Circuits with Resistors and Inductors (LR Circuits) - AP Physics C Electricity and Magnetism

SECTION 1

Historical Context & Motivation

The story of LR circuits begins with the nineteenth-century race to understand the relationship between electricity and magnetism. When Michael Faraday demonstrated in 1831 that a changing magnetic flux through a loop of wire produces an electromotive force (EMF), he laid the groundwork for understanding how coils of wire—later called inductors—store energy in their magnetic fields and resist sudden changes in current. Joseph Henry independently discovered self-induction around the same time, observing dramatic sparking when circuits carrying large currents through coils were suddenly broken. These observations revealed a new circuit element whose behavior depended not on the magnitude of the current but on its rate of change, fundamentally distinguishing inductors from resistors and capacitors.

1831 • Faraday's Law of Induction

Michael Faraday demonstrates that a time-varying magnetic flux through a conducting loop induces an EMF, establishing the physical basis for inductance.

1832 • Henry's Discovery of Self-Induction

Joseph Henry observes that a coil carrying current produces a back-EMF when the current changes, identifying the phenomenon of self-inductance and the unit later named in his honor.

1886 • Heaviside's Operational Calculus

Oliver Heaviside develops mathematical techniques to solve differential equations governing LR and RLC circuits, foreshadowing Laplace-transform methods used in modern circuit analysis.

1893 • Steinmetz and AC Circuit Analysis

Charles Proteus Steinmetz introduces phasor representation for AC circuits, enabling engineers to handle inductors and resistors together using complex impedances.

The central question that LR circuit analysis addresses is deceptively simple: when a DC voltage source is suddenly connected to a series combination of a resistor and an inductor, how does the current evolve with time? Because the inductor opposes instantaneous changes in current, the circuit cannot jump to its steady-state value immediately—instead, an exponential transient process unfolds, governed by the ratio L/R. Understanding this transient behavior is essential not only for the AP Physics C exam but also for grasping how inductors function in power supplies, relay circuits, electromagnetic actuators, and countless other applications.

SECTION 2

Core Principles & Definitions

An LR circuit is any circuit containing an inductor (inductance L, measured in henrys) and a resistor (resistance R, measured in ohms) connected to a source of EMF. The inductor's defining property is that it produces a voltage proportional to the time derivative of the current flowing through it: V_L = −L(dI/dt). This relationship, rooted in Faraday's law applied to a coil's own changing magnetic flux, is what gives LR circuits their characteristic exponential behavior. The following grid summarizes the foundational ideas you must internalize before diving into the mathematics.

Self-Inductance (L)

A measure of a coil's ability to oppose changes in current. Defined as L = N Φ_B / I, where N is the number of turns and Φ_B is the magnetic flux through each turn. The SI unit is the henry (H).

Back-EMF

When current through an inductor changes, the inductor generates a voltage (ε = −L dI/dt) that opposes the change. This is a direct consequence of Lenz's law and is the mechanism behind transient behavior in LR circuits.

Time Constant (τ = L/R)

The characteristic timescale for current growth or decay in an LR circuit. After one time constant, the current has reached about 63.2% of its final value (growth) or decayed to about 36.8% of its initial value (decay).

Energy Stored in an Inductor

An inductor carrying current I stores magnetic field energy U = ½LI². This energy must be accounted for when analyzing circuits—it is dissipated as heat in the resistor during current decay.

Kirchhoff's Voltage Law (KVL)

The sum of voltage drops around any closed loop equals zero. For an LR circuit with EMF ε: ε − IR − L(dI/dt) = 0. This first-order ODE governs the transient response.

✦

KEY TAKEAWAY

Think of an inductor as the electrical analog of a massive flywheel in a mechanical system. Just as a heavy flywheel resists sudden changes in rotational speed—it takes time and sustained torque to spin it up, and it keeps spinning after the drive is removed—an inductor resists sudden changes in current. The "inertia" is the inductance L, and the "friction" slowing things down is the resistance R. A large L/R ratio means the electrical "flywheel" is heavy relative to the friction, so transients last longer.

