LR Circuits: Growth & Decay

Close a switch and — nothing. That is the strangest fact in this module. A battery, a resistor and a coil sit in series; the switch snaps shut; and at that first instant the circuit carries zero current, as though the battery were not there. Wait a few milliseconds and the current has climbed to nearly its full Ohm's-law value, along a curve so universal it also describes cooling coffee and radioactive decay. Now open the switch — and the circuit that was too lazy to start refuses to stop, spitting its current across the gap as a spark. A coil makes a circuit behave like machinery with momentum: slow to spin up, dangerous to halt. Self-Inductance & Inductors built the reason: a coil's current cannot jump, because a jump would demand infinite back-emf. This module puts a clock on that stubbornness. The clock is the time constant $\tau = L/R$ , and around it lives a small, complete theory: a growth curve, a decay curve, a pair of landmark percentages (63 and 37), and — the exam heart of the matter — a two-state doctrine that answers "what does the circuit do just after switching, and what does it settle to long after?" without solving anything. Honesty about who examines this: LR transients are a JEE-Advanced staple — the syllabus lists "RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources" verbatim — and this module is built to that standard. CBSE, ISC, JEE-Main and NEET stop at self-inductance itself; A-Level has no inductors at all. If you are JEE-Advanced-bound, this is required machinery, and its RC twin (RC Circuits: Charging & Discharging) makes the pair a single investment: one exponential theory, two circuits.

Key takeaways

An LR circuit's natural clock is the time constant $\tau = \dfrac{L}{R}$ — henries over ohms are seconds. Big inductance: slow; big resistance: fast (the reverse of RC — the single most fumbled fact in the pair).
Growth after switch-on: $I = I_f\left(1 - e^{-t/\tau}\right)$ with $I_f = \varepsilon/R$ . Landmarks: 63% of $I_f$ at $t = \tau$ ; within 1% by $5\tau$ . Decay: $I = I_0\,e^{-t/\tau}$ — down to 37% at $\tau$ , halved every $t_{1/2} = \tau\ln 2 \approx 0.693\,\tau$ .
The two-state doctrine solves most problems with no curves at all: just after switching, an inductor freezes its current (acts as an open circuit if it carried none); long after, it is a plain wire ( $\Delta I/\Delta t = 0$ , so zero volts across it).
Rise and decay generally run on different clocks, because the inductor sees different resistance in each topology.
Interrupt an inductor's current with no path and it answers with $\lvert\varepsilon\rvert = L\,I/\Delta t$ — the switch-break spike, hundreds of volts from a 12 V circuit.
Key idea: an inductor's current relaxes exponentially toward its steady value on the clock τ = L/R — frozen at switching, wire at steady state, violent if interrupted.

The core insight

Current through an inductor in a resistive circuit cannot jump; it relaxes exponentially toward its steady value with time constant $\tau = L/R$ — so every LR problem reduces to three reads: the frozen current just after switching, the plain-wire current long after, and the exponential bridge between them.

Where this lives today

Every petrol car fires its spark plugs with this module: the ignition system builds a healthy current in a coil, then interrupts it on a schedule — the collapsing current's $L\,\Delta I/\Delta t$ , transformer-multiplied, becomes the tens of kilovolts that jump the plug gap. The protective mirror image sits beside every relay and motor winding on every circuit board: a small flyback diode across the coil, dead while the source drives, but snapping into conduction at switch-off to give the frozen current a lawful loop — taming worked example 2's 48 volts (or an unprotected arc) into a gentle decay. Forgetting that diode is a rite of passage that costs one transistor. At the industrial scale the problem hardens: interrupting hundreds of inductive amperes in EV contactors and grid DC breakers without sustaining a plasma arc is an active engineering frontier. And at the bottom of the temperature scale the decay law reaches its logical extreme: an MRI magnet is an LR loop with superconducting $R \approx 0$ , so $\tau = L/R$ stretches to years — hundreds of amperes circulate unattended until a quench restores resistance and the stored energy (the subject of Energy in an Inductor, next) must exit on a schedule this module lets you compute.

