Mistake Master
Student view — seeing the site as a student does
CED objectives

Circuits with Resistors and Inductors (LR Circuits)

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletLR Switch Lab · commit to what the coil does at the switching instant, then walk the transient in units of L/R

An inductor holds its current across any switching instant, so at $t = 0^+$ in a fresh circuit it is an open branch and long after, with $dI/dt = 0$, it is a wire: the exact mirror of the capacitor's (wire, open). The loop rule $\varepsilon - IR - L\,dI/dt = 0$ integrates to $I = (\varepsilon/R)(1 - e^{-t/\tau})$ with $\tau = L/R$, so more resistance settles the current faster, toward a smaller final value. Building the current banks $\tfrac12 LI^2$ in the field on top of the resistor's heat, and at shutdown that energy is paid out through whatever path exists, $\int I^2R\,dt$ recovering it exactly.

Four errors do most of the damage. Reversing the endpoints, so full current flows at switch-on and the coil blocks it at steady state, usually by importing the capacitor's roles unmirrored. Building the time constant as $LR$ or scaling it the wrong way with $R$, which the units and the $RC$ contrast both catch. Letting the inductor current jump when a switch opens, when in fact it continues and, offered only an air gap, arcs across it. And treating the field's $\tfrac12 LI^2$ as free at start-up or as vanishing at shutdown, when the ledger has to close at both ends.

ε = 12 V, L = 2.0 H: doubling R halves the time constant I t R = 4 Ω: 3.0 A, τ = 0.50 s R = 8 Ω: 1.5 A, τ = 0.25 s 0.25 s 0.50 s both start at slope ε/L = 6 A/s and I = 0: the coil is open at t = 0, a wire at the end
The higher-resistance circuit reaches its final current sooner. In an RC circuit the same doubling of R would slow the charging; here it speeds the settling.
the coil's current crosses the switching instant unchanged I_L t switch opens 3.0 A, steady still 3.0 A at 0+, then decays through 20 Ω 20 Ω reads 3.0 × 20 = 60 V, above the 12 V battery NOT a drop to zero: that needs infinite L dI/dt no other path: the EMF climbs until the switch gap arcs; the arc is the current continuing
Freeze the coil's current across the switching moment, then let the new circuit evolve it. The overshoot on the 20 ohm resistor is what a real inductive load does to a switch.

The work

3 ways in · any order
Lesson
Circuits with Resistors and Inductors (LR Circuits)

Sets the inductor's two endpoint roles against the capacitor's, derives the growth curve and its time constant from the loop rule, carries the coil's current across every switching instant, and closes the energy ledger on the field.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: swapping the inductor's switch-on and long-time roles, building the time constant as L times R or scaling it the wrong way with resistance, letting the inductor's current jump when a switch opens, and losing the field's energy at build-up or shutdown. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions