is the next module's whole business; here we only mark the spot on the graphs.

Figure

Z and I versus frequency — the impedance valley and the current peak, minimum and maximum at f₀

The special cases boards love: RL and RC in series

Delete one element and the triangle simplifies but never disappears — these two-element circuits are explicit ISC syllabus items and everyone else's warm-up questions.

Series RL (a real coil, in fact: winding resistance plus inductance in one package): no capacitor, so $X_C = 0$ and $Z = \sqrt{R^2 + X_L^2}$ , $\tan\varphi = X_L/R$ with the voltage always leading — the circuit is inductive at every frequency. Series RC: $Z = \sqrt{R^2 + X_C^2}$ , $\tan\varphi = -X_C/R$ , current always leading. The four pure and semi-pure circuits now line up as one family:

circuit	$Z$	phase angle $\varphi$	personality
pure $R$	$R$	$0$	in phase
pure $L$	$X_L$	$+90°$ (v leads)	inductive
pure $C$	$X_C$	$-90°$ (i leads)	capacitive
series RL	$\sqrt{R^2 + X_L^2}$	$\tan\varphi = X_L/R$ , $0<\varphi<90°$	always inductive
series RC	$\sqrt{R^2 + X_C^2}$	$\tan\varphi = -X_C/R$	always capacitive
series LCR	$\sqrt{R^2 + (X_L - X_C)^2}$	$\tan\varphi = (X_L - X_C)/R$	depends on $f$ : capacitive below $f_0$ , inductive above

Every row is the same triangle with legs relabelled — learn the construction once and the table is free.

Common trapElement voltages in series add up to the supply voltage

In an AC series circuit they do not — not as numbers. $V_R + V_L + V_C$ can exceed the supply several times over (40 + 90 + 60 = 190 against a 50 V supply, in this module's opening audit), and an individual element voltage can dwarf the supply on its own. The only addition the physics permits is tip-to-tail: $V^2 = V_R^2 + (V_L - V_C)^2$ . Three separate exam injuries flow from ignoring this. First, "find the supply voltage" problems: adding the three meter readings scores zero; the Pythagorean audit scores full. Second, "the voltmeter across L reads more than the source — is something wrong?" — no; arrows at 90° don't compete along a line, and $V_L > V$ is routine near resonance. Third, the sneaky inverse: given the supply and two element voltages, the third comes from the triangle, not from subtraction. One discipline prevents all three: never add AC voltages without first asking what angle they hold between them. Instantaneous voltages do add plainly ( $v = v_R + v_L + v_C$ at every moment — Kirchhoff survives); it is the amplitudes and rms values, the numbers meters report, that live on a triangle.

Worked examples

Board level — the full triangle, on mains numbers (CBSE/ISC pattern, 5 marks)

A series circuit contains $R = 40\ \Omega$ , $L = 254.6$ mH and $C = 63.66\ \mathrm{\mu F}$ , driven by a 230 V, 50 Hz supply. Find (a) the reactances and impedance, (b) the current, (c) the phase angle and the circuit's nature, (d) the rms voltage across each element, verifying the phasor sum.

Step 1. Reactances first, always: $X_L = 2\pi fL = 314.16 \times 0.2546 = 80\ \Omega$ ; $X_C = \dfrac{1}{2\pi fC} = \dfrac{1}{314.16 \times 63.66\times10^{-6}} = 50\ \Omega$ .

Step 2. Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \boxed{50\ \Omega}$ . (A 3-4-5 triangle scaled by 10 — setters recycle these proportions endlessly; learn to spot them.)

Step 3. Current: $I = V/Z = 230/50 = \boxed{4.6\ \mathrm{A}}$ .

Step 4. Phase: $\tan\varphi = (80 - 50)/40 = 0.75$ , so $\varphi = 36.9°$ ; $X_L > X_C$ , hence the circuit is inductive — the supply voltage leads the current by 36.9°.

