An ideal inductor is an inductor with negligible resistance, so the only potential difference across it comes from a changing current (V = L·dI/dt); once the current is constant at steady state, dI/dt = 0 and the ideal inductor behaves like a bare wire with zero voltage across it.
An ideal inductor is the simplified inductor model you use in AP Physics C: E&M (Topic 13.4). It has inductance L but zero internal resistance. That one assumption changes everything about how you analyze it. Since there's no resistance, the only voltage across an ideal inductor comes from self-induced emf, which depends on how fast the current is changing. In equation form, the magnitude of the potential difference is L·dI/dt.
This gives the ideal inductor two famous limiting behaviors in circuits. Right after a switch closes, the inductor fights any sudden change in current, so the current through it starts at whatever it was the instant before (often zero), making it act momentarily like an open switch. A long time later, when the current has settled to a constant value, dI/dt = 0, so the voltage across the inductor drops to zero and it acts like a plain piece of wire. The inductor also stores energy in its magnetic field, U = ½LI², and it stores or releases that energy without dissipating any of it (no resistance means no heat).
The ideal inductor lives in Topic 13.4 (Inductance) and is the backbone of every RL circuit problem in Unit 13. Almost every inductor question on the exam assumes the inductor is ideal unless the problem explicitly gives it a resistance. The two limiting cases (open circuit at t = 0, bare wire at steady state) let you solve for initial and final currents in an RL circuit using nothing but Kirchhoff's rules, before you ever touch the exponential solution. The energy expression U = ½LI² also connects inductors to the broader E&M theme of energy stored in fields, mirroring how capacitors store energy in electric fields. If you can't state what an ideal inductor does at t = 0 and at t → ∞, RL circuit FRQs become nearly impossible.
Keep studying AP® Physics C: E&M Unit 13
dI/dt (Unit 13)
The whole personality of an ideal inductor is captured by dI/dt. Its voltage is L·dI/dt, so a fast-changing current means a big voltage and a constant current means zero voltage. Released questions love flipping this around, like giving you the emf and asking for the rate of change of current (dI/dt = ε/L).
Energy stored in an inductor (Unit 13)
An ideal inductor stores U = ½LI² in its magnetic field and never wastes any of it as heat, since it has no resistance. The rate of energy storage is P = LI·dI/dt, so if current grows linearly as I = βt, the power being stored grows linearly with time too. Constant current means the stored energy just sits there unchanged.
Ideal capacitor steady-state behavior (Unit 11)
Capacitors and inductors are mirror images at steady state. A fully charged capacitor blocks current and acts like an open circuit, while an ideal inductor with constant current acts like a closed wire. Keeping these two straight is one of the highest-payoff memorizations in E&M.
Kirchhoff's rules in RL circuits (Units 11 & 13)
Treating the ideal inductor as a wire at steady state lets you collapse an RL circuit into a pure resistor circuit and solve it with the same Kirchhoff loop and junction analysis from Unit 11. That's exactly the move the 2024 FRQ on an RL circuit rewarded.
The 2024 FRQ Q2 put an ideal inductor in series with a battery and parallel resistors and asked for the resistance R, which hinges on knowing the inductor acts like a wire once current is steady. Multiple-choice questions hit the same ideas from different angles. You might be asked what happens to the stored energy when current is constant (it stays fixed, since U = ½LI² doesn't change), how the storage rate P compares at time t versus 3t when I = βt (it triples, because P = Lβ²t), how dI/dt scales when the emf doubles (it doubles, from dI/dt = ε/L), or how to derive the equivalent inductance of parallel inductors (they combine like parallel resistors, since each shares the same dI/dt-driving voltage). The skill being tested is always the same. Translate "ideal" into "V = L·dI/dt and nothing else," then apply the right limiting case.
Both are ideal circuit elements that store energy, but their steady-state behaviors are opposites. An ideal inductor with constant current has zero voltage across it and acts like a wire, while a fully charged ideal capacitor has zero current through it and acts like an open circuit. At t = 0 they swap roles. The inductor briefly acts like an open circuit (current can't jump instantly) while an uncharged capacitor briefly acts like a wire (voltage can't jump instantly). If you mix these up, you'll get initial and final currents exactly backwards on RL and RC circuit problems.
An ideal inductor has zero resistance, so the only voltage across it is the self-induced emf with magnitude L·dI/dt.
At steady state the current is constant, dI/dt = 0, and an ideal inductor acts like a plain wire with zero potential difference across it.
Immediately after a switch closes, the current through an inductor cannot change instantly, so it momentarily acts like an open circuit.
An ideal inductor stores energy U = ½LI² in its magnetic field and dissipates none of it as heat, and the energy stays constant whenever the current is constant.
The power flowing into an ideal inductor is P = LI·dI/dt, so a linearly increasing current I = βt means the storage rate grows linearly with time.
Ideal inductors in parallel combine like resistors in parallel (1/L_eq = 1/L₁ + 1/L₂) because each one sees the same voltage.
It's an inductor modeled with inductance L but zero internal resistance. Its voltage is entirely the induced emf L·dI/dt, it stores energy U = ½LI² in its magnetic field, and at steady state it has zero potential difference and acts like a bare wire.
No. At steady state the current is constant, so dI/dt = 0 and the voltage L·dI/dt is zero. The inductor behaves exactly like a piece of wire, which is the key step in solving for final currents in RL circuit FRQs like the 2024 exam's RL problem.
They're opposites at the two limits. At t = 0, an inductor acts like an open circuit (current can't jump) while an uncharged capacitor acts like a wire (voltage can't jump). At steady state the inductor becomes a wire and the capacitor becomes an open circuit.
No. With zero resistance there's no I²R heating. It stores energy U = ½LI² in its magnetic field while current increases and returns that energy to the circuit when current decreases. If the current is constant, the stored energy doesn't change at all.
Use the defining relationship ε = L·dI/dt and solve for dI/dt = ε/L. So if the emf across the inductor doubles from ε to 2ε at fixed L, the rate of change of current doubles too, a comparison exam questions ask about directly.
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