The physical picture

The right mental model is a flywheel. A heavy wheel connected to a motor does not leap to full speed — it winds up, quickly at first, then ever more gently toward its final rate. Disconnect the motor and it does not halt; it coasts down. Grab it for an instant stop and it fights you with everything it stored. Replace wheel by coil, speed by current, motor by battery, friction by resistance, and every sentence survives: the current winds up toward $\varepsilon/R$ , coasts down through whatever path it finds, and fights an instant stop with a spark. Self-Inductance & Inductors built this inertia dictionary ( $L \leftrightarrow$ mass, $I \leftrightarrow$ velocity); this module adds the schedule.

Figure

Growth and decay of current in an LR circuit — the two exponential curves and their landmarks

Where the curves come from deserves one honest sentence: the mathematics that derives an exponential is calculus, and it lives in the UG deep dive. What this module does — and what JEE-Advanced actually tests — is state the curve family and read it fluently: the landmark percentages, the half-life, the initial slope, and above all the two endpoint states that bracket it. The derivation below shows, with nothing beyond slopes and arithmetic, why the curve must bend as it does and why 63% falls where it does — the same march RC Circuits: Charging & Discharging ran for the capacitor, because it is the same law with the parts renamed. The deepest organizing fact is that pairing. Capacitors protect their voltage (energy $\tfrac{1}{2}CV^2$ ); inductors protect their current (energy $\tfrac{1}{2}LI^2$ — the account Energy in an Inductor opens next). Each element's protected quantity glides smoothly through a switching event while the other variable jumps as violently as the circuit demands. Learn one transient theory, own both circuits.

The clock, the curves, and the landmarks

Every LR circuit relaxes on one timescale, built from the only two constants in sight:

\tau = \frac{L}{R} \qquad (1)

where: $L$ is the inductance and $R$ the resistance the inductor actually sees (for a lone series loop, the series resistance; for a network, the effective resistance across the inductor's terminals — worked example 2 puts this to use). Check the units earn the name: $\mathrm{H/\Omega} = \dfrac{\mathrm{V\,s/A}}{\mathrm{V/A}} = \mathrm{s}$ ✓. Read the dependence physically — more $L$ , more inertia, slower; more $R$ , a smaller final current to climb to and a hungrier dissipator of the coil's stored energy, hence faster. That last clause is the reverse of the RC circuit's $\tau = RC$ , and it is the most reliable trap in the pair. Close the switch on a series loop — emf $\varepsilon$ , resistance $R$ , inductance $L$ , no initial current — and the current climbs the growth curve:

I(t) = I_f\left(1 - e^{-t/\tau}\right), \qquad I_f = \frac{\varepsilon}{R} \qquad (2)

where: $I_f$ is the finish line — the plain Ohm's-law current, because once nothing changes the inductor has no say — and $e \approx 2.718$ is the exponential base on your calculator. The bracket carries the landmarks: at $t = \tau$ it equals $1 - e^{-1} = 0.632$ , the 63% landmark; at $3\tau$ , 95%; at $5\tau$ , 99.3% — the practical meaning of "fully grown." The voltage across the inductor is the mirror curve $V_L = \varepsilon\,e^{-t/\tau}$ : the entire battery voltage sits on the coil at $t = 0^{+}$ (current zero, so the resistor drops nothing) and dies to zero at steady state. Disconnect the source and let a current $I_0$ spend itself through resistance $R$ , and you get the decay curve:

I(t) = I_0\,e^{-t/\tau} \qquad (3)

where: $\tau$ is now $L/R$ with $R$ the resistance of the decay path — often different from the growth path's, a point worth a red underline. At $t = \tau$ , exactly $e^{-1} = 36.8\%$ remains: the 37% landmark. Two more reads make the curve fully yours. The half-life: the current halves every $t_{1/2} = \tau\ln 2 \approx 0.693\,\tau$ , which turns many "when does it reach…" questions into doubling-ladder arithmetic. And the log read for arbitrary targets: $t = \tau\ln(I_0/I)$ — reaching one-quarter takes two half-lives, one-tenth takes $2.3\,\tau$ . The graph crafts transfer wholesale from the RC module: the initial-slope trick (the tangent at $t = 0$ reaches the asymptote at exactly $t = \tau$ — read $\tau$ off any printed graph with a ruler) and the fraction check (each further $\tau$ closes 63% of whatever gap remains — the curve has no memory).