SECTION 3

Visual Explanation — The Series LR Circuit

A series LR circuit with EMF source ε (gold), inductor L (violet coil), resistor R (pink), and switch S (green). When the switch closes at t = 0, current I(t) (cyan arrow) begins to rise exponentially. The voltage across the inductor V_L starts at ε and decays to zero, while V_R starts at zero and rises to ε. Kirchhoff's voltage law (shown at bottom) yields the governing first-order ODE.

The diagram above represents the prototypical series LR circuit that appears on nearly every AP Physics C exam. Before the switch closes, no current flows and no energy is stored in the inductor. The instant the switch closes at t = 0, the full EMF ε appears across the inductor because the current—and therefore the voltage drop IR across the resistor—is initially zero. The inductor's back-EMF exactly matches the source EMF, so dI/dt is at its maximum value of ε/L. As current builds, more voltage appears across R and less across L, causing dI/dt to decrease. This self-regulating feedback loop produces the characteristic exponential approach to the steady-state current I_max = ε/R, at which point dI/dt = 0 and the inductor behaves like an ideal wire.

SECTION 4

Mathematical Framework

Deriving the Current Growth Equation

Applying Kirchhoff's voltage law around the single loop of the series LR circuit gives ε − IR − L(dI/dt) = 0. Rearranging, we obtain the first-order linear ODE: L(dI/dt) + IR = ε. This is separable. Dividing both sides by L and using the substitution u = ε/R − I, the equation transforms to du/dt = −(R/L)u. Integration with the initial condition I(0) = 0 yields the exponential growth equation below.

CURRENT GROWTH (ENERGIZING)

I(t) = (ε / R)(1 − e^(−Rt/L)) = I_max(1 − e^(−t/τ))

where ε is the EMF of the source, R is the total resistance, L is the inductance, τ = L/R is the inductive time constant, and I_max = ε/R is the final steady-state current.

Current Decay (De-Energizing)

When the EMF source is suddenly removed (or the circuit is switched to a path containing only R and L), the current cannot drop to zero instantaneously because the inductor sustains the current via its stored magnetic energy. The loop equation becomes IR + L(dI/dt) = 0, which gives a pure exponential decay.

CURRENT DECAY (DE-ENERGIZING)

I(t) = I₀ e^(−Rt/L) = I₀ e^(−t/τ)

where I₀ is the current at the moment the source is removed (often I₀ = ε/R if the circuit was in steady state). The current decays to 1/e ≈ 36.8% of I₀ after one time constant τ.

VOLTAGE ACROSS THE INDUCTOR

V_L(t) = L(dI/dt) = ε · e^(−t/τ) (growth); V_L(t) = −I₀R · e^(−t/τ) (decay)

During growth, V_L starts at ε and decays exponentially to zero. During decay, V_L can briefly exceed the original source EMF in magnitude (negative spike), which is why opening inductive circuits can produce dangerous arcs.

ENERGY STORED IN THE INDUCTOR

U = ½ L I²

Energy U (in joules) is stored in the magnetic field of the inductor. Compare with U = ½CV² for a capacitor. During decay, all stored energy is dissipated as heat in R.

💡

AP EXAM TIP

The AP Physics C exam frequently asks about initial and final conditions. Remember: at t = 0⁺ in a growth scenario, the inductor acts like an open circuit (I = 0, V_L = ε). At t → ∞, the inductor acts like a short circuit (I = ε/R, V_L = 0). These limiting cases let you check your answers quickly.

SECTION 5

Current and Voltage Transient Curves

The exponential growth and decay equations produce characteristic curves that you must be able to sketch, interpret, and extract information from on the AP exam. The following diagram plots both the current I(t) and the inductor voltage V_L(t) during the energizing phase, with key time-constant milestones marked.

During energizing, the current I(t) (cyan) rises exponentially toward ε/R while the inductor voltage V_L(t) (pink) decays from ε toward zero. At t = τ, I has reached 63.2% of its maximum and V_L has fallen to 36.8% of ε. After approximately 5τ, the circuit is effectively in steady state.

