Unit 13 · Topic 13.5

13.5 Circuits with Resistors and Inductors (LR Circuits)

In an LR circuit, an inductor keeps the current from jumping, so the current grows or decays exponentially with time constant $\tau = \dfrac{L}{R}$ . This topic covers the loop-rule equation, its solutions, and the quick rules for what the inductor does right after a switch changes and a long time later.

Key terms

LR circuit
time constant
exponential growth and decay
steady state
back emf

Setting up the equation

Connect a battery of emf $\mathcal{E}$ , a resistor R and an inductor L in series, and close the switch at t = 0. Walking around the loop in the direction of the current, the inductor's potential drop is $L\dfrac{dI}{dt}$ , so the loop rule gives $\mathcal{E} - IR - L\dfrac{dI}{dt} = 0$ .

Separating variables and integrating from I = 0 at t = 0 gives:

$I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R}$

The inductor's potential difference starts at $\mathcal{E}$ and decays: $\Delta V_L = \mathcal{E}e^{-t/\tau}$ . The resistor's potential difference rises to match, $\Delta V_R = IR$ , and the two always add to $\mathcal{E}$ .

Here's another way to read τ. Right after the switch closes, the current grows at $\dfrac{\mathcal{E}}{L}$ . If it kept growing at that starting rate, it would reach its final value $\dfrac{\mathcal{E}}{R}$ in exactly $\dfrac{L}{R}$ , one time constant. On an I-versus-t graph, the tangent line at t = 0 hits the final-value line at t = τ.

Decay

If the battery is removed so that a current $I_0$ flows around a loop of just L and R, the loop rule is $-IR - L\dfrac{dI}{dt} = 0$ , and the current decays: $I(t) = I_0e^{-t/\tau}$ . The inductor's stored energy, $\tfrac{1}{2}LI_0^2$ , all ends up as thermal energy in the resistor.

As with RC circuits, after one τ the current has reached 63% of its final value when growing, or fallen to 37% when decaying. The half-time is $\tau\ln 2$ .

Right after and long after

These limits answer many questions without any exponentials:

The current through an inductor can't change instantly. Just after a switch changes, each inductor carries the same current it had just before.
An inductor that had no current acts like a break (open circuit) right after a switch closes.
A long time after any change, the current is steady, so the inductor has no potential difference and acts like a plain wire.
When a switch opens, an inductor's current keeps flowing through whatever path remains, and the potential difference can become much larger than the battery's emf.

Element	Right after a switch closes (starting empty)	Long after
Capacitor	acts like a wire	acts like a break
Inductor	acts like a break	acts like a wire

RC versus LR

The two kinds of circuit mirror each other. A capacitor resists sudden changes in its potential difference; an inductor resists sudden changes in its current. A bigger R makes an RC circuit slower (τ = RC) but an LR circuit faster (τ = L/R).

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Current growth in an LR circuit
A 12 V ideal battery, a 6.0 Ω resistor and a 0.30 H inductor are connected in series, and the switch is closed at t = 0. Find the time constant, the final current, the current at t = 0.050 s, and the potential difference across the inductor at t = 0.
Show the solutionHide the solution
1. Step 1: $\tau = \dfrac{L}{R} = \dfrac{0.30}{6.0} = 0.050\text{ s}$ .
2. Step 2: Long after, the inductor acts like a wire: $I_{final} = \dfrac{\mathcal{E}}{R} = \dfrac{12}{6.0} = 2.0\text{ A}$ .
3. Step 3: At t = 0.050 s, which is one τ: $I = 2.0\left(1 - e^{-1}\right) \approx 1.26\text{ A}$ .
4. Step 4: At t = 0 the current is zero, so the resistor has no potential difference and the inductor has the battery's full 12 V across it.
Answer: τ = 0.050 s, final current 2.0 A, I(0.050 s) ≈ 1.26 A, 12 V across the inductor at t = 0
Example 2Calculator allowed
Opening a switch next to an inductor (classic trap)
A 10 V ideal battery and a switch are in series with R₁ = 5.0 Ω. After R₁, a 0.50 H inductor and R₂ = 20 Ω are connected in parallel. The switch has been closed for a long time. It's then opened. Find the current in each branch just before opening, and the potential difference across R₂ just after.
Show the solutionHide the solution
1. Step 1: Long after closing, the inductor acts like a wire, which shorts out R₂. So R₂ carries no current and the inductor carries 10 V / 5.0 Ω = 2.0 A.
2. Step 2: When the switch opens, the battery and R₁ are cut off. The inductor's current can't change instantly, so it keeps 2.0 A flowing, now around the only loop left: through R₂.
3. Step 3: Potential difference across R₂ just after: (2.0 A)(20 Ω) = 40 V, four times the battery's emf.
4. Step 4: The current then decays with τ = L/R₂ = 0.50/20 = 0.025 s. The trap is assuming every current drops to zero the moment the battery is disconnected.
Answer: Before: 2.0 A through the inductor, 0 A through R₂. Just after: 2.0 A through R₂, giving 40 V across it

Common mistakes

Letting the inductor's current jump when a switch changes. Find the current just before; that's the current just after.
Using $\tau = RL$ or $\tau = \dfrac{R}{L}$ . It's $\dfrac{L}{R}$ , which has units of seconds.
Swapping the capacitor and inductor rules for right after and long after.
Using the wrong resistance in τ after a switch changes. Use the resistance in the loop the inductor's current actually flows through.

On the exam

Free-response questions often ask you to write the loop equation, solve it for I(t), and sketch I and $\Delta V_L$ against t. Label the asymptote and the starting value.
Many multiple-choice questions only need the limits. Redraw the circuit with each inductor as a break or a wire, depending on the moment.

Connected topics

Videos

Topic 13.5 - LR Circuits
Lessons With LondotWatch on YouTube (opens in a new tab)
LR Circuit Basics
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C - RL Circuits
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
RL Circuits Explained Clearly ⚡AP Physics C: E&M - Unit 13 - Lesson 5
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Physics 47 Inductance (12 of 20) The R-L Circuit Analyzed
Michel van BiezenWatch on YouTube (opens in a new tab)
LR Circuit Equation Derivations
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

5 questions on 13.5 Circuits with Resistors and Inductors (LR Circuits). Pick an answer to see if you got it, and why.

Question 1 of 5Calculator allowed

A 0.30 H inductor is connected in series with a 15 Ω resistor. What is the time constant of the circuit?

An ideal 18 V battery, a switch and resistor R₁ = 3.0 Ω are in series. After R₁ the circuit splits into two parallel branches that rejoin before the battery: one branch is resistor R₂ = 6.0 Ω, and the other is an ideal inductor L with no resistance. There is no current anywhere before the switch is closed.

Described circuit

Question 2 of 5Calculator allowed

What is the current in R₁ immediately after the switch is closed?

Question 3 of 5Calculator allowed

After the switch has been closed for a long time, what is the current in R₂?

Question 4 of 5Calculator allowed

After a long time, the switch is opened. What is the magnitude of the current in R₂ immediately afterward?

Question 5 of 5Calculator allowed

Which expression gives the time constant for the current’s decay after the switch is opened?

0 of 5 answered