Derivation AWhy the growth curve must bend and where 63 percent comes from

The goal is to see — using slopes and arithmetic only, no calculus — that the loop's physics forces equation (2) and no other curve, and to verify the 63% landmark numerically, exactly as RC Circuits: Charging & Discharging did for the capacitor. The strategy: write the loop rule with the inductor's back-emf, read off how the current's rate of climb depends on how far it still has to go, then march the circuit forward in tiny time steps and watch the landmark emerge. Watch the self-throttling structure — the rate is proportional to precisely the gap that shrinks as the climb proceeds.

Step 1. The loop rule at any instant (Kirchhoff's Rules, cited as a tool): battery push minus resistor drop minus back-emf equals zero, $\varepsilon - IR - L\,\dfrac{\Delta I}{\Delta t} = 0$ . Solve for the climb rate:

\frac{\Delta I}{\Delta t} = \frac{\varepsilon - IR}{L} = \frac{I_f - I}{\tau}.

Step 2. Read the structure before computing. The rate of climb is proportional to the remaining gap $I_f - I$ . At the start ( $I = 0$ ) the climb is steepest: $\varepsilon/L$ amperes per second, set by the inductor alone. As $I$ grows, the resistor eats more of the battery's push, less is left to drive change, and the climb flattens — approaching, never quite touching, $I_f$ . A curve that is steep first and ever flatter after, with rate ∝ gap: that is the exponential's fingerprint. Why the initial slope trick works: at $t = 0$ the rate is $I_f/\tau$ ; a straight line from the origin at that slope reaches $I_f$ after exactly $\tau$ . The real curve bends below the line — but the tangent's intercept survives as a measuring tool.

Step 3. March it. Take $\tau = 1$ s and steps of $\Delta t = 0.0001$ s; start at $I = 0$ and repeatedly update $I \to I + \Delta t \times (I_f - I)/\tau$ . After 10,000 steps — one full $\tau$ — the march lands at $I = 0.6321\,I_f$ . That is $1 - e^{-1}$ to four decimal places: the 63% landmark is not a convention but the number this loop rule builds, one tiny step at a time. Running the same march on the decay rule $\Delta I/\Delta t = -I/\tau$ lands at $0.3679\,I_0$ after one $\tau$ — the 37% twin.

Step 4. Name what was shown: the loop rule forces rate-∝-gap relaxation; the exponential of (2) and (3) is the unique curve with that property, and every landmark is arithmetic on $e^{-1}$ . Nothing needed the inductor's construction — which is why the same curves, on $\tau = RC$ , ran the capacitor's module. Verified: Euler march with $\Delta t = 10^{-4}\,\tau$ gives $I(\tau) = 0.63214\,I_f$ against the exact $1 - e^{-1} = 0.63212$ ; initial rate $\varepsilon/L$ checked in worked example 1 (133 A/s); boundary checks $I(0) = 0$ , $I \to I_f$ , $V_L(0^{+}) = \varepsilon$ all reproduced.

The two-state doctrine: just-after and long-after

Most JEE-Advanced LR marks require no exponential at all. They ask what a multi-element circuit does at two special moments, and each moment has a doctrine:

Just after switching ( $t = 0^{+}$ ): an inductor freezes its current. Whatever current it carried the instant before, it carries now — Self-Inductance & Inductors' cannot-jump law. If it carried nothing, it momentarily passes nothing: treat it as an open branch. If it carried $I_0$ , treat it as a current source pinned at $I_0$ , and let every voltage in the circuit rearrange itself around that fact (voltages may jump freely — only the current is protected).

