An LR circuit is a loop with a resistor (R) and an inductor (L) in series (often with a battery and switch). When you close the switch, the inductor resists changes in current by producing an induced emf (back emf), so current grows exponentially rather than instantaneously. Using Kirchhoff’s loop rule you get the governing differential equation E = L dI/dt + IR. The solution is exponential with time constant τ = L / R_eq: starting from zero current the inductor current approaches its steady (asymptotic) value I(∞)=E/R with ~63% reached at t = τ. If the inductor has initial current and you open the circuit, the current decays exponentially (≈37% of initial at t = τ) and the resistor dissipates the inductor’s stored magnetic energy. After many τ the inductor behaves like a short (wire). These are exactly the CED ideas you’ll be tested on (derive the differential equation, interpret τ, sketch transients). For the AP topic study guide and extra practice, see the LR circuits study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu), unit overview (https://library.fiveable.me/ap-physics-c-e-m/unit-13), and practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

When you flip the switch the current tries to change quickly, and the inductor “fights” that change because a changing current changes the magnetic flux through the coil. By Faraday/Lenz, that changing flux induces an emf that opposes the change in current (a “back emf”). Mathematically from Kirchhoff’s loop rule for an LR series circuit: E = L (dI/dt) + IR—the L(dI/dt) term is the induced emf. Right after the switch flips dI/dt is large, so the back emf can nearly cancel the battery and keep the current from jumping instantly. Physically the circuit stores (or releases) energy in the inductor’s magnetic field; the resistor then dissipates that energy as current relaxes. The speed of the change is set by the time constant τ = L / R—bigger L or smaller R means the inductor resists changes longer. For the AP exam, be ready to use E = L dI/dt + IR, explain back emf/Lenz’s law, and sketch the exponential transient (see the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu)). For extra practice try the unit practice set (https://library.fiveable.me/practice/ap-physics-c-e-m).

Start with Kirchhoff + the inductor law: E = L (dI/dt) + IR (CED 13.5.A.2). Rearranging for a charging LR circuit gives dI/dt + (R/L) I = E/L. That’s a first-order linear ODE whose standard solution has the form I(t) = I(∞) + [I(0) − I(∞)] e^(−t/τ). The coefficient multiplying I in the ODE is R/L, so the exponent is −(R/L) t. By definition τ is the reciprocal of that rate, so τ = L/R (CED 13.5.A.3.i). Physically: L is the “inertia” for electric current (it resists changes in I by producing a back-emf ∝ dI/dt), while R is the dissipative term that slows changes. So time to change current scales with how big the inertia (L) is and inversely with how strong the damping (R) is. That’s why τ = L/R—larger L → slower change; larger R → faster damping of transients. (See Topic 13.5 study guide: https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu and practice problems at https://library.fiveable.me/practice/ap-physics-c-e-m.)

Short version: inductors store energy in a magnetic field (∝ I^2) and resist changes in current; capacitors store energy in an electric field (∝ V^2) and resist changes in voltage. That leads to different circuit roles and transients. Key differences you should know for AP C E&M (Topic 13.5 and related CED items): - Energy/storage: inductor energy = 1/2 L I^2 (magnetic); capacitor energy = 1/2 C V^2 (electric). - Steady state (after a long time): ideal inductor → behaves like a short (wire, conducts DC); ideal capacitor → behaves like an open circuit (no DC current). (CED: 13.5.A.4.iii) - Transients & time constants: RL growth/decay ∝ e^(−t/τ) with τ = L/Req; RC uses τ = RC. RL differential equation: E = L dI/dt + I R (CED 13.5.A.2). - Continuity: inductor current cannot change instantaneously (current is continuous); capacitor voltage cannot change instantaneously (voltage is continuous). - Opposing change: an inductor generates a back emf (−L dI/dt) opposing current change; a capacitor develops voltage as charge changes opposing further charge flow. For AP prep, review the LR equations and sketch transient I(t)/V(t) curves (CED skills 1.C, 2.A). More on Topic 13.5: (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu). Practice problems: (https://library.fiveable.me/practice/ap-physics-c-e-m).