Step 5. Element voltages: $V_R = IR = 184$ V, $V_L = IX_L = 368$ V, $V_C = IX_C = 230$ V. Verify: $\sqrt{184^2 + (368-230)^2} = \sqrt{184^2 + 138^2} = 230$ V ✓ — and note without alarm that $V_L = 368$ V exceeds the 230 V supply.

Competitive level — the anchor triangle under interrogation (JEE-Main pattern, ~3 minutes)

A series LCR circuit with $R = 30\ \Omega$ , $X_L = 70\ \Omega$ and $X_C = 30\ \Omega$ is connected to a 200 V (rms) AC source. Find (a) the current, (b) the rms voltage across each element and across the LC pair taken together, (c) by how much the source frequency would need to conceptually change for the circuit to look purely resistive — qualitatively, which way?

Step 1. $Z = \sqrt{30^2 + (70-30)^2} = \sqrt{900 + 1600} = 50\ \Omega$ , so $I = 200/50 = \boxed{4\ \mathrm{A}}$ , with $\tan\varphi = 40/30 = 4/3$ ( $\varphi = 53.1°$ , inductive — the 3-4-5 family again).

Step 2. Element voltages: $V_R = 120$ V, $V_L = 280$ V, $V_C = 120$ V. The LC pair together: their arrows oppose, so the combination shows only $|V_L - V_C| = \boxed{160\ \mathrm{V}}$ — a voltmeter across both L and C reads 160 V, less than the 280 V across L alone. Audit: $\sqrt{120^2 + 160^2} = 200$ V ✓.

Step 3. Purely resistive means $X_L = X_C$ . Here $X_L > X_C$ , and lowering the frequency lowers $X_L$ while raising $X_C$ — so the frequency must decrease until the two meet. (Computing that frequency is the resonance module's opening act.) Timing note: 3 minutes. The mark-earning subtleties are in (b): the LC-pair voltage is a subtraction, not an addition — a voltmeter across both reactive elements reads less than across one of them, which feels wrong and is right. And keep $V_L = 280 > 200 = V$ moving without hesitation; near-resonant circuits do this as a matter of course.

Exam traps

Subtract before you square. The reactance leg is $X_L - X_C$ , sign and all — never $X_L + X_C$ . A circuit with $X_L = X_C = 1000\ \Omega$ and $R = 5\ \Omega$ has $Z = 5\ \Omega$ , not 2000. This is simultaneously the most common and the most expensive slip in the chapter.

$Z$ is never less than $R$ . The hypotenuse cannot undercut a leg. If your computed $Z < R$ , you subtracted wrongly or slipped a unit — a free self-check that costs two seconds.

Voltages add on a triangle. $V \neq V_R + V_L + V_C$ ; use $V^2 = V_R^2 + (V_L - V_C)^2$ . Given two of the three and the supply, solve the triangle — resist ordinary subtraction.

An element voltage above the supply is not an error. $V_L$ or $V_C$ routinely exceed $V$ near resonance. Flagging it as impossible — or "correcting" a right answer — costs marks that were already earned.

Phase angle sign tells the personality. $\varphi > 0$ ( $X_L > X_C$ ): inductive, voltage leads. $\varphi < 0$ : capacitive, current leads. Examiners ask for "the nature of the circuit" as a follow-up mark; answer it from the sign, not from re-derivation.

Frequency moves both reactances, opposite ways. Raising $f$ raises $X_L$ and lowers $X_C$ — students who move only one leg predict the wrong regime. Any frequency question deserves a two-line check of both.