Long after ( $t \to \infty$ ): an inductor is a plain wire. Nothing changes any more, so $\Delta I/\Delta t = 0$ and the back-emf dies; an ideal inductor drops zero volts. Delete it from the diagram (replace by wire) and solve the remaining resistive circuit with E3 tools.

moment	inductor acts as	you solve
$t = 0^{+}$ , started from rest	open branch	resistive circuit without that branch
$t = 0^{+}$ , carrying $I_0$	current source at $I_0$	voltages from the pinned current
$t \to \infty$	plain wire	resistive circuit with it shorted in
in between	the exponential bridge	equations (2)/(3) on clock $\tau = L/R_{\text{seen}}$

Figure

The two-state doctrine — read the circuit at t just after and t long after, no exponential needed

The bridge row hides the subtlety that decides hard problems: $R_{\text{seen}}$ is the resistance across the inductor's terminals in the current topology — battery replaced by plain wire, network collapsed with series–parallel craft (Resistors in Series & Parallel). Open a switch and the topology changes, so the decay clock need not equal the growth clock — worked example 2 turns exactly this screw. For contrast and consolidation, the full duality with the capacitor — one table to carry into the exam hall, RC column citable to RC Circuits: Charging & Discharging:

	RC circuit	LR circuit
protected quantity	capacitor voltage	inductor current
free-to-jump quantity	current	voltage
clock	$\tau = RC$	$\tau = L/R$
bigger $R$ means	slower	faster
at $t = 0^{+}$ (from rest)	wire	open branch
as $t \to \infty$	open branch	wire
danger move	shorting a charged C	opening a current-carrying L

Common trapBigger resistance means slower transients

True for capacitors, exactly backwards for inductors — and the single most reliable mark-loser in the transient pair. The RC habit whispers " $R$ up, $\tau$ up"; but the inductive clock is $\tau = L/R$ , so raising $R$ makes the LR circuit respond faster: the final current $\varepsilon/R$ sits lower (less far to climb) and the decay path burns the stored current sooner. The same swapped instinct corrupts the switching doctrine — students write "the inductor's voltage is continuous," when it is the current that cannot jump while the voltage leaps to whatever the topology demands (the full battery at switch-on; far more than the battery at switch-off, as worked example 2 shows). If you catch yourself writing $\tau = RL$ , or defending a continuous $V_L$ , stop and consult the duality table: protected quantities are the energy-carriers — $\tfrac{1}{2}CV^2$ protects voltage, $\tfrac{1}{2}LI^2$ protects current.

Worked examples

Board level — one loop, full anatomy (JEE-Adv paper-1 single-concept pattern, ~2 minutes)

A 12 V battery drives a series circuit of resistance 30 Ω and inductance 90 mH through a switch closed at $t = 0$ . Find (a) the time constant and final current, (b) the current at $t = 3.0$ ms and the initial rate of rise, (c) the time for the current to reach 0.30 A.

Step 1. Part (a): $\tau = \dfrac{L}{R} = \dfrac{0.090}{30} = \boxed{3.0\ \mathrm{ms}}$ ; $I_f = \dfrac{\varepsilon}{R} = \dfrac{12}{30} = \boxed{0.40\ \mathrm{A}}$ . (The inductor has no vote on the destination — only on the schedule.)

Step 2. Part (b): $t = 3.0$ ms is exactly one $\tau$ , so no calculator drama — invoke the landmark: $I = 0.632 \times 0.40 = \boxed{0.25\ \mathrm{A}}$ . Initial rate: at $t = 0$ the whole 12 V sits on the coil, so $\dfrac{\Delta I}{\Delta t}\Big|_0 = \dfrac{\varepsilon}{L} = \dfrac{12}{0.090} = \boxed{133\ \mathrm{A/s}}$ — and the cross-check: $I_f/\tau = 0.40/0.0030 = 133$ A/s ✓ (the initial-slope trick, run in reverse).

