13.5 Circuits with Resistors and Inductors (LR Circuits)

An LR circuit pairs a resistor with an inductor, and the inductor resists sudden changes in current. When you close or open a switch, current grows or decays exponentially with time constant $\tau = L/R_{eq}$, eventually settling to a steady state where the inductor acts like a wire. In AP Physics C: E&M, this topic connects Kirchhoff's loop rule, transient response, exponential functions, and magnetic energy storage.

Why This Matters for the AP Physics C: E&M Exam

LR circuits combine ideas from circuits and induction, so they test whether you can connect functional relationships across topics. On both the multiple-choice and free-response sections, you may be asked to write or solve the loop-rule differential equation, sketch how current and inductor voltage change over time, identify behavior right after a switch flips versus long after, and reason about energy moving between the inductor's magnetic field and the resistor. Laboratory-style questions about circuits containing a solenoid show up too, so being able to interpret graphs and justify claims with physical principles pays off here.

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Key Takeaways

• Kirchhoff's loop rule gives the LR equation $\mathcal{E} = L\frac{dI}{dt} + IR$, a first-order differential equation whose solutions are exponential in time.
• The time constant $\tau = \frac{L}{R_{eq}}$ sets how fast the circuit reaches steady state. After one $\tau$, a rising current reaches about 63% of its final value; a decaying current drops to about 37% of its starting value.
• Right after a switch flips, the inductor's induced emf is equal in magnitude and opposite in direction to the applied potential difference, so current cannot change instantly.
• After a long time (many time constants), the inductor behaves like a wire with zero resistance, and the final current is $I_{final} = \frac{\mathcal{E}}{R}$.
• Energy stored in the inductor is $U_L = \frac{1}{2}LI^2$, and the resistor dissipates that energy as current changes, following conservation of energy.
• Current, inductor voltage, and stored energy all change exponentially toward asymptotes set by the circuit's initial conditions.

Resistor-Inductor Circuit Properties

Energy Dissipation in Resistors

When current flows through a resistor in an LR circuit, electrical energy is converted into thermal energy. This is how energy stored in an inductor eventually leaves the circuit.

• Resistors transform the magnetic energy stored in an inductor into heat as current flows.
• The power dissipated in a resistor follows $P = I^2R$, so the energy loss rate depends on both the current and the resistance.
• As current decreases in an LR circuit, the rate of energy dissipation also decreases, following an exponential pattern.

Kirchhoff's Loop Rule for LR Circuits

Kirchhoff's loop rule says the sum of all voltage changes around any closed loop equals zero. Applying it to a series LR circuit with a battery of emf $\mathcal{E}$ gives:

$\mathcal{E} = IR + L\frac{dI}{dt}$

This equation connects:

• The battery's emf ($\mathcal{E}$)
• The voltage drop across the resistor ($IR$), where $I$ is the current and $R$ is the resistance
• The voltage across the inductor ($L\frac{dI}{dt}$), where $L$ is the inductance and $\frac{dI}{dt}$ is the rate of change of current

Because this is a first-order differential equation, its solutions are exponential functions of time.

Time Constant in LR Circuits

The time constant tells you how quickly the circuit responds to changes, like connecting or disconnecting a battery. It acts as a measure of the circuit's "electrical inertia."

$\tau = \frac{L}{R_{eq}}$

Where:

• $L$ is the inductance (in henries)
• $R_{eq}$ is the equivalent resistance (in ohms)

Key properties of the time constant:

• It represents the time the circuit would take to reach steady state if it kept changing at its initial rate.
• For an inductor with zero initial current, after one time constant the current reaches about 63% of its final value.
• For an inductor with an initial current, after one time constant the current falls to about 37% of its initial value.
• After about five time constants, the circuit is very close to steady state (within roughly 1% of the final value).

Steady State Behavior of Inductors

Steady state occurs after the transient response dies out and the circuit stabilizes.

• In steady state, the current through an inductor becomes constant, so $\frac{dI}{dt} = 0$.
• When current is constant, the inductor's voltage drop is zero: $V_L = L\frac{dI}{dt} = 0$.
• The inductor then acts like a conducting wire with zero resistance, letting current flow freely.
• The final current in a series LR circuit with a battery depends only on the emf and the resistance: $I_{final} = \frac{\mathcal{E}}{R}$.

Transient vs Steady State in LR Circuits

LR circuit behavior splits into two phases: the transient state and the steady state.

Transient state characteristics:

• Occurs right after a change in the circuit, like closing or opening a switch.
• Immediately after a switch is closed or opened in a branch containing an inductor, the induced emf across the inductor is equal in magnitude and opposite in direction to the applied potential difference across that branch. This is why current through an inductor cannot change instantly.
• Current and voltage change rapidly according to exponential functions.
• For a circuit with a battery and initially no current, the current follows $I(t) = \frac{\mathcal{E}}{R}(1-e^{-Rt/L})$.
• For a circuit with initial current but no battery, the current follows $I(t) = I_0e^{-Rt/L}$.
• The inductor opposes changes in current by generating an induced emf.
• The inductor voltage is also exponential in time. For a charging series LR circuit, $V_L(t) = \mathcal{E}e^{-t/\tau}$, so it starts at $\mathcal{E}$ and approaches $0$. For a discharging LR circuit with initial current $I_0$, the magnitude of $V_L$ also decays exponentially toward $0$.
• The energy stored in the inductor, $U_L = \frac{1}{2}LI^2$, changes with time and approaches an asymptote set by the initial conditions: for charging it approaches $\frac{1}{2}L\left(\frac{\mathcal{E}}{R}\right)^2$, and for discharging it approaches $0$.