Time (multiples of τ)	I(t) / I_max	$V_L$ (t) / ε	$V_R$ (t) / ε
0	0	1.000	0
1τ	0.632	0.368	0.632
2τ	0.865	0.135	0.865
3τ	0.950	0.050	0.950
5τ	0.993	0.007	0.993

Transient values during LR circuit energizing. Note that $V_R$ + $V_L$ = ε at every instant (KVL).

A crucial feature of the table is that the sum V_R(t) + V_L(t) = ε at every instant—this is simply Kirchhoff's voltage law in action. The practical rule of thumb is that after five time constants, the transient is more than 99% complete, and the circuit is effectively in its DC steady state. For the decay scenario, the same percentages apply in reverse: I(τ) = 0.368 I₀, I(2τ) = 0.135 I₀, and so on.

SECTION 6

Worked Example

RL Energizing Circuit — Finding Current, Voltage, and Energy

1
Step 1 — Identify Given Values

A 12.0 V battery is connected in series with a 4.00 Ω resistor and a 20.0 mH inductor via a switch that closes at t = 0. The circuit was previously open, so I(0) = 0. We are asked to find (a) the time constant, (b) the current at t = 10.0 ms, and (c) the energy stored in the inductor at t = 10.0 ms.

ε = 12.0 V, R = 4.00 Ω, L = 20.0 × 10⁻³ H
2
Step 2 — Calculate the Time Constant

The inductive time constant is τ = L/R = (20.0 × 10⁻³ H) / (4.00 Ω) = 5.00 × 10⁻³ s = 5.00 ms. This tells us that 10.0 ms corresponds to exactly 2τ.

τ = 5.00 ms
3
Step 3 — Find the Steady-State Current

The maximum (steady-state) current is I_max = ε/R = 12.0 V / 4.00 Ω = 3.00 A. This is the current the circuit will approach as t → ∞.

I_max = 3.00 A
4
Step 4 — Calculate Current at t = 10.0 ms

Using the growth equation: I(t) = I_max(1 − $e^{−t/τ}$ ). Substituting t = 10.0 ms and τ = 5.00 ms: I(10.0 ms) = 3.00 A × (1 − e⁻²) = 3.00 A × (1 − 0.1353) = 3.00 A × 0.8647 = 2.59 A.

I(10.0 ms) = 2.59 A
5
Step 5 — Calculate Stored Energy

The energy stored in the magnetic field of the inductor is U = ½LI² = ½ × (20.0 × 10⁻³ H) × (2.59 A)² = ½ × 0.0200 × 6.71 = 0.0671 J ≈ 67.1 mJ.

U ≈ 67.1 mJ
6
Step 6 — Verify with Limiting Cases

At t = 0, our equation gives I = 0 (correct, switch just closed). At t → ∞, I → 3.00 A = ε/R (correct, inductor acts as wire). At t = 2τ, we expect roughly 86.5% of I_max: 0.865 × 3.00 = 2.59 A ✓. The maximum possible stored energy would be ½ × 0.0200 × 9.00 = 90.0 mJ, and 67.1 mJ < 90.0 mJ ✓.

SECTION 7

Comparing LR and RC Circuits

LR circuits and RC circuits are the two fundamental first-order transient circuits in physics. Both exhibit exponential behavior, but the roles of voltage and current are essentially swapped. Recognizing the structural analogy between them is a powerful tool for the AP exam, because if you can solve one type you can solve the other by pattern-matching.

Property	LR Circuit	RC Circuit
Energy storage element	Inductor (L) — stores energy in magnetic field	Capacitor (C) — stores energy in electric field
Time constant	τ = L/R	τ = RC
Quantity that grows exponentially	Current I(t)	Charge Q(t) / Voltage $V_C$ (t)
At t = 0⁺ (charging)	I = 0; $V_L$ = ε (inductor blocks current)	I = ε/R; $V_C$ = 0 (capacitor is uncharged)
At t → ∞ (charging)	I = ε/R; $V_L$ = 0 (inductor acts as wire)	I = 0; $V_C$ = ε (capacitor fully charged)
Stored energy	U = ½LI²	U = ½CV²
Effect of increasing R	Decreases τ → faster transient	Increases τ → slower transient

Side-by-side comparison of LR and RC circuit properties

✦

KEY TAKEAWAY

Notice the inverse relationship between R and τ in the two circuit types: in an LR circuit, increasing R makes the transient faster (more "friction" damps the inertial current sooner), while in an RC circuit, increasing R makes the transient slower (charge flows through a narrower "pipe"). This distinction is a common source of exam errors—keep the flywheel vs. water-tank analogies separate in your mind.