Boards and depth. CBSE: series LCR by phasor diagram ("phasors only" is the syllabus's own phrase), impedance, the resonance condition as the special case — this module plus the next. ISC: adds the explicit $V_m^2 = V_{Rm}^2 + (V_{Lm} - V_{Cm})^2$ derivation route, the RL and RC special cases, and the $Z$ – $f$ and $I$ – $f$ graphs — all drawn here. JEE-Main/NEET: triangle numericals, element-voltage audits, personality questions. JEE-Advanced: the same machinery inside multi-part problems with d.c./a.c. sources mixed. 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 minArthur Edwin Kennelly and how this was discovered

Arthur Edwin Kennelly

1861–1939 · Edison's lab, New Jersey → Harvard & MIT

Kennelly is the engineer who gave this module's triangle its name. He arrived sideways: an Irish-born telegraph operator who never finished school, he became Thomas Edison's chief electrical assistant — DC royalty — and then watched the AC systems he'd been hired to argue against win the war. Rather than go down with the ship, he crossed over and, in 1893, read to the American Institute of Electrical Engineers a paper titled simply "Impedance": treat resistance and reactance as the two legs of one right triangle, he proposed, and the maddening bookkeeping of AC circuits collapses into geometry any engineer can draw. (At the same 1893 congress, Steinmetz pushed the algebra further — that half of the story is told in the Phasors module.) The word and the triangle stuck; every $Z = \sqrt{R^2 + X^2}$ computed since is Kennelly's memo executed. He spent the rest of his career at Harvard and MIT, and in 1902 made a second immortal guess: that a conducting layer high in the atmosphere was bouncing Marconi's radio waves around the Earth's curve — the Kennelly–Heaviside layer, confirmed a generation later. One man, two legacies: the triangle in every AC problem, and the mirror in the sky that made long-range radio work.

History

1886

— the Westinghouse AC system begins commercial service; AC circuit design becomes an urgent industrial problem

1893

— Kennelly reads "Impedance" to the AIEE: resistance and reactance as perpendicular legs of one triangle

1893

— at the same Chicago congress, Steinmetz systematizes the phasor algebra that makes triangle bookkeeping routine

1902

— Kennelly (and independently Heaviside) predicts the ionospheric layer that reflects radio waves

1912

— impedance methods are standard curriculum for electrical engineers; the triangle enters the textbooks for good

Going further

Texts. NCERT Physics Class 12, Ch. 7 (§7.7) · HC Verma Vol. 2, Ch. 39 · SL Arora, ISC Physics XII, series LCR circuit · Kennelly, "Impedance" (AIEE, 1893), for the original triangle.

Provenance. All results standard. A verifier should recompute: $X_L = 314.16 \times 0.2546 = 80.0\ \Omega$ , $X_C = 1/(314.16 \times 63.66\times10^{-6}) = 50.0\ \Omega$ , $Z = \sqrt{1600+900} = 50\ \Omega$ , $I = 4.6$ A, $\varphi = \arctan(0.75) = 36.9°$ , $V_R, V_L, V_C = 184, 368, 230$ V, $\sqrt{184^2+138^2} = 230$ V ✓; anchor set $Z = 50$ , $I = 4$ , $120/280/120$ V, $\sqrt{120^2+160^2} = 200$ V ✓; teaser audit $\sqrt{40^2 + 30^2} = 50$ V ✓.

Prerequisites. AC Through an Inductor and AC Through a Capacitor (the two 90° arrows) · Phasors: The Rotating-Arrow Toolkit (tip-to-tail addition). Leads to. Resonance & the Q-Factor (the collapsed triangle, next) · Power in AC Circuits & Power Factor (the angle priced in watts) · ARCHETYPE: The LCR Numericals Ladder (the pipeline drilled) · Alternating-Current Circuits & Phasors (UG deep dive).

Why is the current chosen as the reference arrow and not the supply voltage?

Because the current is the one thing every element shares. In a series circuit the same $i(t)$ threads R, L and C, so each element's voltage rule ("in phase", "+90°", "−90°") is stated relative to the current — hanging all three arrows off the current arrow uses each rule exactly as written. If you started from the supply voltage instead, you would need to know $\varphi$ before drawing anything, and $\varphi$ is the answer, not the input. (In a parallel circuit the logic flips: the shared quantity is the voltage, and current arrows hang off it — a UG-side story.) The habit to build: identify what is shared, make it the reference, hang the rules off it.