Steady state characteristics:

• Occurs after roughly five time constants.
• Current and voltage settle to constant values.
• The inductor behaves like a simple wire with negligible resistance.
• All the battery's emf appears across the resistor.
• The magnetic energy stored in the inductor stays constant.

How to Use This on the AP Physics C: E&M Exam

Problem Solving

• Start by writing Kirchhoff's loop rule for the circuit. For a series LR circuit, that is $\mathcal{E} = L\frac{dI}{dt} + IR$. Being able to set this up from scratch is the foundation for almost every LR problem.
• Identify initial and final conditions before plugging into exponential formulas. At $t = 0$ for a switch just closed, current is what it was an instant before (often zero) because current cannot jump. As $t \to \infty$, $\frac{dI}{dt} = 0$.
• Compute the time constant $\tau = \frac{L}{R_{eq}}$ early; it appears in every exponential expression and in graph features.

Free Response

• When asked to derive, show a clear logical path: write the loop equation, separate variables or recognize the standard solution, and state your initial condition. Skipping the differential equation setup can cost reasoning points.
• For "describe" or "justify" parts, connect math to physics. For example, explain that the inductor opposes the change in current because a changing current changes the magnetic flux, which induces an opposing emf (Lenz's law).
• Use energy conservation when a question asks where the inductor's stored energy goes: the resistor dissipates it as $P = I^2R$.

Graphing

• Practice sketching current versus time for both charging (rising exponential toward $\frac{\mathcal{E}}{R}$) and discharging (decaying exponential toward zero).
• Inductor voltage curves usually mirror the opposite shape of the current curve, so sketch them with the correct starting value and asymptote.
• Label the value at $t = \tau$ using the 63% rise or 37% decay rule to show you understand the time constant.

Common Trap

• Treating an inductor as a wire at $t = 0$. Right after a switch flips, an inductor blocks sudden current change; it acts like a wire only after a long time.

Common Misconceptions

• "Current jumps instantly when the switch closes." Current through an inductor changes continuously because an instant change would require an infinite $\frac{dI}{dt}$ and infinite induced emf. It builds up or decays exponentially.
• "An inductor always acts like a short circuit." It behaves like a zero-resistance wire only at steady state, long after the switch flips. Right after switching, it strongly opposes the change in current.
• "The time constant is the time to reach the final value." After one time constant the current only reaches about 63% (rising) or drops to 37% (decaying). It takes roughly five time constants to get very close to the final value.
• "Voltage across the inductor stays constant." Inductor voltage changes exponentially with time, starting at its maximum right after switching and decaying toward zero as current stabilizes.
• "Energy in the inductor disappears." Stored magnetic energy $U_L = \frac{1}{2}LI^2$ is conserved; it is transferred to the resistor as heat during current changes, following conservation of energy.
• "The time constant is $\tau = RC$ here." That is for RC circuits. For LR circuits, $\tau = \frac{L}{R_{eq}}$.

Practice Problem 1: LR Circuit Time Constant

Solution

First, find the time constant of the LR circuit: $\tau = \frac{L}{R}$

Substituting the given values: $\tau = \frac{0.8 \text{ H}}{40 \text{ Ω}} = 0.02 \text{ s}$

By definition, after one time constant the current reaches 63% of its maximum value. So it takes 0.02 seconds for the current to reach 63% of its maximum.

To find the maximum current, use the steady-state condition where the inductor acts like a wire: $I_{max} = \frac{\mathcal{E}}{R} = \frac{12 \text{ V}}{40 \text{ Ω}} = 0.3 \text{ A}$

The maximum current that will eventually flow is 0.3 amperes.

Practice Problem 2: Current in an LR Circuit

Solution

For an LR circuit with a battery and the switch closed at t = 0, the current is: $I(t) = \frac{\mathcal{E}}{R}(1-e^{-Rt/L})$

First, calculate the time constant: $\tau = \frac{L}{R} = \frac{0.3 \text{ H}}{60 \text{ Ω}} = 0.005 \text{ s}$

Now find the current at t = 0.01 seconds: $I(0.01) = \frac{24 \text{ V}}{60 \text{ Ω}}(1-e^{-(60 \text{ Ω})(0.01 \text{ s})/(0.3 \text{ H})})$ $I(0.01) = 0.4 \text{ A} \times (1-e^{-2})$ $I(0.01) = 0.4 \text{ A} \times (1-0.135)$ $I(0.01) = 0.4 \text{ A} \times 0.865$ $I(0.01) = 0.346 \text{ A}$

The current in the circuit at t = 0.01 seconds is 0.346 amperes.

Related AP Physics C: E&M Guides

• 13.2 Electromagnetic Induction
• 13.3 Induced Currents and Magnetic Forces
• 13.4 Inductance
• 13.6 Circuits with Capacitors and Inductors (LC Circuits)
• 13.1 Magnetic Flux