SECTION 8

Connection to Advanced Theory — RLC Circuits and AC Analysis

The series LR circuit is the simplest inductive circuit, but it serves as the gateway to richer phenomena. When a capacitor is added to form an RLC circuit, the governing equation becomes a second-order ODE, and the transient response can exhibit oscillatory (underdamped), critically damped, or overdamped behavior depending on the relative magnitudes of R, L, and C. This is analogous to a mass-spring-damper system in mechanics, where L plays the role of mass, 1/C the role of spring constant, and R the role of damping coefficient.

Feature	LR Circuit (This Lesson)	RLC Circuit (Advanced)
Order of ODE	First-order	Second-order
Transient behavior	Pure exponential growth/decay	Oscillatory, critically damped, or overdamped
Characteristic frequency	None (no oscillation)	ω₀ = 1/√(LC)
AC steady-state analysis	Inductive reactance $X_L$ = ωL; impedance Z = R + jωL	Full complex impedance with resonance at ω₀
AP Physics C coverage	Core topic — expect FRQ and MCQ	LC oscillation tested; full RLC less common

Progression from first-order LR to second-order RLC circuit analysis

In AC circuit analysis, the inductor's impedance is purely imaginary: Z_L = jωL, where ω is the angular frequency of the AC source. This means the voltage across an inductor leads the current by 90°. Combined with a resistor (Z_R = R), the total impedance magnitude is |Z| = √(R² + ω²L²) and the phase angle is φ = arctan(ωL/R). While full AC phasor analysis extends beyond the typical AP Physics C E&M syllabus, the transient DC analysis you have learned here provides the conceptual foundation—understanding how V_L = L(dI/dt) works in the time domain is essential before extending to the frequency domain.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A series LR circuit is connected to an ideal battery. A long time after the switch is closed, the current through the circuit is I₀. If the inductance L is doubled while R and ε remain unchanged, what happens to the final steady-state current and the time to reach that steady state?

A. The steady-state current doubles, and the time constant doubles.
B. The steady-state current is unchanged, and the time constant doubles.
C. The steady-state current is halved, and the time constant is unchanged.
D. The steady-state current is unchanged, and the time constant is halved.

Show answer (B)

The correct answer is B. The steady-state current is I_max = ε/R, which depends only on ε and R, not on L. Therefore doubling L does not change the final current. However, the time constant τ = L/R doubles when L doubles (with R fixed), meaning the circuit takes twice as long to approach steady state. Choice A is wrong because the steady-state current is independent of L. Choice C is wrong because neither the current nor the time constant behave as stated. Choice D is wrong because the time constant increases (not decreases) with L.

Question 2: BASIC CALCULATION

A 50.0 mH inductor is connected in series with a 200 Ω resistor and a 10.0 V battery. What is the current in the circuit 0.500 ms after the switch is closed?

A. 3.16 × 10⁻² A
B. 5.00 × 10⁻² A
C. 4.32 × 10⁻² A
D. 6.77 × 10⁻³ A

Show answer (C)

The correct answer is C. First calculate τ = L/R = 0.0500 H / 200 Ω = 2.50 × 10⁻⁴ s = 0.250 ms. Then t/τ = 0.500 ms / 0.250 ms = 2.00. The steady-state current is I_max = ε/R = 10.0/200 = 0.0500 A. So I(0.500 ms) = 0.0500 × (1 − e⁻²) = 0.0500 × 0.8647 = 0.0432 A = 4.32 × 10⁻² A. Choice A (3.16 × 10⁻²) corresponds to evaluating the current at t = τ instead of t = 2τ, giving I_max × (1 − e⁻¹) = 0.0500 × 0.6321 = 0.0316 A. Choice B gives I_max itself, which would only be reached at t → ∞. Choice D (6.77 × 10⁻³) corresponds to incorrectly using I = I_max × e⁻² = 0.0500 × 0.1353 = 6.77 × 10⁻³ A, which describes a decaying current rather than a growing one.