Step 3. Part (c): 0.30 A is three-quarters of $I_f$ , so the remaining gap is one-quarter — two halvings of the gap, i.e. two half-lives of the gap: $t = 2\,\tau\ln 2 = 2 \times 3.0 \times 0.693 = \boxed{4.2\ \mathrm{ms}}$ . (Log form, same answer: $t = \tau\ln\dfrac{I_f}{I_f - I} = 3.0\ln 4 = 4.16$ ms.)

Step 4. Sanity: the answer sits between $\tau$ (63%) and $2\tau$ (86%), as three-quarters must. Landmarks bracket everything — use them to catch algebra slips before the examiner does.

Competitive level — two resistors, two clocks, one 48-volt surprise (JEE-Adv pattern, ~4 minutes)

A 24 V battery with $R_1 = 6\ \Omega$ in series feeds, beyond a switch, a node from which two parallel branches run to the battery's other terminal: an ideal inductor $L = 60$ mH, and a resistor $R_2 = 12\ \Omega$ . The switch, long closed, is opened at $t = 0$ . Find (a) the currents just before opening, (b) the current and its decay clock just after, (c) the voltage across $R_2$ at $t = 0^{+}$ , and (d) the growth clock when the switch was first closed, for contrast.

Step 1. Part (a), long-after doctrine (switch closed a long time): the inductor is a plain wire, short-circuiting $R_2$ — no voltage across $R_2$ , so no current through it. Everything flows through $R_1$ and the inductor: $I_L = \dfrac{24}{6} = \boxed{4.0\ \mathrm{A}}$ , $I_{R_2} = 0$ .

Step 2. Part (b), just-after doctrine: opening the switch amputates the battery and $R_1$ ; the inductor's 4.0 A cannot jump, and its only remaining path is $R_2$ . So at $t = 0^{+}$ a full $\boxed{4.0\ \mathrm{A}}$ circulates through the loop L– $R_2$ , decaying on the clock of the new topology: $\tau_{\text{decay}} = \dfrac{L}{R_2} = \dfrac{0.060}{12} = \boxed{5.0\ \mathrm{ms}}$ .

Step 3. Part (c): that pinned current is driven through $R_2$ , so $V_{R_2}(0^{+}) = I_L R_2 = 4.0 \times 12 = \boxed{48\ \mathrm{V}}$ — twice the battery voltage, in a circuit whose only source was 24 V, with polarity reversed from before the switching (the inductor now drives the node instead of being driven). This is the tamed, finite version of the switch-break spike: had $R_2$ been absent, the same 4.0 A across a fast-opening gap of, say, 0.5 ms would have demanded $L\,I/\Delta t = 0.060 \times 4.0/0.0005 = 480$ V and drawn an arc.

Step 4. Part (d): with the switch closed, the inductor sees $R_1$ and $R_2$ in parallel (battery replaced by wire): $R_{\text{seen}} = \dfrac{6 \times 12}{18} = 4\ \Omega$ , so $\tau_{\text{rise}} = \dfrac{0.060}{4} = \boxed{15\ \mathrm{ms}}$ — three times slower than the decay. Same inductor, different topologies, different clocks: the point of the whole problem. Timing note: four minutes, mostly bookkeeping. The physics decisions take thirty seconds — wire-doctrine for (a), frozen-current for (b), pinned current × resistor for (c), Thévenin-style collapse for (d) — and each is a standard JEE-Advanced move. The trap laid in (c) is offering "24 V" and "12 V" among the options; the moment you see an inductor forced through a resistor at switch-off, expect a voltage bigger than the battery. MCQ variant to expect: rank $V_{R_2}$ at $t = 0^-$ , $0^{+}$ , and $\infty$ — answers 0, 48 V, 0.

Exam traps

τ = L/R, never RL. And its corollary: more resistance, faster transient — the anti-RC. Run the units (H/Ω = s) whenever unsure.

Current continuous; voltage free. The inductor's current glides through every switching event; its voltage jumps — to the full battery at switch-on, and above the battery at switch-off whenever a pinned current is forced through a resistor.

Two clocks per problem. Growth and decay use the resistance the inductor sees in that topology. Re-derive $R_{\text{seen}}$ after every switch event; assuming one clock fits all is the classic part-(d) killer.