Start with the LR solution from Kirchhoff’s loop rule: for a battery E, L dI/dt + RI = E, the current is I(t) = I∞(1 − e^{−t/τ}) with I∞ = E/R and τ = L/R. Plug in t = τ: I(τ) = I∞(1 − e^{−1}) ≈ I∞(1 − 0.3679) ≈ 0.632 I∞ (≈63%). Why that number physically: the inductor initially opposes changes in current by producing a back emf, so current grows slowly. The math gives an exponential approach to the steady value; e^{−1} shows how much of the initial “gap” remains after one time constant. τ = L/R sets the timescale (larger L or smaller R → slower rise). This exponential behavior and the 63% rule are exactly what the CED describes for LR transients (see 13.5.A.2–.3). For a quick review, check the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

Step-by-step (series LR with battery ℰ, switch closed at t=0): 1. Write Kirchhoff’s loop rule / DE (CED 13.5.A.2): ℰ = L dI/dt + R I. 2. Put in standard form: dI/dt + (R/L) I = ℰ/L. This is a first-order linear ODE. 3. Solve homogeneous part: dI_h/dt + (R/L)I_h = 0 → I_h = C e^{-(R/L)t}. Time constant τ = L/R (CED 13.5.A.3.i). 4. Find particular (steady-state) solution: set dI/dt = 0 → I_p = ℰ/R = I_∞ (final current). 5. Full solution = I(t) = I_∞ + C e^{-t/τ}. Use initial condition to get C. If inductor current is zero at t=0 (common case), I(0)=0 ⇒ C = −I_∞, so I(t) = I_∞(1 − e^{-t/τ}), an exponential rise to steady state (≈63% at t=τ). 6. For switch opening (disconnect battery) with initial current I0: DE becomes 0 = L dI/dt + R I → I(t) = I0 e^{-t/τ} (exponential decay; 37% at t=τ). Remember inductors store energy and resistor dissipates it during transients (CED 13.5.A.1, A.4). For more worked examples and AP-style practice, see the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and the practice problem bank (https://library.fiveable.me/practice/ap-physics-c-e-m).

When the current through an inductor changes, the energy stored in its magnetic field (U = 1/2 L I^2) changes too. If the current increases, work must be done against the induced (back) emf—energy supplied by the battery (or source) is stored in the inductor. If the current decreases, that stored magnetic energy is released and dissipated as heat in the circuit’s resistor(s) (this is exactly EK 13.5.A.1: a resistor dissipates energy that was stored in an inductor). During the transient the voltage across the inductor and the inductor’s energy follow exponential behavior with time constant τ = L / Req; after many τ the circuit reaches steady state and the inductor acts like a short (steady current, no changing energy). For AP review, see the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and unit overview (https://library.fiveable.me/ap-physics-c-e-m/unit-13).

Right after you close the switch the inductor resists changes in current by producing a back emf L dI/dt. That back emf makes the inductor behave like an element with a voltage drop proportional to the rate of change of current, so current starts from its initial value and changes exponentially. The math from Kirchhoff’s loop rule (E = L dI/dt + I R) gives the solution I(t) → I∞ = E/R as t → ∞ with time constant τ = L/Req (CED 13.5.A.2–3). After a time much greater than τ, dI/dt ≈ 0, so L dI/dt → 0 and the inductor no longer produces back emf. With no back emf it just lets the steady current flow and therefore acts like a short (a conducting wire with essentially zero voltage drop) while any energy changes happen in the resistor (CED 13.5.A.1, 13.5.A.4.iii). For worked examples and practice problems on LR transients, see the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and lots more practice at (https://library.fiveable.me/practice/ap-physics-c-e-m).

Think of the inductor as resisting changes in current. When you suddenly apply a battery, the current wants to increase; the changing current produces a changing magnetic flux, and by Lenz’s law the induced emf (back emf) has the polarity that opposes that change. Mathematically in the LR loop Kirchhoff gives E = L (dI/dt) + I R. The term L (dI/dt) comes from the induced emf—its sign is such that it subtracts from the battery’s emf when dI/dt is positive, so the inductor “pushes back” against the rise in current. Physically that’s why current grows exponentially with time constant τ = L/R instead of jumping instantly: energy is temporarily stored in the inductor’s magnetic field and the back emf controls the rate of change. This idea (back emf opposing the applied potential) is exactly what Topic 13.5 emphasizes—see the LR study guide for a quick review (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu). For extra practice on these concepts, check the AP E&M practice set (https://library.fiveable.me/practice/ap-physics-c-e-m).

You know an LR circuit reached steady state when the inductor’s current stops changing (dI/dt ≈ 0). Practically that means: - The back emf L dI/dt → 0, so the inductor behaves like a perfect wire (zero voltage drop) and the steady current is I∞ = ℰ / R_eq (CED 13.5.A.2, 13.5.A.4.iii). - Use the time constant τ = L / R_eq (CED 13.5.A.3.i). For a step turn-on with I(0)=0, I(t)=I∞(1 − e^(−t/τ)). Rule of thumb: t ≈ 3τ → ~95% of I∞, t ≈ 5τ → ~99.3% (so t ≫ τ is “steady”). - For turn-off/decay, I(t)=I0 e^(−t/τ); after a few τ the current (and stored energy) is essentially gone. AP tip: be ready to use the differential equation ℰ = L dI/dt + IR and the τ = L/R_eq idea on FRQs (Topic 13.5). For a quick review see the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