Do the instantaneous voltages add up normally, or is Kirchhoff's law suspended in AC?

Kirchhoff survives intact — instant by instant, $v_{\text{supply}}(t) = v_R(t) + v_L(t) + v_C(t)$ at every moment, no triangle involved. What fails is adding the peaks: the three element voltages peak at different times (the resistor's with the current, the inductor's a quarter cycle early, the capacitor's a quarter cycle late), so the supply's peak is not the sum of their peaks — a maximum of a sum is not the sum of maxima when the summands refuse to peak together. The triangle is precisely the machine that adds peaks (or rms values) honestly, by carrying each one's timing as an angle. So: instantaneous values — plain addition; amplitudes and rms values — tip to tail, always.

How can the voltage across the inductor be larger than the supply voltage without violating energy conservation?

Because voltage is not energy. The inductor and capacitor arrows oppose each other: while the inductor demands 280 V of quadrature voltage, the capacitor supplies 120 V of exactly opposite timing, so the pair costs the supply only 160 V. The large individual voltages reflect large energy swaps between L and C — energy sloshing back and forth through the shared current, borrowed and repaid every half cycle without the supply's involvement. The supply only ever furnishes what the resistor burns plus the net quadrature difference. Energy conservation is enforced cycle by cycle in the power ledger (the next-but-one module); no element creates energy by hosting a big voltage. Near resonance the swap grows spectacular — hundreds of volts across L and C from a modest source — and that "voltage magnification" becomes a feature, exploited deliberately in radio tuning.

Is Z a resistance? It's in ohms and sits in V = IZ, so what's the difference?

$Z$ has the units and the Ohm's-law seat, but it differs in two respects. First, only the $R$ inside it dissipates: a circuit with $Z = 50\ \Omega$ made of $R = 30$ and net reactance $40$ converts to heat only what $R$ eats; the reactive part shuttles energy in and out without loss. Feeding $Z$ into $P = I^2 \times (\text{anything but } R)$ is wrong — the honest power story needs the phase angle and is the Power module's business. Second, $Z$ is frequency-dependent: change $f$ and both reactance legs move (opposite ways), so $Z$ is a property of the circuit at a stated frequency, not a constant of the hardware. A resistor's $R$ is the same on any supply; a circuit's $Z$ must be recomputed the moment the frequency changes.

What happens to the triangle in a circuit with L and C but almost no R?

The base shrinks toward zero and the triangle degenerates into a vertical line: $Z \to |X_L - X_C|$ and $\varphi \to \pm90°$ . Away from the crossover frequency the circuit still limits current (the leftover reactance does it alone, without dissipation), but at the crossover something dramatic happens: with $X_L = X_C$ and $R \approx 0$ , the impedance collapses toward zero and the current grows enormous — limited only by whatever tiny resistance reality provides. That is why resonance in low-resistance circuits is violent, why the resonance module's curves get taller and sharper as $R$ drops, and why real tuned circuits always budget carefully for the small residual resistance of their coil. The triangle's base, however thin, is what keeps the resonant current finite.

Under exam pressure, how do I rebuild tan φ = (X_L − X_C)/R if I blank on it?

Redraw the thirty-second picture. Current arrow east. $V_R = IR$ along it. $V_L = IX_L$ north (inductor's voltage leads). $V_C = IX_C$ south. Net vertical: $I(X_L - X_C)$ . The supply arrow is the diagonal of the rectangle with legs $IR$ east and $I(X_L - X_C)$ north; the angle it makes with the current satisfies $\tan\varphi = \text{opposite}/\text{adjacent} = I(X_L - X_C)/IR$ , and the $I$ cancels. Pythagoras on the same legs gives $Z$ . Every formula in this module is that one sketch read twice — which is why the sketch, not the formulas, is the thing to memorize. Drawing it also scores the diagram marks that CBSE and ISC award explicitly.