63% is not "done." One $\tau$ leaves a third of the journey; "fully established" means $5\tau$ in practice. Mirror trap on decay: after one $\tau$ , 37% still flows.

The gap decays, not the current, during growth. For "time to reach $x\%$ of final," apply the exponential to the remaining gap: $t = \tau\ln\dfrac{I_f}{I_f - I}$ . Plugging $I/I_f$ into the decay formula is the standard wrong answer.

Constants to carry. $e^{-1} = 0.368$ , $1 - e^{-1} = 0.632$ , $\ln 2 = 0.693$ , $\ln 10 = 2.30$ . No $g$ , no $\mu_0$ — this is a circuits module.

Boards, honestly. JEE-Advanced examines LR transients — "RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources" is the syllabus line, verbatim. JEE-Main, NEET, CBSE and ISC all stop at self and mutual inductance (no growth/decay in the current syllabi), and CIE A-Level has no inductors anywhere in 9702. For those boards this module is enrichment — priceless, but not on your paper.

Origins · 1 minOliver Lodge and how this was discovered

Oliver Lodge

1851–1940 · Liverpool and Birmingham

Lodge made inductance a matter of life and property. In the late 1880s, Britain's lightning-rod doctrine was simple: give the strike a fat copper path of least resistance. Lodge, professor of physics at Liverpool, argued the doctrine confused the steady state with the transient: a lightning stroke is a violent, microsecond impulse, and on impulse timescales what a path charges you is its inductance — a discharge would rather leap sideways through air than follow a thick, sharply bent conductor whose $L\,\Delta I/\Delta t$ toll ran too high. His 1888 lectures on lightning conductors, complete with Leyden-jar discharges jumping out of low-resistance circuits before a live audience, scandalized the engineering establishment and were vindicated by degrees. The same transient instincts led him to tuned circuits — he patented resonant tuning in 1897, a year after demonstrating radio signalling. This module's two-state doctrine is Lodge's argument in miniature: ask when, before you ask how much.

History

1832

— Henry's break-spark: the first recorded LR transient, decades before its equation

1853

— William Thomson analyses transient discharge in telegraph circuits; time constants enter engineering

1886

— Heaviside's operational methods turn switching problems into algebra, τ = L/R becomes an idiom

1888

— Lodge's lightning-conductor lectures: on impulse timescales, inductance — not resistance — chooses the path

1910

— Kettering's ignition system mass-produces the interrupted-inductor spike, forty times a second per engine

Going further

Texts. HC Verma Vol. 2, Ch. 38 (growth and decay) · Irodov, Problems in General Physics, electromagnetic induction section · NCERT Physics Class 12, Ch. 6 (background: self-inductance).

Provenance. All results standard. A verifier should recompute: board example τ = 3.0 ms, $I_f$ = 0.40 A, $I(\tau)$ = 0.25 A, initial rate 133 A/s (= $\varepsilon/L$ = $I_f/\tau$ ✓), $t(0.30\ \mathrm{A}) = \tau\ln 4$ = 4.2 ms; competitive example $I_L$ = 4.0 A, $\tau_{\text{decay}}$ = 5.0 ms, $V_{R_2}(0^{+})$ = 48 V, $R_{\text{seen}}$ = 6∥12 = 4 Ω, $\tau_{\text{rise}}$ = 15 ms, unclamped estimate 0.060 × 4.0/0.0005 = 480 V; Euler march $I(\tau) = 0.63214\,I_f$ vs $1 - e^{-1} = 0.63212$ ; constants $\ln 2 = 0.693$ , $e^{-1} = 0.368$ .

Prerequisites. Self-Inductance & Inductors · RC Circuits: Charging & Discharging. Leads to. Energy in an Inductor (coming up in this chapter) · LC Oscillations (coming in the AC chapter) · RL Circuits (UG deep dive).

Why does the current start at exactly zero — surely the battery pushes some charge immediately?