The time constant τ = L / R_eq tells you how fast the circuit’s transient dies away—it’s the characteristic time for the inductor current (and the inductor’s voltage and stored energy) to move from its initial value toward the steady state. Physically: because the inductor resists changes in current (back emf ∝ L dI/dt) and the resistor dissipates energy, τ sets the competition between inductive inertia and resistive damping. If the inductor starts with zero current, after one τ the current has reached ≈63% of its final steady value; if it starts at some I0 and is decaying, after one τ it’s ≈37% of I0. You can also think of τ as “how long the circuit would take to reach steady state if the initial rate of change stayed constant” (CED 13.5.A.3.i–iii). On the AP exam this shows up as solving the differential equation E = L dI/dt + IR and interpreting exponential behavior (Topic 13.5; see the study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and unit overview (https://library.fiveable.me/ap-physics-c-e-m/unit-13)). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-e-m).

Because an LR circuit is governed by a first-order linear differential equation from Kirchhoff’s loop rule, its solutions are exponential. For a series LR circuit with a battery emf ℰ you get ℰ = L (dI/dt) + I R Rearrange to dI/dt = (ℰ/L) − (R/L) I. That’s of the form dI/dt = constant − (1/τ) I, where τ = L/R. Mathematically the rate of change of I depends linearly on I itself, so integrating gives I(t) = I∞ + (I0 − I∞) e^(−t/τ)—an exponential approach to the steady state I∞ = ℰ/R. Physically: the inductor’s back emf (−L dI/dt) resists changes in current, while the resistor dissipates energy. The competition makes the current change at a rate proportional to how far it is from its final value, which produces exponential growth/decay with time constant τ = L/R (see CED 13.5.A.2–A.3). For AP review, the Topic 13.5 study guide covers derivation and time-constant meaning (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu). For extra practice, try problems at https://library.fiveable.me/practice/ap-physics-c-e-m.

Kirchhoff’s loop rule still applies in LR circuits, but you must include the inductor’s induced (back) emf, εL = −L dI/dt, as a voltage drop in the loop. For a series LR with battery emf ℰ, following the loop in the passive sign convention gives the standard differential equation ℰ = L(dI/dt) + IR. The −L dI/dt term reflects Lenz’s law: when I is changing the inductor produces an emf that opposes that change (at the moment a switch closes the induced emf can nearly equal ℰ). Solve that first-order ODE to get exponential transients with time constant τ = L/R; after many τ the inductor behaves like a short (steady state, dI/dt → 0). For AP review see the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and unit overview (https://library.fiveable.me/ap-physics-c-e-m/unit-13). For practice, try problems at (https://library.fiveable.me/practice/ap-physics-c-e-m).

If you missed the lab, expect simple exponential transients when you plot current vs time. - Closing the switch (initial I = 0): current rises as I(t) = I∞(1 − e^(−t/τ)), where I∞ = E/Req and τ = L/Req. At t = τ the current ≈ 0.63 I∞. The inductor initially opposes change (large back-emf), so the curve starts flat and then curves up toward the steady value. - Opening the switch (initial I = I0): current decays as I(t) = I0 e^(−t/τ). At t = τ it’s ≈37% of I0. After many τ the inductor behaves like a short (steady state) or the resistor dissipates stored energy during decay. Practical tips: use an oscilloscope or fast data logger with sampling >>1/τ, measure several τ to see asymptote, and plot semi-log to get a straight line (slope = −1/τ). This matches CED 13.5.A (Kirchhoff, differential eqn, τ = L/R). Review the Topic 13.5 study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and try practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).

Right after you change the circuit (switch closes or opens), the inductor fights any change in current by producing a back emf: L(dI/dt) is large, so the current changes slowly. For example, if the inductor has zero initial current and you close a battery+R, the initial current is I(0)=0 and the inductor’s voltage equals the battery (it acts like an open circuit to sudden change). The transient current then follows the solution of E = L dI/dt + I R and rises exponentially with time constant τ = L/Req (about 63% of final value after one τ). After a long time (t >> τ) the transient dies out: dI/dt → 0, the inductor’s back emf → 0, it acts like an ideal conducting wire (short) so I → E/R and the potential across L is ~0. The energy stored in L is (1/2) L I^2. These points map directly to the AP CED items (Eqs, exponentials, τ = L/R). For a quick refresher, check the LR study guide (https://library.fiveable.me/ap-physics-c-e-m/unit-6/5-circuits-with-resistors-and-inductors-lr-circuits/study-guide/pX2sUTu1DFkqKZNu) and practice problems (https://library.fiveable.me/practice/ap-physics-c-e-m).