Question 3: INTERMEDIATE

An LR circuit with L = 0.40 H and R = 8.0 Ω has been connected to a 24 V battery for a long time. At t = 0 the battery is suddenly short-circuited (removed from the loop, leaving only L and R). At what time does the energy stored in the inductor equal one-quarter of its initial value?

A. 17.3 ms
B. 34.7 ms
C. 50.0 ms
D. 25.0 ms

Show answer (B)

The correct answer is B. The time constant is τ = L/R = 0.40/8.0 = 0.050 s = 50.0 ms. The initial current is I₀ = ε/R = 24/8.0 = 3.0 A, so the initial energy stored in the inductor is U₀ = ½LI₀² = ½(0.40)(3.0)² = 1.80 J. During decay, the current follows I(t) = I₀

e^{−t/τ}

, so the energy as a function of time is U(t) = ½L[I₀

e^{−t/τ}

]² = U₀

e^{−2t/τ}

. Setting U(t) = U₀/4 gives

e^{−2t/τ}

= 1/4. Taking the natural logarithm of both sides: −2t/τ = −ln 4, so t = (τ/2) ln 4 = (50.0 ms / 2)(1.386) = 34.7 ms.

Question 4: APPLIED

4 pts

A solenoid with inductance L = 0.10 H and internal resistance r = 2.0 Ω is connected in series with an external resistor R = 8.0 Ω and a 20 V battery. The switch closes at t = 0. (a) Determine the time constant of the circuit. (1 pt) (b) Write an expression for the current I(t) as a function of time. (1 pt) (c) Determine the voltage across the inductor at t = 5.0 ms. (1 pt) (d) Calculate the power dissipated in the external resistor R at t = 5.0 ms. (1 pt)

Show answer

Rubric (4 points): (a) [1 pt] The total resistance in the circuit is R_total = R + r = 8.0 + 2.0 = 10.0 Ω. The time constant is τ = L/R_total = 0.10 H / 10.0 Ω = 0.010 s = 10.0 ms. (b) [1 pt] I(t) = (ε/R_total)(1 −

e^{−t/τ}

) = (20/10)(1 −

e^{−t/0.010}

) = 2.0(1 −

e^{−100t}

) A. (c) [1 pt] First find I at t = 5.0 ms = 0.5τ: I(5.0 ms) = 2.0(1 −

e^{−0.5}

) = 2.0(1 − 0.6065) = 2.0(0.3935) = 0.787 A. The voltage across the inductor is

V_L

= ε − I(R + r) = L(dI/dt) = ε

e^{−t/τ}

= 20 ×

e^{−0.5}

= 20 × 0.6065 = 12.1 V. (Note:

V_L

here includes the back-EMF due to changing current; if the inductor has internal resistance r, one must distinguish between

V_L

= L dI/dt and the terminal voltage across the solenoid which is

V_L

+ Ir.) (d) [1 pt]

P_R

= I²R = (0.787)² × 8.0 = 0.619 × 8.0 = 4.95 W ≈ 5.0 W.

Question 5: CRITICAL THINKING

5 pts

A student sets up an experiment to measure the inductance of an unknown inductor. The inductor is connected in series with a known resistance R = 100 Ω and a 6.0 V battery. Using a current probe and data logger, the student records the current as a function of time after the switch is closed. The data shows that the current reaches 63% of its maximum value at t = 2.0 ms. (a) Explain how the student can determine the inductance L from the experimental data. (1 pt) (b) Calculate the inductance of the unknown inductor. (1 pt) (c) The student notices the final current is only 54 mA instead of the expected 60 mA. Propose a physical explanation for this discrepancy and describe how it would affect the time constant measurement. (1 pt) (d) Derive an expression showing that the total energy delivered by the battery during the growth phase (t = 0 to t → ∞) is exactly twice the energy stored in the inductor at steady state. (2 pts)