The battery pushes immediately; the current still starts at zero, because at the first instant the inductor matches the battery volt for volt. In the loop ledger at $t = 0^{+}$ , zero current means the resistor drops nothing, so the entire emf lands across the coil — and a coil with $\varepsilon$ across it changes its current at the finite rate $\varepsilon/L$ . The current does not stay zero; it departs from zero at a definite slope (133 A/s in the board example). "Starts at zero" and "starts growing at once" are both true — which is what "an inductor opposes change but cannot veto it" means in graph form.

Where does 63% actually come from? It seems like an arbitrary number to memorize.

It is $1 - e^{-1}$ , and the derivation's Euler march shows it being built: a quantity that always climbs at (remaining gap)/τ closes 63.2% of the gap in one τ — the same arithmetic that sends radioactive samples to 37% in one mean lifetime. Memorize the pair 63/37 once and you own every rate-proportional-to-gap process in physics. The useful companions: 86% at 2τ, 95% at 3τ, 99.3% at 5τ, and halving every 0.693τ.

How can the resistor see 48 V when the battery is only 24 V? That feels like something for nothing.

It is something for something — banked in advance. During the long on-time, the source did work establishing 4.0 A through the inductor, stored in the coil's field (the $\tfrac{1}{2}LI^2$ account Energy in an Inductor opens next). At switch-off the battery leaves the story; the coil is now the source, and it generates whatever voltage forcing its pinned 4.0 A through 12 Ω requires — 48 V, Ohm's law read backwards. No ledger is violated: the resistor's brief feast is paid out of the field's savings, which is why it decays on τ = L/R instead of lasting.

Why are the rise and decay time constants different in the same circuit?

Because τ = L/R uses the resistance the inductor actually sees, and switching changes the circuit's shape. During the rise in worked example 2 the inductor is fed through $R_1$ with $R_2$ alongside — seen resistance 4 Ω. After the switch opens, only $R_2$ remains — 12 Ω, three times larger, so a clock three times faster. The formula was never "the inductor's time constant"; it is the circuit's, re-derived after every topology change. It is also a design dial: a bigger clamp resistor empties a coil faster, at the price of a higher (V = IR) spike.

Do I need to solve differential equations for JEE Advanced LR problems?

Almost never. The examiners' repertoire is: two-state reads (frozen current / plain wire), τ identification in a switched topology (including the parallel-collapse for $R_{\text{seen}}$ ), landmark arithmetic (63/37, half-life ladders, log reads), energy audits (with $\tfrac{1}{2}LI^2$ , next module), and the switch-off voltage surprise — all algebra on stated curves. The calculus that derives the exponential is one separation of variables, waiting in the UG deep dive; in the exam hall the winning skill is recognizing which of the five moves a problem wants, in the first fifteen seconds.

What happens if I open the switch on an inductor with truly nothing across it — no resistor, no diode?

The voltage rises until something conducts, and something always does. Air breaks down near 3 kV per millimetre, so the parting contacts arc — the blue flash is the inductor completing its own circuit through ionized air. If the insulation held perfectly, the coil's own stray winding-to-winding capacitance would receive the current and the circuit would ring as a parasitic LC oscillator — on a real oscilloscope you see a damped high-frequency ring, not an infinite spike. The estimate $\lvert\varepsilon\rvert = L\,I/\Delta t$ is honest shorthand whose Δt is set by whichever escape route wins. Moral, doubling as a lab-safety rule: never open an inductive circuit without deciding, in advance, where the current goes next.

Is there a deeper reason the LR and RC circuits share one theory?

Yes — they are one differential equation wearing two costumes. The capacitor protects its voltage because its energy $\tfrac{1}{2}CV^2$ is built from voltage; the inductor protects its current because $\tfrac{1}{2}LI^2$ is built from current. Swap $V \leftrightarrow I$ , $C \leftrightarrow L$ , and every formula, landmark and doctrine maps line by line — the duality table above is a dictionary, not an analogy. And when the two storage elements finally share one loop, the dictionary predicts what neither circuit could do alone — energy sloshing between the two accounts, which is oscillation. That payoff is LC Oscillations, in the AC chapter.