Show answer

Rubric (5 points): (a) [1 pt] The current in an LR circuit grows as I(t) = (ε/R)(1 −

e^{−t/τ}

), where τ = L/R. At t = τ, the current reaches (1 − e⁻¹) ≈ 63.2% of its maximum value I_max = ε/R. Therefore, the time at which the current reaches 63% of its final value directly gives the time constant τ. Once τ is known, the inductance is L = τR. (b) [1 pt] From the data, τ = 2.0 ms = 2.0 × 10⁻³ s. Therefore L = τR = (2.0 × 10⁻³ s)(100 Ω) = 0.20 H. (c) [1 pt] The expected final current is ε/R = 6.0 V / 100 Ω = 0.060 A = 60 mA, but the observed steady-state current is 54 mA. This indicates additional resistance in the circuit, most likely the internal (winding) resistance r of the inductor. From I_max = ε/(R + r): r = ε/I_max − R = 6.0/0.054 − 100 ≈ 11 Ω. This additional resistance reduces the true time constant to τ_actual = L/(R + r), which is smaller than L/R. The 63% point therefore occurs earlier than it would for an ideal inductor, and the inductance calculated in part (b) is an underestimate of the true value. (d) [2 pts] Start with Kirchhoff's voltage law for the LR circuit: ε = IR + L(dI/dt) Multiply both sides by I to obtain the power equation: εI = I²R + LI(dI/dt) This states: power from battery = power dissipated in R + rate of energy storage in L. Substitute I(t) = I_max(1 −

e^{−t/τ}

) where I_max = ε/R and τ = L/R, and integrate each term from t = 0 to t → ∞. Since the steady-state (DC) contributions diverge, we isolate the transient contributions by noting that after the current is fully established, the inductor plays no role. The energy associated with building the current from zero is found by integrating only the transient parts. Define δI(t) = I_max − I(t) = I_max

e^{−t/τ}

. The transient energy delivered by the battery (the energy required to drive current through the circuit above what a zero-current state would require) is: W_batt = ε ∫₀^∞ I dt − (steady-state baseline) More directly, the energy the battery must supply to establish the magnetic field can be computed by noting that the transient current deficit δI satisfies the same KVL relation. The total energy the battery delivers to "charge" the inductor is: W_batt,trans = ε ∫₀^∞ δI dt = ε · I_max ∫₀^∞

e^{−t/τ}

dt = ε · I_max · τ Substituting I_max = ε/R and τ = L/R: W_batt,trans = ε · (ε/R) · (L/R) = L(ε/R)² = LI_max² The energy stored in the inductor at steady state is:

U_L

= ½LI_max² Therefore: W_batt,trans = LI_max² = 2 × ½LI_max² = 2U_L ✓ We can verify this with energy conservation. The transient energy dissipated in the resistor is:

W_R

,trans = R ∫₀^∞ δI² dt = R · I_max² ∫₀^∞

e^{−2t/τ}

dt = R · I_max² · (τ/2) = ½LI_max² =

U_L

Check: W_batt,trans =

W_R

,trans +

U_L

→ LI_max² = ½LI_max² + ½LI_max² ✓ Thus the total energy delivered by the battery during the current growth phase is exactly twice the energy stored in the inductor. Half is stored in the magnetic field; half is dissipated as heat in the resistor.

SUMMARY

Summary — LR Circuits

Summary

A series LR circuit combines a resistor R and an inductor L. The inductor produces a back-EMF proportional to dI/dt, which causes the current to grow or decay exponentially with a time constant τ = L/R. During energizing, the current rises as I(t) = (ε/R)(1 − $e^{−t/τ}$ ), approaching a steady-state value of ε/R. During de-energizing, the current decays as I(t) = I₀ $e^{−t/τ}$ . After approximately five time constants, the transient is effectively complete.

The energy stored in an inductor is U = ½LI², analogous to ½CV² for a capacitor. At t = 0⁺ during growth, the inductor acts like an open circuit (I = 0); at t → ∞, it acts like a short circuit (V_L = 0). LR circuits contrast with RC circuits in that increasing R decreases the LR time constant but increases the RC time constant. Mastery of these exponential transient equations and their limiting-case behavior is essential for success on the AP Physics C: E&M exam.